# NCERT Solutions for Class 10 Maths Chapter 6 - Triangles

## NCERT Solutions for Class 10 Mathematics Chapter 6 Free PDF Download

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Exercise 6.1

1. Fill in the blanks using the correct word given in brackets.

(i) All circles are ........... . (congruent, similar)

(ii) All squares are ......... . (similar, congruent)

(iii) All ........ triangles are similar. (isosceles, equilateral)

(iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are ......... and (b) their corresponding sides are ........ . (equal, proportional).

Sol. (i) All circles are similar because they have similar shape but their radii could be different. So they are similar.

(ii) All squares are similar because they have similar shape but not of the same size.

(iii) All equilateral triangles are similar because they have similar shape but not of same size as their side lengths could be different.

(iv) Two polygons of the same number of sides are similar, if

(a) their corresponding angles are equal.

(b) their corresponding sides are proportional.

2. Give two different examples of pair of

(i) similar figures.

(ii) non-similar figures.

Sol. (i) (a) Pair of circles are similar figures.

(b) Pair of squares are similar figures.

(ii) (a) A triangle and a rectangle form a pair of non-similar figures.

(b) A circle and a square form a pair of non-similar figures.

3. State whether the following quadrilaterals are similar or not :

Sol. The two quadrilaterals shown in figure are not similar because their corresponding angles are not equal. It is clear from the figure that, ∠A is 90° but ∠P is not equal to 90°. (less than 90°)

Exercise 6.2

1. In figures (i) and (ii), DE || BC. Find EC in figure (i) and AD is figure (ii).

Sol. (i) In figure (i), in DABC, DE || BC

$$\therefore\space\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}\\\text{(By basic proportionality theorem)}\\\Rarr\space\frac{1.5}{3}=\frac{1}{\text{EC}}\\{(\because \text{AD =} 1.5 \text{cm, DB = 3 cm and AE = 1 cm)}}\\\Rarr\space\text{EC}=\frac{3}{1.5}=2\text{cm}$$

∴ EC = 2 cm

(ii) In figure (ii), in ΔABC, DE || BC

$$\therefore\space\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}\\\text{(By basic proportionality theorem)}\\\Rarr\space\frac{\text{AD}}{7.2}=\frac{\text{1.8}}{5.4}\\(\because \text{AE = 1.8 cm, EC = 5.4 cm and DB = 7.2 cm)}\\\Rarr\space\text{AD}=\frac{1.8×7.2}{5.4}=2.4\space\text{cm}$$

2. E and F are points on the sides PQ and PR respectively of a DPQR. For each of the following cases, state whether EF || QR :

(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm.

(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm.

(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm

Sol. (i) In figure,

$$\frac{\text{PE}}{\text{EQ}}=\frac{3.9}{3}=1.3\\\text{and}\space\frac{\text{PF}}{\text{FR}}=\frac{3.6}{2.4}=\frac{3}{2}=1.5\\\Rarr\space\frac{\text{PE}}{\text{EQ}}\neq\frac{\text{PF}}{\text{FR}}$$

⇒ EF is not parallel QR because converse of basic proportionality theorem is not satisfied.

(ii) In figure,

$$\frac{\text{PE}}{\text{EQ}}=\frac{4}{4.5}=\frac{40}{4.5}=\frac{8}{9}\\\text{and}\space\frac{\text{PF}}{\text{FR}}=\frac{8}{9}\\\Rarr\space\frac{\text{PE}}{\text{EQ}}=\frac{\text{PF}}{\text{FR}}$$

⇒ EF || QR because converse of basic proportionality of theorem is satisfied.

(iii) In figure,

EQ = PQ – PE

= 1.28 – 0.18

= 1.10 cm

and FR = PR – PF

= 2.56 – 0.36

= 2.20 cm

$$\frac{\text{PE}}{\text{EQ}}=\frac{0.18}{1.10}=\frac{9}{55}\\\text{and}\space\frac{\text{PF}}{\text{FR}}=\frac{0.36}{2.20}=\frac{9}{55}\\\Rarr\space\frac{\text{PE}}{\text{EQ}}=\frac{\text{PF}}{\text{FR}}$$

⇒ EF || QR because converse of basic proportionality theorem is satisfied.

3. In figure, if LM || CB and LN || CD. Prove that

$$\frac{\textbf{AM}}{\textbf{AB}}=\frac{\textbf{AN}}{\textbf{AD}}$$

Sol. Given : LM || CB and LN || CD

$$\textbf{Proof :}\space\text{In ΔACB,}\\\text{LM|| CB (Given)}\\\text{So, by basic proportionality theorem}\\\Rarr\space\frac{\text{AM}}{\text{MB}}=\frac{\text{AL}}{\text{LC}}\space\text{...(i)}$$

In ΔACD, LN || CD (Given)

By basic proportionality theoerm,

$$\Rarr\space\frac{\text{AN}}{\text{AD}}=\frac{\text{AL}}{\text{LC}}\space\text{...(ii)}$$

From equation (i) and (ii), we get

$$\frac{\text{AM}}{\text{MB}}=\frac{\text{AN}}{\text{ND}}\\\Rarr\space\frac{\text{MB}}{\text{AM}}=\frac{\text{ND}}{\text{AN}}\\\Rarr\space\frac{\text{MB}}{\text{AM}}=\frac{\text{ND}}{\text{AN}}\\\Rarr\space\frac{\text{MB}}{\text{MB}}=\frac{\text{ND}}{\text{AN}}\\\Rarr\space\frac{\text{MB}}{\text{AM}}+1=\frac{\text{ND}}{\text{AN}}+1\\\text{((On adding both sides by 1))}\\\Rarr\space\frac{\text{MB+AM}}{\text{AM}}=\frac{\text{ND+AN}}{\text{AN}}\\\Rarr\space\frac{\text{AM}}{\text{AM+MB}}=\frac{\text{AN}}{\text{AN+ND}}\\\Rarr\space\frac{\text{AM}}{\text{AB}}=\frac{\text{AN}}{\text{AD}}\space\textbf{Hence Proved.}$$

4. In figure, DE || AC and DF || AC. Prove that

$$\frac{\textbf{BF}}{\textbf{FE}}=\frac{\textbf{BE}}{\textbf{EC}}$$

Sol. Given : DE || AC and DE || AC.

$$\textbf{To prove :}\space\frac{\text{BF}}{\text{FE}}=\frac{\text{BE}}{\text{EC}}\\\textbf{Proof}\text{: In} \Delta\text{BAC, DE || AC (Given)}\\\Rarr\space\frac{\text{BF}}{\text{FE}}=\frac{\text{BD}}{\text{DA}}\space\text{...(i)}\\\text{(by basic proportionality theorem)}\\\text{In} \Delta\text{BAE, DF || AE (Given)}\\\Rarr\space\frac{\text{BF}}{\text{FE}}=\frac{\text{BD}}{\text{DA}}\space\text{...(ii)}$$

(by basic proportionality theorem)

From equation (i) and (ii), we get

$$\frac{\text{BF}}{\text{FE}}=\frac{\text{BE}}{\text{EC}}\space\textbf{Hence Proved.}$$

5. In figure, DE || OQ and DF || OR. Show that EF || QR.

Sol. Given : DE || OQ

and DF || OR

To prove : EF || QR

Proof : In ΔPQO, we have

DE || OQ

(by basic proportionality Theorem)

$$\therefore\space\frac{\text{PE}}{\text{EQ}}=\frac{\text{PD}}{\text{DO}}\space\text{...(i)}\\\text{and in} \Delta \text{POR, DF || OR}\\\therefore\space\frac{\text{PF}}{\text{FR}}=\frac{\text{PD}}{\text{DO}}\space\text{...(ii)}$$

(by basic proportionality Theorem)

From equation (i) and (ii), we get

$$\frac{\text{PE}}{\text{EQ}}=\frac{\text{PF}}{\text{FR}}\\\text{Now, in} \Delta\text{PQR, we have proved that}\\\frac{\text{PE}}{\text{EQ}}=\frac{\text{PF}}{\text{FR}}$$

∴ EF || QR Hence Proved.

Play Video about Chapter 6 Triangles in One Shot

6. In figure A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.

Sol. Given : AB || PQ and AC || PR

To Prove : BC || QR

Proof : In given figure,

∴ AB || PQ (Given)

$$\Rarr\space\frac{\text{OA}}{\text{AP}}=\frac{\text{OB}}{\text{BQ}}\space\text{...(i)}$$

(Basic proportionality theorem)

Also, in figure, AC || PR (Given)

$$\Rarr\space\frac{\text{OA}}{\text{AP}}=\frac{\text{OC}}{\text{CR}}\space\text{...(ii)}$$

(Basic proportionality theorem)

From equations (i) and (ii), w
e get

$$\Rarr\space\frac{\text{OB}}{\text{BQ}}=\frac{\text{OC}}{\text{CR}}$$

∴ BC || QR Hence Proved.

7. Using Theorem, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).

Sol. Given : A ΔABC, in which D is the mid-point of AB and l || BC.

To Prove : E is the mid-point of AC.

Proof : In ΔABC, D is the mid-point of AB.

$$\text{i.e.}\space\frac{\text{AD}}{\text{DB}}=1\space\text{...(i)}$$

As straight line l || BC.

Line l is drawn through D and it meets AC at E.

By basic proportionality theorem, we get

$$\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}\\\Rarr\space\frac{\text{AE}}{\text{EC}}=1\space\text{[From equation (i)]}$$

⇒ AE = EC

$$\Rarr\space\frac{\text{AE}}{\text{EC}}=1$$

⇒ E is the mid-point of AC. Hence Proved.

