# NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

**Q. ABCD is a rhombus. Show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.**

**Ans. Given :** ABCD is a rhombus.

AB = BC = CD = AD

Now, in △ ABC and △ ADC,

BC = AD [Given]

AC = AC [Common side]

AB = DC [Given]

△ ABC **≅** △ ADC [By SSS]

∠BAC = ∠DAC

and ∠ACB = ∠ACD [CPCT]

Hence, AC bisects ∠A as well as ∠C.

Similarly , △ BAD **≅** △ BCD

∠ABD = ∠DBC and ∠ADB = ∠BDC

Hence, BD bisects ∠B as well as ∠D.

**Q. D, E and F are respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC. Prove that DDEF is also an equilateral triangle.**

**Ans. Given :** ABC is an equilateral triangle in which D, E, F are the mid-points of side, BC, AC, AB, respectively.**To prove :** DEF is an equilateral triangle.**Proof :** We have, F and E are the mid-points of sides AB and AC, respectively.

$$\text{By mid-point theorem,} \\FE=\frac{1}{2}BC…(i) \\\text{Similarly , E and D are the mid-points of AC and BC, respectively.} \\ ED=\frac{1}{2}AB…(ii) \\ \text{And, F and D are the mid-points of AB and BC respectively.} \\ FD=\frac{1}{2}AC…(iii) \\ \text{Now, ABC is an equilateral triangle} \\ AB = BC = CA \\or\space \frac{1}{2}AB=\frac{1}{2}BC=\frac{1}{2}CA \\ ⇒ ED = FE = FD [Using (i), (ii) and (iii)] \\ \bigtriangleup DEF\space text{is an equilateral triangle. Hence Proved.} $$

**Q. In figure, ABCD is a trapezium in which side AB is parallel to side DC and E is the mid-point of side AD. F is a point on the side BC such that the segment EF is parallel to side DC.**

**Ans. Given :**ABCD is a trapezium with AB || DC. Also E is the mid-point of AD such that EF || DC.

AB || EF || DC.

$$\text{To prove :}EX=\frac{1}{2}(AB+DC)\\ \text{Construction : Join BD such that it intersect EF at X.}\\ \text{Proof : In DABD, E is the mid-point of AD and EX || AB.}\\ \text{By the converse of mid-point theorem,}\\ \text{X is the mid-point of DB.}\\ \text{and}\space EX=\frac{1}{2}AB….(i)\\ \text{Similarly, in DBDC, X is the mid-point of BD and XF || DC.}\\ \text{F is the mid-point of BC.}\\ \text{and}\space XX=\frac{1}{2}DC….(ii)\\ \text{Adding (i) and (ii), we get}\\ EX + XF =\frac{1}{2}AB+\frac{1}{2}AD\\ ⇒ EF =\frac{1}{2}(AB+DC)$$

**Q. Prove that the angles bisectors of a parallelogram form a rectangle.**

**Ans.** Consider a parallelogram ABCD in which AP, BP, CR and

DR are bisectors at ∠A, ∠B, ∠C and ∠D respectively.**To prove :** PQRS is a rectangle.**Proof :** We know, adjacent angles of a parallelogram are supplementary.

$$∠A + ∠D = 180°\\ or\space \frac{1}{2}\angle A+\frac{1}{2}\angle D=\frac{1}{2}×180° [Dividing both sides by 2]\\ ⇒ ∠QAD + ∠ADQ = 90° …(i)\\ Now, in \bigtriangleup ADQ\\ ∠ADQ + ∠QAD + ∠AQD = 180° [Angle sum property of a \bigtriangleup]\\ ⇒ 90° + ∠AQD = 180° [Using (i)]\\ ⇒ ∠AQD = 180° – 90° = 90°\\ So,∠PQR = ∠AQD = 90° [Vertically opposite angles]\\ Similarly, ∠PSR = ∠QPS = ∠QRS = 90°\\ \text{So, in quadrilateral PQRS,}\\ ∠Q = ∠S = 90° and ∠P = ∠R = 90°\\ \text{i.e., opposite angles are equal.}\\ \text{Hence, PQRS is a parallelogram.}\\ Also, ∠P = ∠Q = ∠R = ∠S = 90°\\ \text{So, PQRS is a parallelogram with each angle equal to 90°.}\\ \text{PQRS is a rectangle.}$$

**Q. In given DABC, lines are drawn through A, B and C parallel respectively to the sides BC, CA and AB forming DPQR. Show that BC = 1/2QR.**

