QUADRILATERALS

 

QUADRILATERALS

Property 1: The sum of the angles of a quadrilateral is 360°.

Proof: ABCD is a quadrilateral.

To prove: ∠A + ∠B + ∠C+ ∠D = 360°.

Construction: AC joined.

From ∆ABC, we get,

∠BAC + ∠B + ∠ACB = 180° ………………… (1)

From ∆ACD, we get,

∠DAC + ∠D + ∠ACD = 180° ………………… (2)

Now, (1) + (2)

∠BAC + ∠B + ∠ACB +∠DAC + ∠D + ∠ACD = 180°+180°

∠BAC +∠DAC + ∠B + ∠ACB + ∠ACD + ∠D =360°

∠A + ∠B + ∠C+ ∠D = 360°

⸫The sum of the angles of a quadrilateral is 360°.

Theorem 8.1: A diagonal of a parallelogram divides it into two congruent triangles.

Solution: ABCD is a parallelogram and AC is the diagonal.

To prove: ∆ABC ∆CDA

Proof:

From ∆ABC and ∆CDA, we have

AB || CD, AC transversal

⸫ 

AD || BC, AC transversal

⸫

AC=AC [Common side]

By ASA congruence rule,

∆ABC ∆CDA

⸫Diagonal AC divides parallelogram ABCD into two congruent triangles ∆ABC and ∆CDA.

Theorem 8.2: In a parallelogram, opposite sides are equal.

Solution: ABCD is a parallelogram and AC is the diagonal.

To prove: ∆ABC ∆CDA

Proof:

From ∆ABC and ∆CDA, we have

AB || CD, AC transversal

⸫

AD || BC, AC transversal

⸫

AC=AC [Common side]

By ASA congruence rule,

∆ABC ∆CDA

⸫AB=CD [CPCT]

AD=BC [CPCT]

⸫In a parallelogram, opposite sides are equal.

Theorem 8.3: If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram.

Solution: ABCD is a quadrilateral where AB=CD and AD=BC.

To prove: ABCD is a parallelogram.

Construction: AC joined.

Proof:

From ∆ABC and ∆CDA, we have 

AC=AC [Common side]

AB=CD [Given]

AD=BC [Given]

By SSS congruence rule,

∆ABC ∆CDA

⸫ [CPCT]

⸫AB || CD. [⸪Alternate interior angles are equal]

[CPCT]

⸫AD || BC [⸪Alternate interior angles are equal]

⸫ ABCD is parallelogram.

Theorem 8.4: In a parallelogram, opposite angles are equal.

Solution: ABCD is a parallelogram and AC is the diagonal.

To prove: ∆ABC ∆CDA

Proof:

From ∆ABC and ∆CDA, we have

AB || CD, AC transversal

AD || BC, AC transversal

AC=AC [Common side]

By ASA congruence rule,

∆ABC ∆CDA

⸫ [CPCT]

Again,

 

⸫ In a parallelogram, opposite angles are equal.

Theorem 8.5: If in a quadrilateral, each pair of opposite angles is equal, then it is a parallelogram.

Solution: ABCD is a quadrilateral where 

. 

To prove, ABCD is a parallelogram

Proof:

We know that,

⸫AD || BC [⸪Adjacent angles are supplementary]

Similarly, we can prove that, AB || CD.

⸫ABCD is a parallelogram

Theorem 8.6: The diagonals of a parallelogram bisect each other.

Solution: ABCD is a parallelogram where AC and BD diagonals. 

To prove AO=BO and CO=DO.

Proof:

From ∆AOB and ∆COD, we have

AB || CD, AC transversal

AB=CD [Opp. sides of a parallelogram]

AB || CD, BD transversal

⸫By ASA congruence rule,

∆AOB ∆COD

⸫ AO=BO and CO=DO [CPCT]

Theorem 8.7: If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.

Solution: ABCD is a quadrilateral where AC and BD diagonals intersects at O such that AO=BO and CO=DO. 

To prove ABCD is a parallelogram.

Proof:

From ∆AOB and ∆COD, we have

AO=CO

BO=DO

⸫By SAS congruence rule,

∆AOB ∆COD

⸫AB || CD [⸪Alternate interior angles are equal]

Similarly, we can prove that, AD || BC.