8. Using Theorem, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.

(Recall that you have done it in Class IX).

Sol. Given : A DABC in which D and E are the mid-points of sides D and E.

To Prove : DE || BC

Proof : In ΔABC, D and E are mid-points of side AB and AC, respectively.

$$\Rarr\space\frac{\text{AD}}{\text{DB}}=1\text{and}\space\frac{\text{AE}}{\text{EC}}=1\space\text{(Given)}\\\Rarr\space\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}$$

∴ DE || BC

(By converse of basic proportionality theorem)

9. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O.

$$\textbf{Show that}\frac{\textbf{AO}}{\textbf{BO}}=\frac{\textbf{CO}}{\textbf{DO}}$$

Construction : Draw EOF || AB

Proof :

In ΔACD, OE || CD

$$\Rarr\space\frac{\text{AE}}{\text{ED}}=\frac{\text{AO}}{\text{OC}}$$

(by basic proportionality theorem)

In ΔABD, OE || BA

$$\Rarr\space\frac{\text{DE}}{\text{ED}}=\frac{\text{DO}}{\text{OB}}$$

(by basic proportionality theorem)

$$\text{or}\space\frac{\text{AE}}{\text{ED}}=\frac{\text{OB}}{\text{OD}}\space\text{...(i)}$$

From equation (i) and (ii), w
e get

$$\text{or}\space\frac{\text{AE}}{\text{ED}}=\frac{\text{OB}}{\text{OD}}\space\text{...(i)}$$

From equation (i) and (ii), w
e get

$$\frac{\text{AO}}{\text{OC}}=\frac{\text{OB}}{\text{OD}}\\\text{i.e.\space}\frac{\text{AO}}{\text{BO}}=\frac{\text{CO}}{\text{DO}}\space\textbf{Hence Proved.}$$

10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that

$$\frac{\textbf{AO}}{\textbf{BO}}=\frac{\textbf{CO}}{\textbf{DO}}$$

show that ABCD is a trapezium.

Sol. Given : ABCD is a quadrilateral, in which diagonals intersect at O, such that

$$\frac{\text{AO}}{\text{OB}}=\frac{\text{OC}}{\text{OD}}$$

To Prove : ABCD is a trapezium

Proof : In figure,

$$\frac{\text{AO}}{\text{BO}}=\frac{\text{CO}}{\text{DO}}$$

$$\frac{\text{AO}}{\text{BO}}=\frac{\text{CO}}{\text{DO}}\space\text{(Given)}\\\Rarr\frac{\text{AO}}{\text{OC}}=\frac{\text{BO}}{\text{OD}}\space\text{...(i)}$$

Through O, we draw

OE || BA

In ΔDAB, EO || AB

$$\Rarr\space\frac{\text{DE}}{\text{EA}}=\frac{\text{DO}}{\text{OB}}\\\text{or}\space\frac{\text{AE}}{\text{ED}}=\frac{\text{BO}}{\text{OD}}\space\text{..(ii)}$$

From equation (i) and (ii), w
e get

$$\frac{\text{AO}}{\text{OC}}=\frac{\text{AE}}{\text{ED}}$$

⇒ OE || CD

(by converse of basic proportionality theorem)

Now, we have

BA|| OE and OE || CD

Therefore, AB || CD

Hence Proved.

Exercise 6.3

1. State which pairs of triangles in figure are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form :

Sol. (i) In ΔABC and ΔPQR,

∠A = ∠P = 60°, ∠B = ∠Q = 80°
and ∠C = ∠R = 40°

Here, corresponding angles are equal.
Therefore, ΔABC ~ ΔPQR

(By AAA similarity criterion)

(ii) In DABC and ΔPQR,

$$\frac{\text{AB}}{\text{QR}}=\frac{\text{2}}{\text{4}}=\frac{\text{1}}{\text{2}},\frac{\text{BC}}{\text{RP}}=\frac{\text{2.5}}{\text{5}}=\frac{1}{2}\space\text{and}\space\frac{\text{CA}}{\text{PQ}}=\frac{\text{3}}{\text{6}}=\frac{\text{1}}{\text{2}}$$

Here, ratio of all corresponding sides are equal.

Therefore, ΔABC ~ ΔQRP

(By SSS similarity criterion)

(iii) In ΔLMP and ΔDEF

$$\frac{\text{MP}}{\text{DE}}=\frac{\text{2}}{\text{4}}=\frac{\text{1}}{\text{2}},\frac{\text{BC}}{\text{RP}}=\frac{2.5}{5}=\frac{1}{2}\\\text{and}\space\frac{\text{CA}}{\text{PQ}}=\frac{3}{6}=\frac{1}{2}$$

Here, ratio of all corresponding sides are equal.

Therefore, ΔABC ~ ΔQRP

(By SSS similarity criterion)

(iii) In ΔLMP and ΔDEF

$$\frac{\text{MP}}{\text{DE}}=\frac{2}{4}=\frac{1}{2}\\\text{and}\space\frac{\text{LP}}{\text{DF}}=\frac{\text{3}}{\text{6}}=\frac{\text{1}}{\text{2}}\\\text{and}\space\frac{\text{LM}}{\text{EF}}=\frac{\text{2.7}}{\text{5}}\neq\frac{1}{2}\\\text{i.e.\space}\frac{\text{MP}}{\text{DE}}=\frac{\text{LP}}{\text{DF}}\neq\frac{\text{LM}}{\text{EF}}$$

Here, ratio of all corresponding sides are not equal.

Thus, the two triangles are not similar.

(iv) In ΔLMN and ΔPQR

∠M = ∠Q = 70°

$$\frac{\text{MN}}{\text{PQ}}=\frac{\text{2.5}}{\text{5}}=\frac{\text{1}}{\text{2}}\\\frac{\text{ML}}{\text{QR}}=\frac{5}{10}=\frac{1}{2}\\\text{i.e.\space}\frac{\text{MN}}{\text{PQ}}=\frac{\text{ML}}{\text{QR}}$$

Here, the ratio of two corresponding adjacent sides are equal and one angle is also equal.

Therefore, ΔMNL ~ ΔQPR

(By SAS similarity criterion)

(v) In ΔABC, ∠A is given but the included side AC is not given.

So, can't decide whether ΔABC and ΔDEF are similar or not.

(vi) In ΔDEF and ΔPQR,

∠D = 70°, ∠E = 80° then ∠F = 30°

(∵ In ΔDEF, ∠D + ∠E + ∠F = 180°)
∠Q = 80°, ∠R = 30°, then ∠P = 70°
(∵ In ΔQPR, ∠Q + ∠P + ∠R = 180°)
Here, ∠D = ∠P, ∠E = ∠Q, ∠F = ∠R

Therefore, ΔDEF ~ ΔPQR

(By AAA similarity criterion)

2. In figure, ΔODC ~ ΔOBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.

Sol. Since, DOB is a straight line.

∴ ∠DOC + ∠COB = 180°

⇒ ∠DOC + 125° = 180°

⇒ ∠DCO + 180° – 125° = 55°

In ΔOC,

∠DCO + ∠CDO + ∠DOC = 180°

(Angle sum property)

∠DCO + 70° + 55° = 180°

⇒ ∠DCO + 125° = 180°

⇒ ∠DCO = 180° – 125° = 55°

Now, it is given that,

ΔODC ~ ΔOBA

∴ ∠OCD = ∠OAB

⇒ ∠OAB = ∠OCD = ∠DCO = 55°

i.e., ∠OAB = 55°

Hence, we have ∠DOC = 55°, ∠DCO = 55° and ∠OAB = 55°.

3. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles,

$$\textbf{show that}\space\frac{\textbf{OA}}{\textbf{OC}}=\frac{\textbf{OB}}{\textbf{OD}}.$$

Sol. Given : ABCD is a trapezium with AB || CD and AC and BD are diagonals intersect at O.

$$\textbf{To Prove :}\space\frac{\textbf{OA}}{\textbf{OC}}=\frac{\textbf{OB}}{\textbf{OD}}.$$

Proof : In figure, AB || DC (Given)

⇒ ∠1 = ∠3, ∠2 = ∠4

(Alternate interior angles)

Also,

∠DOC = ∠BOA

(Vertically opposite angles)

∴ ΔOCD ~ ΔOAB

$$\Rarr\space\frac{\text{OC}}{\text{OA}}=\frac{\text{OD}}{\text{OB}}$$

(Ratios of the corresponding sides of the similar triangles are equal)

$$\Rarr\space\frac{\text{OA}}{\text{OC}}=\frac{\text{OB}}{\text{OD}}\space\text{(taking reciprocals)}\\\textbf{Hence Proved.}$$

4. In figure,$$\frac{\textbf{QR}}{\textbf{OC}}=\frac{\textbf{OB}}{\textbf{OD}}\space\textbf{and}\angle\textbf{1}=\angle\textbf{2},\textbf{show that}\Delta\textbf{PQS}∼\Delta\textbf{TQR.}$$

$$\textbf{Sol.}\space\text{Given}:\frac{\text{QR}}{\text{QS}}=\frac{\text{QT}}{\text{PR}}\space\text{and}\space\angle1=\angle2$$

To Proof : ΔPQS ~ ΔTQR

Proof : In figure,

∠1 = ∠2 (Given)

⇒ PQ = PR

(Sides opposite to equal angle are equal)

$$\text{And,}\space\frac{\text{QR}}{\text{QS}}=\frac{\text{QT}}{\text{PR}}\space(\text{given})\\\Rarr\space\frac{\text{QR}}{\text{QS}}=\frac{\text{QT}}{\text{PQ}}\space(\because \text{PQ=PR})\\\text{or}\space\frac{\text{QS}}{\text{QR}}=\frac{\text{PQ}}{\text{QT}}$$

(Taking reciprocals) ...(i)

Now, in ΔPQS and ΔTQR, we have

∠PQS = ∠TQR = ∠Q

$$\text{and}\space\frac{\text{QS}}{\text{QR}}=\frac{\text{PQ}}{\text{QT}}\space\text{[By equation (i)]}$$

Therefore, by SAS similarity criterion, we have ΔPQS ~ ΔTQR. Hence Proved.

