# Quadrilaterals – Exercise 8.1 – Class IX

1. The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.

Solution: Suppose the measures of four angles are 3x, 5x, 9x and 13x

Then, 3x + 5x + 9x + 13x = 360˚

30x = 360˚

x = 360˚/30 = 12˚

3x = 3*12 = 36˚

5x = 5*12 = 60˚

9x = 9*12 = 108˚

13x = 13*12 = 156˚

⸫ the angles of the quadrilateral are 36˚, 60˚, 108˚ and 156˚.

1. If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Solution: Given: ABCD is a parallelogram in which AC = BD

To prove: ABCD is a rectangle.

Proof:

In ∆ABC and ∆ABD

AB = AB (common)

BC = AD [opposite sides of the parallelogram]

AC = BD (given)

Since,  ABCD is a parallelogram, thus,

⸫ 2∠ABC = 180˚ [From (1) and (2)]

Thus, ABCD is a parallelogram, in which one angle is 90˚

⇒ABCD is a rectangle.

1. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

Solution: Given: A quadrilateral ABCD, in which diagonals AC and BD bisect each other at right angles.

To prove: ABCD is a rhombus.

Proof: In ∆AOB and ∆BOC

AO = OC [diagonals AC and BD bisect each other]

∠AOB = ∠COB = 90˚

BO = BO [common]

⸫∆AOB ≅ ∆BOC [SAS congruence]

AB = BC …………………(1) [CPCT]

Since, ABCD is a quadrilateral in which AB = BC

If the diagonals of a quadrilateral bisect each other, then it is a parallelogram and opposite sides of a parallelogram are equal.

Hence, ABCD is a rhombus.

1. Show that the diagonals of a square are equal and bisect each other at right angles.

Solution: Given: ABCD is a square in which AC and BD are diagonals.

To prove: AC = BD and AC bisects BD at right angles, i.e., AC⊥ BD, OA = OC and OB = OD

Proof:

AB = AB (common)

BC = AD (sides of a square are equal)

∠ABC = ∠BAD = 90˚ (angles of a square)

⸫ ∆ABC ≅ ∆BAD (SAS congruence)

⇒AC = BD (CPCT)

In ∆AOB and ∆COD

AB = Dc (sides of a square)

∠AOB = ∠COD (vertically opposite angles)

∠OAB = ∠OCD (alternate angles)

⸫∆AOB ≅∆COD (AAS congruence)

OA = OC (CPCT)

Similarly, by taking ∆AOD and ∆BOC, we can show that OB = OD

In ∆ABC, ∠BAC + ∠BCA = 90˚ (since ∠ABC = 90˚)

⇒2∠BAC = 90˚ [∠BAC = ∠BCA as BC = AD]

⇒ ∠BAC = 45˚ or ∠BCO = 45˚

⇒∠CBO = 45˚

In∆BCO,

∠BCO + ∠CBO + ∠BOC = 180˚

⇒90˚ + ∠BOC = 180˚

⇒∠BOC = 90˚

⇒BO⊥OC

⇒AC ⊥BD

Hence, AC = BD , AC⊥BD , OA = OC and OB = OD

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

Solution:

To prove: Quadrilateral ABCD is a square.

Proof: First let us prove that ABCD is a parallelgram

In ∆AOD and ∆COB,

AO = OC [given]

OD = OB [given]

∠AOD = ∠COB [vertically opposite angles]

By SAS congruence, we have,

∆AOD ≅∆COB

⇒∠OAD = ∠OCB [ corresponding parts of congruent triangles are equal]

We have AC intersects AD and BC at A and C respectively such that,,

∠OAD = ∠OCB [from (1)] [alternate interior angles are equal]

Similarly, AB||CD

Now, we have to prove that it is a square.

