# NCERT Solutions for Class 10 Maths Chapter 7 Triangles

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.

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 7.3

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 7.4

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.