5. S and T are points on sides PR and QR of DPQR such that ∠P = ∠RTS. Show that DRPQ ~ DRTS.

Sol. Given : A DRPQ such that S and T are points on sides PR and QR and ∠P = ∠RTS.

To Prove : ΔRPQ ~ ΔRTS

Proof : In figure, we have ΔRPQ and ΔRTS in which

∠RPQ = ∠RTS (Given)

∠PRQ = ∠SRT = ∠R

Then, by AA similarity criterion, we have

ΔRPQ ~ΔRTS   Hence Proved.

6. In figure, if ΔABE ≅ ΔACD, show that ΔADE ~ ΔABC.

Sol. Given : ΔABE ≅ ΔACD

To Prove : ΔADE ~ ΔABC

Proof : Since, ΔABE ≅ ΔACD

∴ AB = AC and AE = AD (cpct)

$$\Rarr\space\frac{\text{AB}}{\text{AC}}=1\text{and}\space\frac{\text{AD}}{\text{AE}}=1\\\text{or}\space\frac{\text{AB}}{\text{AC}}=\frac{\text{AD}}{\text{AE}}\space\text{...(i)}$$

Now, in ΔADE and ΔABC, we have

$$\frac{\text{AD}}{\text{AE}}=\frac{\text{AB}}{\text{AC}}\space\text{(from (i))}\\\text{i.e.}\space\frac{\text{AD}}{\text{AB}}=\frac{\text{AE}}{\text{AC}}$$

and also, ∠DAE = ∠BAC = ∠A

(By SAS similarity criterion)

Hence Proved.

7. In figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that :

(i) ΔAEP ~ ΔCDP

(ii) ΔABD ~ ΔCBE

(iv) ΔPDC ~ ΔBEC

Sol. (i) Given : Altitudes AD and CE of ΔABC intersect each other at point P.

In ΔAEP and ΔCDP

∠AEP = ∠CDP = 90°

and ∠APE = ∠CPD

(Vertically opposite angles)

∴ ΔAEP ~ ΔCDP

(By AA similarity criterion)

(ii) In ΔABD and ΔCBE

and ∠ABD = ∠CBE = ∠B

∴ ΔABD ~ ΔCBE

(By AA similarity criterion)

and ∠PAE = ∠DAB (Common angle)

(By AA similarity criterion)

(iv) In ΔPDC and ΔBEC

∠PDC = ∠BEC = 90°

and ∠PCD = ∠BCE (Common angle)

∴ ΔPDC ~ ΔBEC

(By AA similarity criterion)

Hence Proved.

8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ~ ΔCFB.

Sol. Given : A parallelogram ABCD in which a line AD is produced to E and BE is joined.

To Prove : ΔABE ~ ΔCFB

Proof : In parallelogram ABCD,
∠A = ∠C

(opposite angles of parallelogram) ...(i)

Now, in ΔABE and ΔCFB,

we have

∠EAB = ∠BCF [From equation (i)]

∠ABE = ∠BFC

(Pair of alternate angles as AB || FC)

⇒ ΔABE ~ ΔCFB (by AA similarity)

9. In figure, ABC and AMP are two right triangles, right angled at B and M, respectively. Prove that :

(i) ΔABC ~ ΔAMP

$$\textbf{(ii)}\space\frac{\textbf{CA}}{\textbf{PA}}=\frac{\textbf{BC}}{\textbf{MP}}$$

Sol. Given : ΔABC and ΔAMP are two right triangles right angled at B and H.

To Prove : (i) ΔABC ~ ΔAMP

$$\text{(ii)}\space\frac{\text{CA}}{\text{PA}}=\frac{\text{BC}}{\text{MP}}$$

Proof : (i) In ΔABC and ΔAMP

∠ABC = ∠AMP (Each = 90°)

Because in ΔABC and ΔAMP are right angled at B and M, respectively.

Also, ∠BAC = ∠PAM = ∠A

⇒ ΔABC ~ ΔAMP

(By AA similarity criterion)

(ii) Since ΔABC ~ ΔAMP,

$$\therefore\space\frac{\text{AC}}{\text{AP}}=\frac{\text{BC}}{\text{MP}}$$

(Ratio of the corresponding sides of similar triangles are equal)

$$\text{or}\space\frac{\text{CA}}{\text{PA}}=\space\frac{\text{BC}}{\text{MP}}\space\textbf{Hence Proved.}$$

10. CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG, respectively. If ΔABC ~ ΔFEG. Show that :

$$\textbf{(i)}\space\frac{\textbf{CD}}{\textbf{GH}}=\frac{\textbf{AC}}{\textbf{FG}}$$

(ii) ΔDCB ~ ΔHGE

(iii) ΔDCA ~ ΔHGF

Sol. Given : Two ΔABC and ΔEFG and CD and GH are the bisectors of ∠ACB and ∠EGG.

ΔABC ~ ΔFEG

$$\textbf{To Prove : (i)}\frac{\text{CD}}{\text{GH}}=\frac{\text{AC}}{\text{FG}}$$

(ii) ΔDCB ~ ΔHGE

(iii) ΔDCA ~ ΔHGF

Proof : (i) In ΔACD and ΔFGH

$$\bigg[\because\Delta \text{ABC}∼\Delta \text{FEG}\therefore\space\angle\text{CAB}=\angle\text{GFE}\bigg]$$

∠ACD = ∠FGH ...(ii)

$$\begin{bmatrix}\because\space\Delta\text{ABC∼}\Delta\text{FEG}\\\therefore\angle\text{ACB}=\angle\text{FGE}\\\Rarr\space\frac{1}{2}\angle\text{ACB}=\frac{1}{2}\angle\text{FGE}\space(\text{halves of equal angles})\end{bmatrix}$$

From equation (i) and (ii), we get

ΔACD ~ ΔFGH

(by AA similarity criterion)

$$\therefore\space\frac{\text{CD}}{\text{GH}}=\frac{\text{AC}}{\text{FG}}$$

(∵ Corresponding sides of two similar triangles are proportional)

(ii) In ΔDCB and ΔHGE,

∠DBC = ∠HEF ...(iii)

$$\\\begin{cases}\because\Delta\text{ABC}∼\Delta\text{FEG}\\\therefore \angle\text{ABC}=\angle\text{FEG}\end{cases}$$

and ∠DCB = ∠HGE ...(iv)

$$\\\begin{cases}\because\space\Delta\text{ABC}∼\Delta\text{FEG}\\\therefore\angle\text{ABC}=\angle\text{FGE}\\\Rarr\space\frac{1}{2}\angle\text{ACB}=\frac{1}{2}\angle\text{FGE}\space\text{(Halves of equals angles)}\end{cases}$$

From equations (iii) and (iv), we get

ΔDCB ~ ΔHGE

(by AA similarity criterion)

(iii) In ΔCDA and DHGF,

∠DAC = ∠HFG ...(v)

$$\begin{bmatrix}\because\space\Delta\text{ABC}∼\Delta\text{FEG}\\\therefore\space\angle\text{CAB}=\angle\text{GFE}\\\Rarr\space\angle\text{CAD}=\angle\text{GFH}\end{bmatrix}\\\text{∠DCA = ∠HGF ...(vi)}\\\begin{bmatrix}\because\space\Delta\text{ABC}∼\Delta\text{FEG},\space\therefore\angle\text{ACB}=\angle\text{FGE}\\\therefore\frac{1}{2}\angle\text{ACB}=\frac{1}{2}\angle\text{FGE}\space\Rarr\space\angle\text{DAC}=\angle\text{HGF} \end{bmatrix}$$

(Halves of equal angles)

From equation (v) and (vi), we get

ΔDCA ~ ΔHGF

(by AA similarity criterion)

Hence Proved.

11. In figure, E is a point on side CB produced of an isosceles ΔABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ΔABD ~ ΔECF.

Sol. Given : ΔABC is an isosceles triangle in which AB = AC.

AD ⊥ BC and EF ⊥ AC
To Prove : ΔABD ~ ΔECF

Proof : ΔABC is an isosceles triangle.

∴ AB = AC ⇒ ∠B = ∠C ...(i)

∠ABD = ∠ECF [From equation (i)]

and ∠ADB = ∠EFC = 90°

ΔABD ~ ΔECF

(by AA similarity criterion)

Hence Proved.

12. Sides AB and BC are median AD of a ΔABC are respectively proportional to sides PQ and QR and median PM of DPQR. Show that ΔABC ~ ΔPQR.