In ∆AOB and ∆AOD, we haave,

AO = AO [common]

∠AOB = ∠AOD = 90˚

OB = OD [diagonals of a parallelogram bisect each other]

Therefore AB = BC = CD = AD ……………..(2)

Now, in ∆ABD and ∆BAC,, we have,

AB = BA

AD = BC [opposite sides of a parallelogram are equal]

BD = AC [given]

Therefore, by SSS congruence,

∆ABD≅∆BAC

⇒∠DAB = ∠CBA  [ corresponding parts of congruent triangles are equal]

1. Diagonal AC of a parallelogram ABCD bisects A (see Fig. 8.19). Show that

(i) it bisects C also,

(ii) ABCD is a rhombus.

Solution:

Given: A parallelogram ABCD, in which diagonal AC bisects ∠A i.e., ∠DAC = ∠BAC

To prove: (i)Diagonal AC bisects ∠C i.e., ∠DCA = ∠BCA

(ii)ABCD is a rhombus

Proof:

(i) ∠DAC = ∠BCA [alternate angles]

∠BAC = ∠DCA [ alternate angles]

But ∠DAC = ∠BAC [given]

⸫∠BCA = ∠DCA

Therefore, AC bisects ∠DCB

Thus, AC bisects ∠C

(ii) In ∆ABC and ∆CDA

AC = AC [common]

∠BAC = ∠DAC [given]

and ∠BCA = ∠DAC [which is already proved]

⸫ ∆ABC  ≅  ∆ADC [ASA  congruence]

⸫AB = CD

⸫ AB = BC = DC = AD

Therefore, ABCD is a rhombus.

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

Solution: Given: ABCD is a rhombus ie., AB = BC = CD = DA

To prove: (i) diagonal AC bisects ∠A as well as ∠C

(ii) diagonal BD bisects ∠B as well as ∠D

AD = DC [sides of a rhombus are equal]

⇒∠DAC = ∠DCA …………………(1) [angles opposite to equal sides of a triangle are equal]

Now AB||DC and AC intersects them

AB = AD [Sides of a rhombus]

AC = AC [Common]

BC = CD [Sides of a rhombus]

So, ∠DAC = ∠BAC

∠BCA = ∠DCA

Similarly, ∠ADB = ∠CDB and ∠ABD = ∠CBD.

Hence, diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.

1. ABCD is a rectangle in which diagonal AC bisects A as well as C. Show that:

(i) ABCD is a square

(ii) diagonal BD bisects B as well as D.

Solution: Given: Rectangle ABCD where AC bisects ∠A i.e., ∠1 = ∠2 and AC bisects ∠C i.e., ∠3 = ∠4

To prove: (i) ABCD is a square

(ii) diagonal BD bisects ∠ B as well as ∠ D.

Proof:

(i) AB|||DC and tranversal AC intersects them

⸫∠ACD = ∠CAB (alternate interior angles)

⸫AD = CD (sides opposite to equal angles of a triangle are equal)

⸫ABCD is a sqare.

(ii) In ∆BDA and ∆DBC,

BD = DB (common)

DA = BC (sides of a square ABCD)

AB = DC (sides of a square ABCD)

⸫∆BDA ≅ BDC (SSS rule)

⸫∠ABD = ∠CDB (CPCT)

but, ∠CDB = ∠CBD (⸫CB = CD sides of a square ABCD)

⸫∠ABD = ∠CBD ………….(1)

⸫BD bisects ∠B

Now, ∠BAD = ∠CBD (from (1))

∠ABD = ∠ADB (⸫AB = CD)

∠CBD = CDB (⸫CB = CD)

∠CDB = ∠CBD = ∠ABD = ∠ADB

⸫BD bisects ∠D

1. In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see Fig. 8.20). Show that:

(i) ∆ APD ∆ CQB

(ii) AP = CQ

(iii) ∆ AQB ∆ CPD

(iv) AQ = CP

(v) APCQ is a parallelogram

Solution:

Given : ABCD is a parallelogram and P and Q are points on diagonal BD such that DP = BQ.