⸫ ABCD is a parallelogram.

Theorem 8.8: A quadrilateral is a parallelogram if a pair of opposite sides is equal and parallel.

Solution: ABCD is a quadrilateral where AB=CD and AB || CD.

To prove: ABCD is a parallelogram

Construction: AC joined.

Proof:

From ∆ABC and ∆CDA, we get,

AC=AC [Common side]

AB || CD, AC transversal

 

AB=CD [Given]

⸫By SAS congruence rule,

∆ABC   ∆CDA

 

⸫AD || BC [⸪Alternate interior angles are equal]

⸫ABCD is a parallelogram.

Exercise 8.1

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

Solution:

Let the measures of the angles be 3x, 5x, 9x and 13x.

We know that,

⸫The measures angles of the quadrilateral are 36°, 60°, 108° and 156°

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

Solution:

In parallelogram ABCD, AC=BD.

To prove: ABCD is a rectangle.

Proof:

From ∆ABC and ∆BAD, we get,

AB=AB [Common side]

AC=BD [Given]

BC=AD [Opposite sides of a parallelogram]

⸫By SSS congruence rule,

∆ABC ∆BAD

⸫ [CPCT] ………………………… (1)

Again,

[Adjacent angles of a parallelogram are supplementary]

⸫ABCD is a parallelogram with one angle equal to 90°.

⸫ABCD is a rectangle.

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

Solution:

In quadrilateral ABCD, AO = CO, BO = DO and AC ⊥ BD.

To prove: ABCD is a rhombus.

Proof: From ∆AOB and ∆COD, we have

AO=CO [Given]

[Vertically opposite angles]

BO=DO [Given]

By SAS congruence rule,

∆AOB ∆COD

⸫AB=CD [CPCT] …………………… (1)

Similarly, we can prove,

AD=BC ……………………….. (2)

Again, from ∆AOB and ∆AOD, we have

AO=AO [Common side] 

BO=DO [Given] 

By SAS congruence rule,

∆AOB ∆AOD

⸫AB=AD [CPCT] ………………… (3)

⸫ From (1), (2) and (3), we get

AB=BC=CD=AD.

⸫ ABCD is a rhombus.

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

Solution:

ABCD is a square, where AC and BD are diagonals intersect at O.

To prove: AC=BD, AO=BO, CO=DO and AC ⊥ BD.

Proof:

From ∆ABC and ∆BAD, we get,

AB=AB [Common side] 

[Angles of a square]

BC=AD [Sides of a square]

⸫By SAS congruence rule,

∆ABC ∆BAD

⸫AC=BD.

From ∆AOB and ∆COD, we get,

AB || CD, AC transversal,

[Alternate interior angles]

AB=CD [Sides of a square] 

AB || CD, BD transversal,

[Alternate interior angles]

⸫By ASA congruence rule,

∆AOB ∆COD

⸫AO=CO and BO=DO. [CPCT]

From ∆AOB and ∆AOD, we get,

AO=AO [Common side]

BO=DO [⸪∆AOB ∆COD]

AB=AD [Sides of a square]

⸫By SSS congruence rule,

∆AOB ∆AOD

[CPCT]

Again,

⸫AC ⊥ BD

⸫The diagonals of a square are equal and bisect each other at right angles.

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

Solution: ABCD is a quadrilateral, where AC=BD and AC and BD bisect each other at right angles at O. 

To prove: ABCD is a square.

We have, 

AC=BD, AO=CO, BO=DO and

From ∆AOB and ∆COD,

AO=CO [Given, BD bisects AC] 

BO=DO [Given, AC bisects BD]

Therefore, by S-A-S congruence criterion, 

∆AOB∆COD

⸫AB=CD [CPCT]

and

Therefore, AB||CD [Since, alternate interior angles are equal]

Thus, ABCD is a parallelogram.

Again, from ∆AOB and ∆AOD,

AO=AO [Common side]

BO=DO [Given, AC bisects BD]

Therefore, by S-A-S congruence criterion, 

∆AOB∆AOD

⸫AB=AD [CPCT]

From, (1), (2) and (3), we have

AB=BC=CD=AD

Again, from ∆ABC and ∆DCB,

BC=BC [Common side]

AB=CD [From (1)]

AC=BD [Given]

Therefore, by S-S-S congruence criterion, 

∆ABC ∆DCB

Again,

Similarly we can prove that,

Therefore, ABCD is a square.