Sol. Given : In ΔABC and ΔPQR, where AD and PM are median of triangles,

$$\frac{\text{AB}}{\text{PQ}}=\frac{\text{BC}}{\text{QR}}=\frac{\text{AD}}{\text{PM}}$$

To prove : ΔABC ~ ΔPQR

Proof : Since, AD and PM are medians of ΔABC and ΔPQR

$$\therefore\space\text{BD}=\frac{1}{2}\text{BC}\space\text{and QN}=\frac{1}{2}\text{QR}\space\text{...(i)}\\\text{Given that}\\\frac{\text{AB}}{\text{PQ}}=\frac{\text{BC}}{\text{QR}}=\frac{\text{AD}}{\text{PM}}\space\text{...(ii)}$$

∴ From (i) and (ii),

$$\frac{\text{AB}}{\text{PQ}}=\frac{\text{AB}}{\text{QR}}=\frac{\text{AB}}{\text{PM}}\space\text{...(iii)}\\\text{In} \Delta \text{ABD and} \Delta\text{PQM,}\\\frac{\text{AB}}{\text{PQ}}=\frac{\text{BD}}{\text{QM}}=\frac{\text{AD}}{\text{PM}}\space\text{[by (iii)]}$$

∴ By SSS criterion of proportionality

ΔABD ~ ΔPQM

∴ ∠B = ∠Q ...(iv)

In ΔABC and ΔPQR

$$\frac{\text{AB}}{\text{PQ}}=\frac{\text{BC}}{\text{QR}}\space\text{(from (ii))}$$

and ∠B = ∠Q (from iv)

∴ ΔABC ~ ΔPQR (by SAS criterion)

Hence Proved.

13. D is point on the side BC of a ΔABC such that ∠ADC = ∠BAC. Show that CA2 = CB.CD.

Sol. Given : A ΔABC and D is a point on BC.

To Prove : CA2 = CB . CD

Proof : In ΔABC and ΔDAC, we have

and ∠ACB = ∠DCA = ∠C (Common)

∴ ΔABC ~ ΔDAC

(by AA similarity criterion)

$$\therefore\space\frac{\text{AC}}{\text{CB}}=\frac{\text{CD}}{\text{CA}}$$

(Corresponding sides are similar Δs)

$$\Rarr\space\frac{\text{CA}}{\text{CD}}=\frac{\text{CB}}{\text{CA}}$$

⇒ CA × CA = CB × CD

⇒ CA2 = CB × CD Hence Proved.

14. Sides AB and AC and median AD of a DABC are respectively proportional to sides PQ and PR and median PM of another ΔPQR. Show that ΔABC ~ ΔPQR.

Sol. Given : ΔABC and ΔPQR, in which AD and PM are their medians, respectively.

$$\text{Also}\space\frac{\text{AB}}{\text{PQ}}=\frac{\text{AC}}{\text{PR}}=\frac{\text{AD}}{\text{PM}}\qquad\text{...(i)}$$

To prove : ΔABC ~ ΔPQR

Construction : Produce AD to E such that AD = DE and produce PM to N such that PM = MN. Join BE, CE, QN and RN.

Proof: Quadrilaterials ABEC and PQNR are parallelograms because their diagonals bisect each other at D and M, respectively.

⇒ BE = AC

and QN = PR

$$\Rarr\space\frac{\text{BE}}{\text{QN}}=\frac{\text{AC}}{\text{PR}}\\\Rarr\space\frac{\text{BE}}{\text{QN}}=\frac{\text{AB}}{\text{PQ}}\space\text{[By equation (i)]}\\\text{i.e.\space}\frac{\text{AB}}{\text{PQ}}=\frac{\text{BE}}{\text{QN}}\space\text{...(ii)}\\\text{From equation (i),}\\\frac{\text{AB}}{\text{PQ}}=\frac{\text{AD}}{\text{PM}}=\frac{\text{2AD}}{\text{2PM}}=\frac{\text{AE}}{\text{PN}}$$

(∵ Diagonals are bisect each other)

$$\text{i.e.}\space\frac{\text{AB}}{\text{PQ}}=\frac{\text{AE}}{\text{PN}}\space\text{...(iii)}$$

From equations (ii) and (iii), we have

$$\frac{\text{AB}}{\text{PQ}}=\frac{\text{BE}}{\text{QN}}=\frac{\text{AE}}{\text{PN}}\space\text{...(iii)}$$

From equations (ii) and (iii), we have

$$\frac{\text{AB}}{\text{PQ}}=\frac{\text{BE}}{\text{QN}}=\frac{\text{AE}}{\text{PN}}$$

∴ ΔABE ~ ΔPQN (by SSS similarity)

⇒ ∠1 = ∠2 ...(iv)

Similarly, we can prove that

ΔACE ~ ΔPRN

∠3 = ∠4 ...(v)

On adding equations (iv) and (v), w
e have

∠1 + ∠3 = ∠2 + ∠4

⇒ ∠A = ∠P

$$\text{and}\space\frac{\text{AB}}{\text{PQ}}=\frac{\text{AC}}{\text{PR}}\space\text{(by equation (i))}$$

∴ ΔABC ~ ΔPQR

(SAS similarity criterion)

⇒ ∠1 = ∠2 ...(iv)

Similarly, we can prove that

ΔACE ~ ΔPRN

∠3 = ∠4 ...(v)

On adding equations (iv) and (v), we have

∠1 + ∠3 = ∠2 + ∠4

⇒ ∠A = ∠P

$$\text{and}\space\frac{\text{AB}}{\text{PQ}}=\frac{\text{AC}}{\text{PR}}\space\text{(by equation (i))}$$

∴ ΔABC ~ ΔPQR

(SAS similarity criterion)

15. A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

Sol. In figure (i), AB is a pole and behind it a Sun is risen which casts a shadow of length BC = 4 cm and makes an angle q to the horizontal and in figure (ii) PM is a height of the tower and behind a Sun risen which casts a shadow of length,

NM = 28 cm

In ΔACB and ΔPNM,

∠C = ∠N = θ

and  ∠ABC = ∠PMN = 90°

∴ ΔABC ~ ΔPMN

(by AA similarity criterion)

$$\Rarr\space\frac{\text{AB}}{\text{PM}}=\frac{\text{BC}}{\text{MN}}\\\Rarr\space\frac{\text{AB}}{\text{BC}}=\frac{\text{PM}}{\text{MN}}\\\Rarr\space\frac{6}{4}=\frac{h}{28}\\\Rarr\space\frac{6×28}{4}=42\space\text{m}$$

16. If AD and PM are medians of ΔABC and ΔPQR, respectively, where ΔABC ~ ΔPQR, prove that

$$\frac{\textbf{AB}}{\textbf{PQ}}=\frac{\textbf{AD}}{\textbf{PM}}.$$

Sol. Given : ΔABC and ΔPQR and AD and PM are the medians of the ΔABC and ΔPQR respectively.
ΔABC ~ ΔPQR

To Prove :

$$\frac{\text{AB}}{\text{PQ}}=\frac{\text{AD}}{\text{PM}}$$

Proof : ΔABC ~ ΔPQR (Given)

$$\text{and}\space\frac{\text{AB}}{\text{PQ}}=\frac{\text{BC}}{\text{QR}}=\frac{\text{AC}}{\text{PR}}\space\text{...(i)}$$

∠A = ∠P, ∠B = ∠Q, ∠C = ∠R ...(ii)

$$\text{Now,}\space\text{BD = CD =}\frac{1}{2}\text{BC}\space\text{...(iii)}\\\text{and \space QM = RM}=\frac{1}{2}\text{QR}\space\text{...(iv)}$$

(∵ D is mid-point on BC and M is mid-point of QR)

From equation (i),

$$\frac{\text{AB}}{\text{PQ}}=\frac{\text{BC}}{\text{QR}}\\=\frac{\text{AB}}{\text{PQ}}=\frac{\text{2BD}}{\text{2QM}}$$

[By equation (iii) & (iv)]

$$\Rarr\space\frac{\text{AB}}{\text{PQ}}=\frac{\text{BC}}{\text{QM}}\\\text{Thus, we have}\space\frac{\text{AB}}{\text{PQ}}=\frac{\text{BD}}{\text{QM}}$$

and ∠ABD = ∠PQM (∵∠B = ∠Q)

∴ ΔABD ~ ΔPQM

(by SAS similarity criterion)

$$\Rarr\space\frac{\text{AB}}{\text{PQ}}=\frac{\text{AD}}{\text{PM}}\space\textbf{Hence Proved.}$$

Exercise 6.4

1. Let ΔABC ~ ΔDEF and their areas be respectively, 64 cm2 and 121 cm2. If EF = 15.4 cm, find BC

Sol. Given, ΔABC ~ ΔDEF

Since, the ratio of the areas of two similar Δs is equal to the square of the ratio of their corresponding sides.

$$\therefore\space\frac{\text{ar(}\Delta \text{ABC})}{\text{ar(}\Delta \text{DEF})}=\frac{\text{BC}^{2}}{\text{EF}^{2}}\\\Rarr\space\frac{64}{121}=\frac{\text{BC}^{2}}{\text{EF}^{2}}\\\Rarr\space\bigg(\frac{\text{BC}}{\text{EF}}\bigg)^{2}=\bigg(\frac{\text{8}}{\text{11}}\bigg)^{2}\Rarr\frac{\text{BC}}{\text{EF}}=\frac{8}{11}\\\Rarr\space\text{BC}=\frac{8}{11}×\text{EF}\\\Rarr\space\text{BC}=\frac{8}{11}×15.4=11.2\space\text{cm}$$

2. Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2 CD, Find the ratio of the areas of Triangles AOB and Triangle COD.

Sol. ABCD is a trapezium, with AB = 2CD

$$\frac{\text{ar(}\Delta\text{AOB})}{\text{ar(}\Delta\text{COD})}=\frac{\text{AB}^{2}}{\text{CD}^{2}}\\\text{(By property of area of similar triangles)}$$

$$=\frac{(2\text{CD})^{2}}{\text{CD}^{2}}\space\text{(∵ AB = 2CD)}\\=\frac{(2\text{CD})^{2}}{\text{CD}^{2}}=\frac{4}{1}$$

3. In figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O,

$$\textbf{show that}\space\frac{\textbf{ar}\textbf{(ABC)}}{\textbf{ar}\textbf{(DBC)}}=\frac{\textbf{AO}}{\textbf{DO}}.$$

Sol. Given : ΔABC and ΔDBC on the same base BC.