To Prove : (i) ∆APD ≅ ∆CQB

(ii) AP = CQ

(iii) ∆AQB ≅ ∆CPD

(iv) AQ = CP

(v) APCQ is a parallelogram

Proof : (i) In ∆APD and ∆CQB, we have

AD = BC [Opposite sides of a parallelogram]

DP = BQ [Given]

∴ ∆APD ≅ ∆CQB [SAS congruence]

(ii) ∴ AP = CQ [CPCT]

(iii) In ∆AQB and ∆CPD, we have

AB = CD [Opposite sides of a ||gm]

DP = BQ [Given]

∠ABQ = ∠CDP [Alternate angles]

∴ ∆AQB ≅ ∆CPD [SAS congruence]

(iv) ∴ AQ = CP [CPCT]

(v) Since in APCQ, opposite sides are equal, therefore it is a parallelogram.

1. ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. 8.21). Show that

(i) ∆ APB ∆ CQD

(ii) AP = CQ

Solution:

ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on BD.(given)

To Prove : (i) ∆APB ≅ ∆CQD

(ii) AP = CQ

Proof : (i) In ∆APB and ∆CQD, we have

∠ABP = ∠CDQ [Alternate angles]

AB = CD [Opposite sides of a parallelogram]

∠APB = ∠CQD [Each = 90°]

∴ ∆APB ≅ ∆CQD [ASA congruence]

(ii) So, AP = CQ [CPCT]

1. In ∆ ABC and ∆ DEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. 8.22). Show that

(i) quadrilateral ABED is a parallelogram

(ii) quadrilateral BEFC is a parallelogram

(iv) quadrilateral ACFD is a parallelogram

(v) AC = DF

(vi) ∆ ABC ∆ DEF.

Solution:

To Prove : (i) ABED is a parallelogram

(ii) BEFC is a parallelogram

(iv) ACFD is a parallelogram

(v) AC = DF

(vi) ∆ABC ≅ ∆DEF

Proof : (i) In quadrilateral ABED, we have

AB = DE and AB || DE. [Given]

⇒ ABED is a parallelogram.

[One pair of opposite sides is parallel and equal]

(ii) In quadrilateral BEFC, we have

BC = EF and BC || EF [Given]

⇒ BEFC is a parallelogram. [One pair of opposite sides is parallel and equal]

(iii) BE = CF and BE||BECF [BEFC is parallelogram]

(iv) ACFD is a parallelogram. [One pair of opposite sides is parallel and equal]

(v) AC = DF [Opposite sides of parallelogram ACFD]

(vi) In ∆ABC and ∆DEF, we have

AB = DE [Given]

BC = EF [Given]

AC = DF [Proved above]

∴ ∆ABC ≅ ∆DEF [SSS axiom]

1. ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.23). Show that

(i) A = B

(ii) C = D

(iv) diagonal AC = diagonal BD

[Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]

Solution:

Given : In trapezium ABCD, AB || CD and AD = BC.

To Prove : (i) ∠A = ∠B

(ii) ∠C = ∠D

(iv) diagonal AC = diagonal BD

Constructions : Join AC and BD. Extend AB and draw a line through C parallel to DA meeting AB produced at E. Proof : (i) Since AB || DC

⇒ AE || DC …(1)

and AD || CE …(2) [Construction]

⇒ ADCE is a parallelogram [Opposite pairs of sides are parallel]

∠A + ∠E = 180° …(3) [Consecutive interior angles]

∠B + ∠CBE = 180° …(4) [Linear pair]

AD = CE …(5) [Opposite sides of a parallelogram]

⇒ BC = CE [From (5) and (6)]

⇒ ∠E = ∠CBE …(7) [Angles opposite to equal sides]

∴ ∠B + ∠E = 180° …(8) [From (4) and (7)

Now from (3) and (8) we have

∠A + ∠E = ∠B + ∠E

⇒ ∠A = ∠B Proved.

(ii) ∠A + ∠D = 180°

∠B + ∠C = 180°

⇒ ∠A + ∠D = ∠B + ∠C [∵ ∠A = ∠B]

⇒ ∠D = ∠C

(iii) In ∆ABC and ∆BAD, we have

∠A = ∠B [Proved]

AB = CD [Common]

∴ ∆ABC ≅ ∆BAD [ASA congruence]

(iv) diagonal AC = diagonal BD [CPCT]