6. Diagonal AC of a parallelogram ABCD bisects ∠A. Show that

(i) It bisects ∠C also,

(ii) ABCD is a rhombus.

Solution:

Given, ABCD is a parallelogram, where AC bisects. 

To prove:

(i) AC bisects.

(ii) ABCD is a rhombus

We have, AC bisects.

Again, AB||CD, AC transversal.

And, AD||BC, AC transversal.

Therefore, AC bisects.

In ΔABC and ΔADC

∠BAC = ∠DAC [From (1)]

AC = AC [Common side]

∠BCA = ∠DAC [From (3)]

⸫By ASA congruence rule,

ΔABC ≅ ΔADC

⸫ BC = DC ……………... (4) [CPCT]

Also, 

AB = DC ……………….. (5) [Opposite sides of a parallelogram]

AD = BC ……………….. (6) [Opposite sides of a parallelogram]

From (4), (5) and (6), we have,

⸫ AB = BC = DC = AD

Hence, ABCD is a rhombus

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

Solution:

ABCD is a rhombus. We have to prove that AC bisects and and BD bisects and .

From ∆ABC and ∆ADC, we have

AB=AD [Side of a rhombus]

AC=AC [Common side]

BC=CD [Side of a rhombus]

Therefore, by S-S-S congruence criterion, 

∆ABC ∆ADC

Again,

Similarly, we can prove that BD bisects and also.

8. 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:

ABCD is a rectangle in which diagonal AC bisects as well as . We have to prove

(i) ABCD is a square.

From ∆ABC and ∆ADC, we have

Therefore, by A-S-A congruence criterion, 

∆ABC ∆ADC

AB=AD [CPCT]

And BC=CD [CPCT]

Again,

AB=CD [Opposite sides of a rectangle]

Therefore, from (1), (2) and (3), we have

AB=BC=CD=AD

Therefore, ABCD is a square.

(ii) From ∆ABD and ∆CBD, we have

AB=CB [Side of a rhombus]

BD=BD [Common side]

AD=CD [Side of a rhombus]

Therefore, by S-S-S congruence criterion, 

∆ABD ∆CBD

Again,

9. In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP=BQ. Show that:

(i) ΔAPD ≅ ΔCQB

(ii) AP = CQ

(iii) ΔAQB ≅ ΔCPD

(iv) AQ = CP

(v) APCQ is a parallelogram

Solution:

In parallelogram ABCD, DP=BQ.

To prove: 

(i) ΔAPD ≅ ΔCQB 

(ii) AP = CQ

(iii) ΔAQB ≅ ΔCPD

(iv) AQ = CP

(v) APCQ is a parallelogram

Proof:

From ΔAPD and ΔCQB,

DP=BQ [Given]

AD || BC, BD transversal,

⸫∠ADP=∠CBQ [Alternate interior angles]

AD=BC [Opposite sides of a parallelogram]

By SAS congruence rule,

ΔAPD ≅ ΔCQB

⸫AP=CQ [CPCT]

From ΔAQB and ΔCPD,

AB=CD [Opposite sides of a parallelogram]

AB || CD, BD transversal,

⸫∠ABQ=∠CPD [Alternate interior angles]

BQ=DP [Given]

By SAS congruence rule,

ΔAQB ≅ ΔCPD

⸫AQ=CP.

⸪ΔAPD ≅ ΔCQB

⸫∠APD=∠CQB

⸫AP || CQ [Alternate interior angles are equale]

Similarly, we can prove that, AQ || CP.

⸫APCQ is a parallelogram.

10. ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD. Show that

(i) ΔAPB ≅ ΔCQD

(ii) AP = CQ

Solution:

Given, in parallelogram ABCD, AP ⊥ BD and CQ ⊥ BD.