AD and BC intersect at O.

$$\textbf{To Prove :}\space\frac{\text{ar(}\Delta\text{ABC})}{\text{ar(}\Delta\text{DBC})}=\frac{\text{AO}}{\text{OD}}\\\textbf{Construction :} \text{Draw AL} \perp \text{BC and DM} \perp \text{BC(See figure)}$$

Proof : In ΔOLA and ΔOMD

∠ALO = ∠DMO = 90°

and ∠AOL = ∠DOM

(Vertically opposite angle)

∴ ΔOLA ~ ΔOMD

(AAA similarity criterion)

$$\Rarr\space\frac{\text{AL}}{\text{DM}}=\frac{\text{AO}}{\text{DO}}\space\text{...(i)}\\\text{Now,}\space\frac{\text{ar}(\Delta\text{ABC})}{\text{ar}(\Delta\text{DBC})}=\frac{\frac{1}{2}×\text{(BC)}×\text{(AL)}}{\frac{1}{2}×\text{(BC)}×\text{(DM)}}\\=\frac{\text{AL}}{\text{DM}}=\frac{\text{AO}}{\text{DO}}\\\text{[By equation (i)]}\\\therefore\space\frac{\text{ar}(\Delta\text{ABC})}{\text{ar}(\Delta\text{DBC})}=\frac{\text{AO}}{\text{BO}}\space\textbf{Hence Proved.}$$

4. If the areas of two similar triangles are equal, prove that they are congruent.

Sol. Given : ΔABC ~ ΔPQR and ar(ΔABC) = ar(ΔPQR)

To Prove : ΔABC ≅ ΔPQR

Proof : Given,

ar(ΔABC) = ar(ΔPQR)

$$\therefore\space\frac{\text{ar}(\Delta\text{ABC})}{\text{ar}(\Delta\text{PQR})}=1\\\Rarr\space\frac{\text{AB}^{2}}{\text{PQ}^{2}}=\frac{\text{BC}^{2}}{\text{QR}^{2}}=\frac{\text{CA}^{2}}{\text{PR}^{2}}=1$$

(By property of area of similar triangles)

⇒ AB = PQ, BC = QR and CA = PR

(by SSS congruency criterion)

ΔABC ≅  ΔPQR.

5. D, E and F are respectively the mid-points of sides AB, BC and CA of DABC. Find the ratio of the areas of ΔDEF and ΔABC.

Sol. Given : A ΔABC in which D, E and F are mid points of sides AB, BC and AC, respectively. Join them.

By Mid-point theorem, we get

$$\text{DF}=\frac{1}{2}\text{BC},\space\text{DE =}\frac{1}{2}\text{CA}\\\text{and}\space\text{EF}=\frac{1}{2}\text{AB}\space\text{...(i)}\\\Rarr\space\frac{\text{DF}}{\text{BC}}=\frac{\text{DE}}{\text{CA}}=\frac{\text{EF}}{\text{AB}}=\frac{1}{2}$$

∴ ΔDEF ~ ΔCAB

(by SSS similarity criterion)

$$\Rarr\space\frac{\text{ar}(\Delta\text{DEF})}{\text{ar}(\Delta\text{CAB})}=\frac{\text{DE}^{2}}{\text{CA}^{2}}\\\text{(by SSS similarity criterion)}\\\frac{\bigg(\frac{1}{2}\text{CA}\bigg)^{2}}{\text{CA}^{2}}=\frac{1}{4}\\\bigg(\because\space\text{DE}=\frac{1}{2}\text{CA}\bigg)\\\therefore\space\frac{\text{ar}(\Delta\text{DEF})}{\text{ar(}\Delta\text{ABC})}=\frac{1}{4}$$

[∵ ar(ΔCAB) = ar(ΔABC)]

Hence, the required ratio is 1 : 4.

6. Prove that the ratio of the area of two similar triangles is equal to the square of the ratio of their corresponding medians.

Sol. Given : AD is a median of ΔABC and PM is a median of ΔPQR. Therefore, D is mid-point of BC and M is mid-point of QR and ΔABC ~ ΔPQR.

$$\textbf{To Prove :}\frac{\text{ar}(\Delta\text{ABC})}{\text{ar}(\Delta\text{PQR})}=\frac{\text{AD}^{2}}{\text{PM}^{2}}$$

Proof :

∴ ΔABC ~ ΔPQR (given)

⇒ ∠B = ∠Q ...(i)

(Corresponding angles are equal)

$$\text{Also}\space\frac{\text{AB}}{\text{PQ}}=\frac{\text{BC}}{\text{QR}}$$

(Ratio of corresponding sides of similar triangles are equal)

$$\Rarr\space\frac{\text{AB}}{\text{PQ}}=\frac{\text{2BD}}{\text{2QM}}$$

(∵D is mid-point of BC and M is mid-point of QR)

$$\Rarr\space\frac{\text{AB}}{\text{PQ}}=\frac{\text{BD}}{\text{QM}}\space\text{...(i)}$$

In ΔABD and ΔPQM,

∠ABD = ∠PQM [By equation (i)]

$$\text{and}\space\frac{\text{AB}}{\text{PQ}}=\frac{\text{BD}}{\text{QM}}\space\text{[By equation (ii)]}$$

⇒ ΔABD ~ ΔPQM

(by SAS similarity criterion)\

$$\therefore\space\frac{\text{AB}}{\text{PQ}}=\frac{\text{AD}}{\text{PM}}\space\text{...(iii)}$$

(corresponding sides of similar triangles are proportion)

$$\text{Now,}\space\frac{\text{ar(}\Delta\text{ABC})}{\text{ar(}\Delta\text{PQR})}=\frac{\text{AB}^{2}}{\text{PQ}^{2}}\\\text{[From equation (iii)]}\\\text{(Using property of area of similar triangles)}\\\Rarr\space\frac{\text{ar(}\Delta\text{ABC})}{\text{ar(}\Delta\text{PQR})}=\frac{\text{AD}^{2}}{\text{PM}^{2}}$$

[From equation (iii)]

Hence Proved.

7. Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.

Sol. Given : A square ABCD having sides of length = a.

Then, the diagonal,

$$\text{BD}=a\sqrt{2}$$

$$\textbf{To Prove :}\space\text{ar}(\Delta\text{PAB})=\frac{1}{2}\text{ar}(\Delta\text{QBD})$$

Proof : Triangles ΔPAB and ΔQBD are equilaterial triangle.

∴ ΔPAB ~ ΔQBD

(Equilateral triangles are similar)

$$\Rarr\space\frac{\text{ar}(\Delta\text{PAB})}{\text{ar}(\Delta\text{QBD})}=\frac{\text{AB}^{2}}{\text{BD}^{2}}$$

(By property of area of similar triangles)

$$=\frac{a^{2}}{(a\sqrt{2})^{2}}=\frac{1}{2}\\\Rarr\space\text{ar}(\Delta\text{PAB})=\frac{1}{2}\text{ar}(\Delta\text{QBD})\space\textbf{Hence Proved.}$$

Thick the correct answer and justify :

8. ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the area of ΔABC and ΔBDE is :

(a) 2 : 1

(b) 1 : 2

(c) 4 : 1

(d) 1 : 4

Sol. (c) 4 : 1

Explanation :

Here, ΔABC is an equilaterial triangle

∴ AB = BC = CA = a (Say)

$$\text{Now,\space BD}=\frac{1}{2}a\\(\because\text{D is mid-point of BC})$$

Now, ΔABC ~ ΔBDE

(∵ Equilaterial triangles are similar)

$$\Rarr\space\frac{\text{ar}(\Delta\text{ABC})}{\text{ar}(\Delta\text{BDE})}=\frac{\text{AB}^{2}}{\text{BD}^{2}}\\\text{(By property of area of similar to triangles)}\\=\frac{a^{2}}{\bigg(\frac{1}{2}a\bigg)}=\frac{4}{1}$$

i.e. The ratio is 4 : 1.

9. Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio

(a) 2 : 3    (b) 4 : 9    (c) 81 : 16    (d) 16 : 81

Sol. (d) 16 : 81

Explanation : Areas of two similar triangles are in the ratio of the square of their corresponding sides

$$=\bigg(\frac{4}{9}\bigg)^{3}=\frac{16}{81}$$

Exercise 6.5

1. Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse.

(i) 7 cm, 24 cm, 25 cm

(ii) 3 cm, 8 cm, 6 cm

(iii) 50 cm, 80 cm, 100 cm

(iv) 13 cm, 12 cm, 5 cm

Sol. By pythagoras theorem in right triangles, sum of squares of two smalller sides is equal to the square of the third (longest) side (hypotenuse).

(i) Here, (7)2 + (24)2 = 49 + 576 = 625 = (25)2

Therefore, given sides are 7 cm, 24 cm and 25 cm, will make a right triangle and length of its hypotenuse is 25 cm. (longest)

(ii) Here, (3)2 + (6)2 = 9 + 36 = 45 and (8)2 = 64. Both values are not equal.

Therefore, given sides 3 cm, 8 cm and 6 cm does not make a right angled triangle.

(iii) Here, (50)2 + (80)2 = 2500 + 6400 = 8900 and (100)2 = 10000. Both values are not equal.

Therefore, given sides 50 cm, 80 cm and 100 cm does not make a right angled triangle.

(iv) Here, (12)2 + (5)2 = 144 + 25 = 169 = (13)2

Therefore, given sides 13 cm, 12 cm and 5 cm makes a right angled triangle and length of the hypotenuse is 13 cm.

2. PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM2 = QM . MR.

Sol. Given : ΔPQR is a right angled triangle, in which ∠P = 90° and PM  QR

To Prove : PM2 = QM × MR

Proof : In ΔPQR and ΔMPQ,

∠1 + ∠2 = ∠2 + ∠4 (Each = 90°)

⇒ ∠1 = ∠4

Similarly, ∠2 = ∠3

and ∠PMR = ∠PMQ = 90° (given)

∴ ΔQPM ~ ΔPRM (by AA criterion)

$$\Rarr\space\frac{\text{ar(}\Delta\text{QPM})}{\text{ar(}\Delta\text{PRM})}=\frac{\text{PM}^{2}}{\text{RM}^{2}}\\\text{(By property of area of similar triangles)}\\\Rarr\space\frac{\frac{1}{2}(\text{QM}×\text{PM})}{\frac{1}{2}(\text{RM}×\text{PM})}=\frac{\text{PM}^{2}}{\text{RM}^{2}}\\\Rarr\space\frac{\text{QM}}{\text{RM}}=\frac{\text{PM}^{2}}{\text{RM}^{2}}$$

⇒ PM2 = QM × RM

or PM2 = QM × MR Hence Proved.

3. In figure, ABD is a triangle right angled at A and AC ⊥ BD. Show that

(i) AB2 = BC.BD

(ii) AC2 = BC.DC

Sol. Given : DABD is right angled at A and AC ⊥ BD
To Prove : (i) AB2 = BC.BD

(ii) AC2 = BC.DC

By theorem, that if a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then triangles on both sides of the perpendicular are similar to whole triangle and to each other.

∴ ΔABC ~ ΔDAC ~ ΔDBA

(i) Now ΔABC ~ ΔDBA

$$\text{Then}\space\frac{\text{ar}(\Delta\text{ABC})}{\text{ar}(\Delta\text{DBA})}=\frac{\text{AB}^{2}}{\text{DB}^{2}}\\\text{(By property of area of similar triangles)}\\\Rarr\frac{\frac{1}{2}(\text{BC})×(\text{AC})}{\frac{1}{2}(\text{BD})×(\text{AC})}=\frac{\text{AB}^{2}}{\text{DB}^{2}}$$

⇒ AB2 = BC.BD

Hence Proved.

(ii) ΔABC ~ ΔDAC

$$\Rarr\space\frac{\text{ar}(\Delta\text{ABC})}{\text{ar}\Delta(\text{DAC})}=\frac{\text{AC}^{2}}{\text{DC}^{2}}\\\text{(By property of area of similar triangles)}\\\Rarr\space\frac{\frac{1}{2}(\text{BC})×(\text{AC})}{\frac{1}{2}(\text{DC})×\text{(AC)}}=\frac{\text{AC}^{2}}{\text{DC}^{2}}$$

⇒ AC2 = BC. DC

Hence Proved.

(iii) ΔDAC ~ ΔDBA

$$\Rarr\space\frac{\text{ar}(\Delta\text{DAC})}{\text{ar}(\Delta\text{DAC})}=\frac{\text{DA}^{2}}{\text{DB}^{2}}$$

(By property of area of similar triangles)

$$\Rarr\frac{\frac{1}{2}(\text{CD})×(\text{AC})}{\frac{1}{2}(\text{BD})×(\text{AC})}=\frac{\text{AD}^{2}}{\text{BD}^{2}}$$

Hence Proved.

4. ABC is an isosceles triangle right angled at C. Prove that AB2 = 2AC2.

Sol. Given : ΔABC is an isosceles triangle right angled at C.

and AC = BC ...(i)

To Prove : AB2 = 2AC2

Proof : By Pythagoras theorem, in ΔABC,

AB2 = AC2 + BC2

= AC2 + AC2 = 2AC2      [∵ BC = AC]

Hence Proved.

5. ABC is an isosceles triangle with AC = BC. If AB2 = 2AC2, prove that ABC is a right triangle.

Sol. Given : An isosceles DABC with AC = BC and AB2 = 2AC2

To prove : ΔABC is a right angled triangle.

Proof : In ΔABC, we are given that

AC = BC ...(i)

and AB2 = 2AC2 ...(ii)

Now, AC2 + BC2 = AC2 + AC2 [By equation (i)]

= 2AC2 = AB2 [By equation (ii)]

i.e., AC2 + BC2 = AB2

Hence, by the converse of the Pythagoras theorem, we have ΔABC is right angled at C.

6. ABC is an equilateral triangle of side 2a. Find each of its altitudes.

Sol. Given : An equilateral ΔABC, each of side 2a and AD ⊥ BC.

and AB = AC

(sides of equilateral triangles)

$$\therefore\space\text{BD = CD =}\frac{1}{2}\text{BC}=a\space\text{(c.p.c.t.)}$$

Now, in ΔABD by Pythagoras theorem,

⇒ (2a)2 = AD2 + a2

$$\Rarr\space\text{AD}=\sqrt{3a}$$

7. Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

Sol. Given : ABCD is a rhombus in which AB = BC = CD = DA = a

Its diagonal AC and BD bisect each other at right angle at O.

To Prove : AB2 + BC2 + CD2 + AD2 = AC2 + BD2

Proof : In ΔOAB,

∠AOB = 90°

$$\text{and OA =}\frac{1}{2}\text{AC}\space\text{and OB =}\frac{1}{2}\text{BD}$$

In ΔAOB, by Pythagoras theorem

OA2 + OB2 = AB2

(∵ ∠AOB = 90°)

$$\Rarr\space\bigg(\frac{1}{2}\text{AC}\bigg)^{2}+\bigg(\frac{1}{2}\text{BD}\bigg)^{2}=\text{AB}^{2}$$

⇒ AC2 + BD2 = 4AB2

or   4AB2 = AC2 + BD2

⇒ AB2 + BC2 + CD2 + DA2 = AC2 + DB2

(∵ AB = BC = CD = DA)

Hence Proved.

8. In figure, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that

(i) OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2

(ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2

Sol. In ΔABC, from point O join lines OB, OC and OA.

(i) In right angled ΔOFA.

⇒ OA2 = OF2 + AF2

(By Pythagoras theorem)

⇒ OA2 – OF2 = AF2 ...(a)

Similarly, in ΔOBD,

OB2 – OD2 = BD2 ...(b)

and in ΔOCE,

OC2 – OE2 = CE2 ...(c)

On adding equation (a), (b) and (c), we get

OA2 + OB2 + OC2 – OD2 – OE2 – OF2

= AF2 + BD2 + CE2

(ii) From part (i), we get

OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2 ...(d)

Similarly,

OA2 + OB2 + OC2 – OD2 – OE2 – OF2

= BF2 + CD2 + AE2 ...(e)

From equation (d) and (e), we get

AF2 + BD2 + CE2 = AE2 + CD2 + BF2

9. A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.

Sol. Let A be the position of the window and AC be the length of the ladder.

Then, AB = 8 m (Height of window)

AC = 10 m (Length of ladder)

Let BC = x m be the distance of the foot of the ladder from the base of the wall.
Using Pythagoras theorem in DABC, we get

AC2 = AB2 + BC2

∴ 102 = B2 + x2

⇒ x2 = 100 – 64 = 36

⇒ x = 6 i.e., BC = 6 m

10. A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?

Sol. Let AB be the vertical pole of height 18 m. A guy wire is of length AC = 24 m.

Let BC = x m be the distance of the stake from the base of the pole.

Using Pythagoras theorem is DABC, we get

i.e., AC2 = AB2 + BC2

∴ (24)2 = x2 + 182

⇒ x2 = (24)2 – (18)2

= 576 – 324

= 252

$$\Rarr\space x=\sqrt{252}\space \text{m}$$

(∵ we take positive sign becasue cannot be negative)

$$\Rarr\space x=6\sqrt{7}\space \text{m}$$

11. An aeroplane leaves an airport and flies due north at a speed of 1000 kmh–1. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 kmh–1.

$$\textbf{How far apart will be two planes after}\space\textbf{1}\frac{\textbf{1}}{\textbf{2}}?$$

$$\therefore\space\text{Length of BC = 1000}×\frac{3}{2}\text{km}$$

(∵ Distance = speed × time)

= 1500 km

The second plane travels distance BA in the direction of west at a speed of 1200 km/h.

$$\therefore\space\text{Length of BA = 1200}×\frac{3}{2}=1800\space\text{km}$$

In right angled ΔABC,

AC2 = AB2 + BC2

(By Pythagoras theorem)

= (1800)2 + (1500)2

= 3240000 + 2250000

= 5490000

$$\Rarr\space\text{AC}=\sqrt{5490000}\space\text{km}\\\Rarr\space\text{AC}=300\sqrt{61}\space\text{km}\\\text{Hence, the distance between two planes is}\\\space300\sqrt{61}\space\text{km.}$$

12. Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.

Sol. Let BC and AD be the two poles of heights 11 m and 6 m and distance between them is AB = 12 m

Then, CE = BC – BE

= 11 – 6

= 5 m

Now, AB = DE = 12 m

and  AD = EB = 6 m

Let distance between tops of two poles DC = x m

By using Pythagoras theorem in ΔDEC, we get

i.e., DC2 = DE2 + CE2

⇒ x2 = (12)2 + (5)2 = 169

⇒ x = 13

Hence, distance between their tops = 13 m.

13. D and E are points on the sides CA and CB respectively of a ΔABC right angled at C. Prove that AE2 + BD2 = AB2 + DE2.

Sol. Given : DABC right angled at C. D and E are points on the sides CA and BC, respectively.

To Prove : AE2 + BD2 = AB2 + DE2

Construction : Join ED, BD and EA.