$$\text{Proof : In quadrilateral BCQA, we have}\\ \text{BC || AQ and QC || AB}\\ \text{BCQA is a parallelogram.}\\ BC = QA \text{[Opposite sides of a ||gm] …(i)}\\ \text{Similarly, in quadrilateral ARBC, we have}\\ \text{To prove : P is the mid-point of CD i.e., DP = CP.}\\ \text{Proof : We have, ∠DAB = 60° and AP is the bisectors of ∠A}\\ ∠DAP =\frac{1}{2}\angle DAB=\frac{1}{2}×60°= 30°\\ Now, ∠DAB + ∠ABC = 180° \text{[Co-interior angles]}\\ ⇒ 60° + ∠ABC = 180°\\ ⇒ ∠ABC = 180° – 60° = 120°\\ ∠ADC = ∠ABC = 120°\\ In \bigtriangleup ADP,\\ ∠ADP + ∠PAD + ∠APD = 180° text{[Angle sum property of a D]}\\ ⇒ 120° + 30° + ∠APD = 180°\\ ⇒ ∠APD = 180° – 150° = 30°\\ ⇒ ∠DAP = ∠APD = 30°\\ DP = AD …(i) text{(opposite sides of equal angles)}\\ Similarly , ∠PBC =\frac{1}{2}\angle ABC=\frac{1}{2}×120°= 60°\\ ∠PCB = ∠DAB = 60°\\ ∠BPC = 60° [Angle sum property of ∠BCP]\\ ∠PBC = ∠PCB = ∠BPC = 60°\\ \bigtriangleup PBC text{is an equilateral triangle.}\\ CP = BC …(ii)\\ AD = BC [Opposite sides of ||gm ABCD]\\ DP = CP [Using (i) and (ii)]\\ ⇒ P is the mid-point of side CD.$$

**Q. In the given figure, ABCD is a trapezium in which**

AB || CD and AD = BC. Show that :

(i) ∠A = ∠B

(ii) ∠C = ∠D

(iii) diagonal AC = diagonal BD

AB || CD and AD = BC. Show that :

(i) ∠A = ∠B

(ii) ∠C = ∠D

(iii) diagonal AC = diagonal BD

**Ans. Given :**ABCD is a trapezium in which AB || DC and

AD = BC.

To prove : (i) ∠A = ∠B

(ii) ∠C = ∠D

(iii) Diagonal AC = BD

Construction : Draw CE || DA and produce AB which meet CE at the point E.

$$\text{Proof : (i) We have, AE || DC and AD || EC}\\ \text{ADCE is a parallelogram.}\\ ∠D = ∠E \space[Opposite angles of ||gm] …(A)\\ \text{ADCE is a parallelogram.}\\ and, AD = CE \space \text{[Opposite sides of ||gm]}\\ \text{But AD = BC [Given]}\\ BC = CE\\ ∠BEC = ∠CBE …(B)\space\text{[Opposite angles of equal sides]}\\ AD || CE\\ ∠A + ∠E = 180° [Co-interior angles]\\ ⇒ ∠A + ∠CBE = 180° [From (B)] …(C)\\ Also, ∠ABC + ∠CBE = 180° [Linear pair] …(D)\\ Subtracting equation (D) from equation (C), we get\\ ⇒ ∠A – ∠ABC = 0\\ ⇒ ∠A = ∠ABC = ∠B\\ (ii) Now, ∠D = ∠E\\ \text{[Opposite angle of ||gm ADCE]}\\ ⇒ ∠D = ∠CBE \text{[Using (B)]}\\ ⇒ ∠D = ∠BCD \text{[… ∠CBE = ∠BCD (Alternate angles)]}\\ ⇒ ∠D = ∠C\\ (iii) \space In \bigtriangleup ADB and \bigtriangleup ABC\\ ∠A = ∠B [Proved in (i)]\\ AB = AB [common side]\\ AD = BC [Given]\\ DADB \cong\bigtriangleup BCA [SAS congruency rule]\\ BD = AC (CPCT)$$

**Q. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angle, then it is a square.**

**Ans.** Consider quadrilateral ABCD in which AC = BD, AC ⊥ BD and diagonal AC and BD bisect each other.

OA = OC and OB = OD

To prove : ABCD is a square.

Proof : In ΔAOB and ΔCOD,

AO = CO [Given]

BO = DO [Given]

∠AOB = ∠COD

[Vertically opposite angles]

ΔAOB ** ≅** ΔCOD [SAS congruency rule]

AB = CD [CPCT] ...(i)

and ∠ABO = ∠ODC [C.P.C.T.]

Angles ABO and ODC form a pair of alternate angles between the lines AB and DC.

AB || DC ...(ii)

From (i) and (ii),

AB || DC and AB = DC

i.e., a pair of opposite sides is parallel and equal.

ABCD is a parallelogram. ...

(iii) Now, in DAOB and DAOD, AO = AO [Common side]

∠AOB = ∠AOD [Each 90°]

BO = DO [Given]

ΔAOB ** ≅** ΔAOD [SAS congruency rule]

AB = AD [C.P.C.T.]

i.e., in parallelogram ABCD, adjacent sides are equal.

ABCD is a rhombus.

Now, in DABD and DBAC,

AB = AB [Common side]

AD = BC [Proved above]

BD = AC [Given]

ΔABD ** ≅** ΔBAC [SSS congruency rule]

∠BAD = ∠ABC [C.P.C.T.]

But ∠ABC + ∠BAD = 180° [Co-interior angles]

⇒ 2∠ABC = 180°

⇒ ∠ABC = 90°

i.e., An angle of rhombus ABCD is a right angle.

ABCD is a square.

**Q. In the given figure, ABCD is a parallelogram, E and F are the mid-points of the sides AB and CD, respectively. Prove that the line segments AF and CE trisect the diagonal BD.**

**Q. Show that the quadrilateral formed by joining the mid-points of the consecutive sides of a rectangle is a rhombus.**

**Q. In DABC, AD is the median through point A and E is the mid-point of AD. BE produced meets AC is F. Show that**

$$AF=\frac{1}{3}AC$$

**Q. AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD. Show that ∠A > ∠C and ∠B > ∠D.**