To prove: 

(i) ΔAPB ≅ ΔCQD

(ii) AP = CQ

In ΔAPB and ΔCQD, we have,

∠APB = ∠CQD = 90

AB = CD [Opposite sides of a parallelogram]

∠ABP = ∠CDQ [Alternate interior angles]

⸫By AAS congruence rule,

ΔAPB ≅ ΔCQD

This proves (i)

⸫AP = CQ [CPCT]

This proves (ii)

11. In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively.

Show that

(i) Quadrilateral ABED is a parallelogram

(ii) Quadrilateral BEFC is a parallelogram

(iii) AD || CF and AD = CF

(iv) Quadrilateral ACFD is a parallelogram

(v) AC = DF

(vi) ΔABC ≅ ΔDEF.

Solution:

In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF.

To prove: 

 (i) Quadrilateral ABED is a parallelogram

(ii) Quadrilateral BEFC is a parallelogram

(iii) AD || CF and AD = CF

(iv) Quadrilateral ACFD is a parallelogram

(v) AC = DF

(vi) ΔABC ≅ ΔDEF.

AB = DE, AB || DE

⸫ ABED is a parallelogram [⸪A pair of opposite sides is equal and parallel]

This proves (i)

⸪BC = EF and BC || EF

⸫ BEFC is a parallelogram [⸪A pair of opposite sides is equal and parallel]

This proves (ii)

We know that, Opposite sides of a parallelogram are equal and parallel.

AD = BE ………………. (1)

AD || BE ………………. (2) 

CF = BE ……………….(3)

CF || BE ……………….(4)

From (2) and (4), we get,

AD || CF

From (1) and (3), we get,

AD = CF

⸫ AD || CF and AD = CF

This proves (iii)

⸫ ACFD is a parallelogram [⸪A pair of opposite sides is equal and parallel]

This proves (iv)

AC = DF …………………… (5) [Opposite sides of a parallelogram]

This proves (v)

In ΔABC and ΔDEF, we have,

AB = DE

BC = EF

AC = DF [From (5)]

⸫By SSS congruence rule,

ΔABC ≅ ΔDEF.

This proves (vi)

12. ABCD is a trapezium in which AB || CD and AD = BC. Show that

(i) ∠A = ∠B

(ii) ∠C = ∠D

(iii) ΔABC ≅ ΔBAD

(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

(iii) ΔABC ≅ ΔBAD

(iv) Diagonal AC = diagonal BD

Construction: AB is produced to point E, such that, AD || EC.

Now, AD || EC and AE || CD.

⸫AECD is a parallelogram.

AD = EC [Opposite sides of a parallelogram.]

⇒BC = EC [⸪AD = BC]

⇒∠BEC = ∠EBC ……………….. (1) [Opposite angles of equal sides are equal]

⸪AECD is a parallelogram,

⇒∠DAE + ∠BEC =180 [⸪Adjacent angles are supplementary]

Again,

From, (2) and (3),

This proves (i)

(ii) Now, we know that, consecutive angles of a parallelogram are supplementary. 

And 

From (4) and (5), we have,

This proves (ii)

(iii) From, ∆ABC and ∆BAD,

AB=AB [Common sides]

BC=AD [Given]

⸫By SAS congruence rule, 

∆ABC ∆BAD

This proves (iii)

⸫AC=BD [CPCT]

This proves (iv)

EXERCISE 8.2

1. ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig). AC is a diagonal. Show that:

(i) SR || AC and SR = AC

(ii) PQ = SR

(iii) PQRS is a parallelogram.

Solution:

Given, in quadrilateral ABCD, P, Q, R and S are mid-points of the sides AB, BC, CD and DA. 

To prove:

(i) SR || AC and SR = AC

(ii) PQ = SR

(iii) PQRS is a parallelogram.

We know that, line joining midpoints of two sides of a triangle is parallel to the third side and is half of it.

In ∆ACD, S, R midpoints of AD and DC

SR || AC …………………… (1)

and 

SR = AC …………………… (2)

This proves (i)

In ∆ABC, P, Q midpoints of AB and BC

PQ || AC …………………….(3)

and 

PQ = AC …………………….(4)

From (1) and (3),

PQ = SR

This proves (ii)

From (2) and (3),

PQ = SR

⸫PQRS is a parallelogram [⸪A pair of opposite sides is equal and parallel]

This proves (iii)