Proof : In right angled ΔACE,

AE2 = CA2 + CE2 ...(i)

(By Pythagoras theorem)

and in right angled DBCD,

BD2 = BC2 + CD2 ...(ii)

On adding equation (i) and (ii), we get

AE2 + BD2 = (CA2 + CE2) + (BC2 + CD2)

= (BC2 + CA2) + (CD2 + CE2)

(∵ In ΔABC, BA2 = BC2 + CA2 and in ΔECD,
DE2 = CD2 + CE2)

= BA2 + DE2

(By Pythagoras theorem)

∴ AE2 + DB2 = AB2 + DE2 Hence Proved.

14. The perpendicular from A on side BC of a ΔABC intersects BC at D such that DB = 3 CD (see in figure). Prove that 2AB2 = 2AC2 + BC2.

Sol. Given : ΔABD, in which DB = 3CD

To Prove : 2 AB2 = 2AC2 + BC2

Proof : Given, DB = 3CD

$$\Rarr\space\text{CD}=\frac{1}{4}\text{BC}\\\text{and}\space\text{DB}=\frac{3}{4}\text{BC}$$

In ΔABD, by pythagoras theorem

AB2 = DB2 + AD2 ...(ii)

In ΔACD, by pythagoras theorem

AC2 = CD2 + AD2 ...(iii)

On subtracting equations (iii) from equation (ii), we get

AB2 – AC2 = DB2 – CD2

$$=\bigg(\frac{3}{4}\text{BC}\bigg)^{2}-\bigg(\frac{1}{4}\text{BC}\bigg)^{2}\\=\frac{9}{16}\text{BC}^{2}-\frac{1}{16}\text{BC}^{2}=\frac{1}{2}\text{BC}^{2}$$

⇒ 2AB2 – 2AC2 = BC2

⇒ 2AB2 = 2AC2 + BC2 Hence Proved.

15. In an equilateral DABC, D is a point on side BC such that

$$\textbf{BD}=\frac{1}{3}\textbf{BC}.\textbf{Prove that 9AD}^2 = \textbf{7AB}^2.$$

Sol. Given : ΔABC is an equilateral triangle,

$$\text{D is a point on side BC such that BD}=\frac{1}{3}\text{BC.}$$

To Prove : 9AD2 = 7AB2

Consturction : Draw a line AE is perpendicular to BC.

Proof :

AB = BC = CA = a (Say)

(By property of equilateral triangle)

$$\text{and}\space\text{BD}=\frac{1}{3}\text{BC}=\frac{1}{3}a\space\text{(given)}\\\Rarr\space\text{CD}=\frac{2}{3}\text{BC}=\frac{2}{3}\text{a}$$

∵ AE ⊥ BC

∵ In an equilateral triangel, altitude AE is perpendicular bisector of BC.

$$\Rarr\space\text{BE}=\text{EC}=\frac{1}{2}a\\\text{Now,}\space\text{DE = BE – BD =}\frac{1}{2}a-\frac{1}{3}a-\frac{1}{6}a$$

AD2 = AE2 + DE2 ...(i)

In ΔABE,

AB2 = AE2 + BE2

or

AE2 = AB2 – BE2 ...(ii)

Put the value of (ii) in (i),

AD2 = AB2 – BE2 + DE2

$$=a^{2}-\bigg(\frac{1}{2}a\bigg)^{2}+\bigg(\frac{1}{6}a\bigg)^{2}\\=a^{2}-\frac{1}{4}a^{2}+\frac{1}{36}a{2}\\=\frac{(36-9+1)a^{2}}{36}\\=\frac{28}{36}a^{2}\\=\frac{7}{9}\text{AB}^{2}$$

⇒ 9 AD2 = 7 AB2   Hence Proved.

16. In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.

Sol. Given : An equilateral ΔABC is of side a.

To Prove : 3(Side)2 = 4(Altitude)2

Proof :

Let AD = x and AB = BC = AC = a

$$\text{Now,}\space\text{BD = CD =}\frac{1}{2}\text{BC}\frac{1}{2}a$$

(In an equilateral triangle altitude AD is a perpendicular bisector of BC)

In right angled ΔABD,

$$\Rarr\space a^{2}=x^{2}+\bigg(\frac{1}{2}a\bigg)^{2}\\\Rarr\space a^{2}=x^{2}+\frac{1}{4}a^{2}$$

⇒ 4a2 = 4x2 + a2 ⇒ 3a2 = 4x2

Hence Proved.

17. Tick the correct answer and justify : In ΔABC,

$$\textbf{AB}=\textbf{6}\sqrt{\textbf{3}},\textbf{AC = 12 cm and BC = 6 cm. The angle B is :}$$

(a) 120° (b) 60° (c) 90° (d) 45°

Sol. (c) 90°

Explanation : Given BC = 6 cm and

$$\text{AB=}\space 6\sqrt{3}\space\text{and AC = 12 cm}$$

$$\text{Now, AB}^{2}+\text{BC}^{2}=(6\sqrt{3})^{2}+(6)^{2}$$

= 108 + 36 = 144

= (12)2 = (AC)2

So by converse of pythagoras theorem,

⇒ ΔABC is right angled at B.

∴ ∠B = 90°

Exercise 6.6 (Optional)

1. In figure, PS is the bisector of ∠QPR of ΔPQR.

$$\textbf{Prove that}\space\frac{\textbf{QS}}{\textbf{SR}}=\frac{\textbf{PQ}}{\textbf{PR}}.$$

Sol. Given : PS is the bisector of ∠QPR of ΔPQR.

$$\textbf{To Prove :}\space\frac{\text{QS}}{\text{SR}}=\frac{\text{PQ}}{\text{PR}}.$$

Construction : Draw RT || SP to meet QP produced in T.

Proof : ∵ RT || SP and transversal PR intersects them

∴ ∠1 = ∠2

(Pair of alternate interior angle) ...(i)

∵ RT || SP and transversal QT intersects them

∴ ∠3 = ∠4

(Pair of corresponding angle) ...(ii)

But  ∠1 = ∠3 (As PS bisects ∠QPR)

∴ ∠2 = ∠4

[From equations (i) and (ii)]

∴ PT = PR ...(iii)

(∵  Sides opposite to equal angles of a triangle are equal)

Now, in ΔQRT,

PS || RT (By construction)

$$\frac{\text{QS}}{\text{SR}}=\frac{\text{PQ}}{\text{PT}}\\\text{(By basic proportionality theorem)}\\\Rarr\space\frac{\text{QS}}{\text{SR}}=\frac{\text{PQ}}{\text{PR}}$$

[From equations (iii)]

Hence Proved.

2. In figure, D is a point on hypotenuse AC of DABC, such that BD ⊥ AC, DM ⊥ BC and
DN ⊥ AB. Prove that :

(i) DM2 = DN . MC   (ii) DN2 = DM . AN

Sol. Given : Δ is a point of hypotenuse AC of DABC, DM ⊥ BC and DN ⊥ AB.
To Prove : (i) DM2 = DN . MC

(ii) DN2= DM . AN

Construction : Join NM. Let BD and NM intersect at O.

Proof : (i) In ΔDMC and ΔNDM, ∠DMC = ∠NDM = 90° ...(i) Let MCD = ∠1 Then, ∠MDC = 90° – ∠1 (∵ ∠MCD + ∠MDC + ∠DMC = 180°) ∴ ∠ODM = 90° – (90° – ∠1) = ∠1 ⇒ ∠DMN = ∠1 or ∠MCD = ∠DMN ...(ii) From (i) and (ii), ΔDMC ~ ΔNDM (AA similarity criterion) $$\frac{\text{DM}}{\text{ND}}=\frac{\text{MC}}{\text{DM}}$$ (Corresponding sides of the similar triangles are proportional) ⇒ DM2 = MC.ND (ii) In ΔDNM and ΔNAD. ∠NDM = ∠AND (Each equal to 90°) Let ∠NAD = ∠2 Then, ∠NDA = 90° – ∠2 (∵ ∠NDA + ∠DAN + ∠DNA = 180°) ⇒ ∠ODN = 90° – (90° – ∠2) = ∠2 ⇒ ∠ODN = ∠DNM = ∠2 ⇒ ∠DNM = ∠NAD ∠ΔDNM ~ ΔNAD (AA similarity criterion) $$\therefore\space\frac{\text{DN}}{\text{NA}}=\frac{\text{DM}}{\text{ND}}\\\text{(corresponding sides of similar triangles)}\\\Rarr\space\frac{\text{DN}}{\text{AN}}=\frac{\text{DM}}{\text{DN}}$$ ⇒ DN2 = DM × AN

3. In figure, ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced. Prove that AC2 = AB2 + BC2 + 2BC . BD.

Sol. Given : ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced.

To Prove : AC2 = AB2 + BC2 + 2BC.BD
Proof : In ΔADC, ∠D = 90°

∴ AC2 = AD2 + DC2 (By Pythagoras theorem)

= AD2 + (BD + BC)2 [∵ DC = DB + BC]

= (AD2 + DB2) + BC2 + 2 BD.BC

[∵ (a + b)2 = a2 + b2 + 2ab]

= AB2 + BC2 + 2BC.BD

(∵ In right ΔADB with ∠D = 90°, AB2 = AD2 + DB2) Hence Proved.

4. In figure, ABC is a triangle in which ∠ABC < 90° and AD ⊥ BC. Prove that AC2 = AB2 + BC2 – 2BC.BD.

Sol. Given : ABC is a triangle in which ∠ABC < 90° and AD ⊥ BC.

To Prove : AC2 = AB2 + BC2 – 2BC.BD

∴ ∠D = 90°

∴ AC2 = AD2 + DC2

(by pythagoras theorem)

= AD2 + (BC –BD)2 (∵ BC = BD + DC)

= AD2 + BC2 + BD2 – 2BC.BD

[∵ (a – b)2 = a2 + b2 – 2ab]

= (AD2 + BD2) + BC2 – 2BC.BD

= AB2+ BC2 – 2 BC.BD

{∴ In right DADB with ∠D = 90°, AB2 = AD2 + BD2 } Hence Proved.

5. In figure, AD is a median of a DABC and AM ⊥ BC. Prove that

$$\textbf{(i)\space}\textbf{AC}^{2}=\textbf{AD}^{2}+\textbf{BC}.\textbf{DM}+\bigg(\frac{\textbf{BC}}{\textbf{2}}\bigg)^{2}\\\textbf{(ii)}\space\textbf{AB}^{2}=\textbf{AD}^{2}-\textbf{BC.DM}+\bigg(\frac{\textbf{BC}}{\textbf{2}}\bigg)^{2}\\\textbf{(iii)}\space\textbf{AC}^{2}+\textbf{AB}^{2}=2\textbf{AD}^{2}+\frac{\textbf{1}}{\textbf{2}}\textbf{BC}^{2}$$

Sol. Given that, in figure, AD is a median of a ΔABC and AM ⊥ BC.
Proof :

(i) In right ΔAMC,

∵ ∠M = 90°

∴ AC2 = AM2 + MC2

(By Pythagoras theorem)

= AM2 + (MD + DC)2

(∵ MC = MD + DC)

= (AM2 + MD2) + DC2 + 2 MD.DC

[∵ (a + b)2 = a2 + b2 + 2ab]

= AD2 + DC2 + 2DC.MD

[∴ In right ΔAMD with ∠M = 90°,

AM2 + MD2

$$=\text{AD}^{2}+\bigg(\frac{\text{BC}}{2}\bigg)^{2}+2\bigg(\frac{\text{BC}}{2}\bigg)\text{DM}\\\text{[}\therefore \text{2DC = BC(AD is a median of DABC)}]\\\therefore\space\text{AC}^{2}=\text{AD}^{2}+\bigg(\frac{\text{BC}}{2}\bigg)^{2}+\text{BC.DM}\\\text{...(i)}$$

(ii) In right ΔAMB,

∵ ∠M = 90°

∴ AB2 = AM2 + MB2

(By Pythagoras theorem)

= AM2 + (BD – MD)2

[∵ BD = BM + MD]

= AM2 + BD2 + MD2 – 2BD.MD

[∵ (a – b)2 = a2 + b2 – 2ab]

= (AM2 + MD2) + BD2 – 2BD.MD

= AD2 + BD2 – 2 BD.MD

[∵ In right ΔAMD with ∠M = 90°,

(By Pythagoras theorem)]

$$=\text{AD}^{2}-2\bigg(\frac{\text{BC}}{2}\bigg)\text{DM}+\bigg(\frac{\text{BC}}{2}\bigg)^{2}$$

(∵ 2BD = BC, AD is a median of ΔABC)

$$\therefore\space\text{AB}^{2}=\text{AD}^{2}-\text{BC.DM}+\bigg(\frac{\text{BC}} (iii) On adding Equations (i) and (ii), we get$$\text{AC}^{2}+\text{AB}^{2}=2\text{AD}^{2}+\frac{1}{2}\text{(BC)}^{2}$$6. Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides. Sol. Given : ABCD is a parallelogram whose diagonals are AC and BD. To Prove : AC2 + BD2 = AB2 + BC2 + CD2 + AD2 Construction : Draw AM ⊥ DC and BN ⊥ DC (produced). Proof : In right ΔAMD and ΔBNC. AD = BC (Opposite sides of a parallelogram) AM = BN (Both are altitudes of the same parallelogram to the same base) ∴ ΔAMD ≅ ΔBNC (RHS Congruency criterion) ∴ MD = NC (CPCT) ...(i) In right ΔBND, ∵ ∠N = 90° ∴ BD2 = BN2 + DN2 (By Pythagoras theorem) = BN2 + (DC + CN)2 (∵ DN = DC + CN) = BN2 + DC2 + CN2 + 2 DC.CN [∵ (a + b)2 = a2 + b2 + 2ab] = (BN2 + CN2) + DC2 + 2DC.CN ⇒ BD2 = BC2 + DC2 + 2 DC.CN ...(ii) (∵ In right ΔBNC with ∠N = 90°) BN2 + CN2 = BC2 (By Pythagoras theorem) In right ΔAMC, ∠M = 90° ∴ ACM2 = AM2 + MC2 (∵ MC = DC – DM) = AM2 + (DC – DM)2 [∵ (a – b)2 = a2 + b2 – 2ab] = AM2 + DC2 + DM2 – 2DC.DM = (AM2 + DM2) + DC2 – 2DC.DM = AD2 + DC2 – 2DC.DM [∵ In right triangle AMD with ∠M = 90°, AD2 = AM2 + DM2 (By Pythagoras theorem)] ⇒ AC2 = AD2 + AB2 – 2DC.CN ...(iii) [∵ DC = AB, opposite sides of parallelogram and MD = NC from equation (i)] On adding equations (iii) and (ii), we get AC2 + BD2 = (AD2 + AB2) + (BC2 + DC2) = AB2 + BC2 + CD2 + DA2 Hence Proved. 7. In figure, two chords AB and CD intersect each other at the point P. Prove that : (i) ΔAPC ~ ΔDPB (ii) AP.PB = CP.DP Sol. Given : Two chords AB and CD intersects each other at the point P. To Prove : (i) ΔAPC ~ ΔDPB (ii) AP.PB = CP.DP Proof : (i) ΔAPC and ΔDPB ∠APC = ∠DPB (Vertically opposite angles) ∠CAP = ∠BDP (Angles in the same segment) ∴ ΔAPC ~ ΔDPB (AA similarity criterion) (ii) ΔAPC ~ ΔDPB [Proved in (i)]$$\therefore\space\frac{\text{AP}}{\text{DP}}=\frac{\text{PC}}{\text{PB}}$$(∵ Corresponding sides of two similar triangles are proportional) ⇒ AP.BP = CP.DP ⇒ AP.PB = CP.DP 8. In figure, two chords AB and CD of a circle intersect each other at the point P(when produced) outside the circle. Prove that (i) ΔPAC ~ ΔPDB (ii) PA.PB = PC.PD Sol. Given : Two chords AB and CD of a a circle intersects each other at the point P (when produced) out the circle. To Prove : (i) ΔPAC ~ ΔPDB (ii) PA.PB = PC.PD Proof : (i) We know that, in a cyclic quadrilaterals, the exterior angle is equal to the interior opposite angles. Therefore, ∠PAC = ∠PDB ...(i) and ∠PCA = ∠PBD ...(ii) By equations (i) and (ii), we get ΔPAC ~ ΔPDB (By AA similarity criterion) (ii) Since, ΔPAC ~ ΔPDB (By AA similarity criterion) (ii) Since, ΔPAC ~ ΔPDB$$\therefore\space\frac{\text{PA}}{\text{PD}}=\frac{\text{PC}}{\text{PB}}$$(∵ Corresponding sides of the similar triangles are proportional) ⇒ PA.PB = PC.PD 9. In figure, D is a point on side BC of DABC such that$$\frac{\textbf{BD}}{\textbf{CD}}=\frac{\textbf{AB}}{\textbf{AC}}.$$Prove that AD is the bisector of ∠BAC. Sol. Given : D is a point on side BC of ΔABC such that$$\frac{\text{BD}}{\text{CD}}=\frac{\text{AB}}{\text{AC}}.$$To Proof : AD is bisector of ∠BAC. Construction : Produce BA to E such that AC = AE. Join CE.$$\textbf{Proof :}\space\frac{\text{BD}}{\text{CD}}=\frac{\text{AB}}{\text{AC}}\space\text{(Given)}\\\Rarr\space\frac{\text{BD}}{\text{CD}}=\frac{\text{AB}}{\text{AE}}

[∵ AC = AE (by construction)]

∴ In ΔBCE,

(By converse of basic proportionality theorem)

(Pair of corresponding angle) ...(i)

and

(Pair of alternate interior angle) ...(ii)

∵ AC = AE (By construction)

∴ ∠AEC = ∠ACE ...(iii)

(Angles opposite to equal sides of a triangle are equal)

Using equations (i), (ii) and (iii), we get

i.e., AD is the bisector of ∠BAC.

Hence Proved.

10. Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out? If she pulls in the string at the rate of 5 cms–1, what will be the horizontal distance of the fly from her after 12 s?

Sol. Let AB be the height of the top of the fishing rod from the water surface. Let BC be the horizontal distance of the fly from the tip of the fishing rod. Then, AC is the length of the string.

In ΔABC, by pythagoras theorem,

AC2 = AB2 + BC2

= (1.8)2 + (2.4)2

= 3.24 + 5.76

= 9

AC = 3 m

Thus, length of string out is 3 m.
Length of string is pulled at the rate of 5 cm/s.

∴ String pulled in 12 sec = 12 × 5 = 60 cm = 0.6 m

Let, fly be at D after 12 seconds. Length of string out after 12 second is AD.

⇒ AD = 3 – 0.6 = 2.4 m

⇒ (1.8)2 + BD2 = (2.4)2

⇒ BD2 = 5.76 – 3.24

= 2.52

BD = 1.587 m

Horizontal distance of fly = BD + 1..2 m

Horizontal distance of fly = 1.587 + 1.2

Horizontal distance of fly = 2.79 m