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**Question 1. The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral. Solution: **Let the angles of the quadrilateral be 3x, 5x, 9x and 13x.

∵ Sum of all the angles of quadrilateral = 360º

∴ 3x + 5x + 9x + 13x = 360º ⇒ 30x = 360º

⇒ x=(360

∴ 3x = 3 x 12° = 36º 5x

= 5 x 12° = 60º 9x = 9 x 12°

= 108º 13x = 13 x 12° = 156º

⇒ The required angles of the quadrilateral are 36º, 60º, 108º and 156º.

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

**Solution: **A parallelogram ABCD such that AC = BD

In ΔABC and ΔDCB, AC = DB [Given]

AB = DC [Opposite sides of a parallelogram]

BC = CB [Common]

ΔABC ≌ ΔDCB [SSS criteria]

∴ Their corresponding parts are equal.

⇒ ∠ABC = ∠DCB …(1)

∵ AB || DC and BC is a transversal. [∵ ABCD is a parallelogram]

∴ ∠ABC + ∠DCB = 180º [Interior opposite angles are supplementary] …(2)

From (1) and (2), we have ∠ABC = ∠DCB = 90º

i.e. ABCD is parallelogram having an angle equal to 90º.

∴ ABCD is a rectangle.

**Question 3. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus. Solution:** We have a quadrilateral ABCD such that the diagonals AC and

BD bisect each other at right angles at O.

∴ In ΔAOB and ΔAOD, we have AO = AO [Common]

OB = OD [Given that O in the mid-point of BD]

∠AOB = ∠AOD [Each = 90°]

∴ ΔAOB ≌ ΔAOD [SAS criteria]

⇒ Their corresponding parts are equal. ∴ AB = AD …(1)

Similarly, AB = BC …(2) BC = CD …(3)

CD = AD …(4)

∴ From (1), (2), (3) and (4),

we have AB = BC = CD = DA Thus, the quadritateral ABCD is a rhombus.

**Question 4. Show that the diagonals of a square are equal and bisect each other at right angles. Solution: **We have a square ABCD such that its diagonals AC and BD intersect at O.

(i) To prove that the diagonals are equal, i.e. AC = BD In ΔABC and ΔBAD, we have

AB = BA [Common]

BC = AD [Opposite sides of the square ABCD]

∠ABC = ∠BAD [All angles of a square are equal to 90º]

∴ DABC ≌ DBAD [SAS criteria]

⇒ Their corresponding parts are equal.

⇒ AC = BD …(1)

(ii) To prove that ‘O’ is the mid-point of AC and BD.

∵ AD || BC and AC is a transversal. [∵ Opposite sides of a square are parallel]

∴ ∠1 = ∠3 [Interior alternate angles]

Similarly, ∠2= ∠4 [Interior alternate angles]

Now, in ΔOAD and ΔOCB, we have AD = CB [Opposite sides of the square ABCD]

∠1= ∠3 [Proved] ∠2= ∠4 [Proved]

∴ ΔOAD ≌ ΔOCB [ASA criteria]

∴ Their corresponding parts are equal.

⇒ OA = OC and OD = OB ⇒ O is the mid-point of AC and BD,

i.e. the diagonals AC and BD bisect each other at O. …(2)

(iii) To prove that AC ⊥ BD.

In ΔOBA and ΔODA, we have OB = OD [Proved]

BA = DA [Opposite sides of the square]

OA = OA [Common]

∴ ΔOBA ≌ ΔODA [SSS criteria]

⇒ Their corresponding parts are equal.

⇒ ∠AOB = ∠AOD

But ∠AOB and ∠AOD form a linear pair.

∴ ∠AOB + ∠AOD = 180º

⇒ ∠AOB = ∠AOD = 90º

⇒ AC ⊥ BD …(3)

From (1), (2) and (3), we get AC and BD are equal and bisect each other at right angles.

**Question 5. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square. Solution:** We have a quadrilateral ABCD such that ‘O’ is the mid-point of AC and BD. Also AC ⊥ BD.

Now, in ΔAOD and ΔAOB, we have

AO = AO [Common]

OD = OB [∵ O is the mid-point of BD]

∠AOD = ∠AOB [Each = 90º]

∴ ΔAOD ≌ ΔAOB [SAS criteria]

∴ Their corresponding parts are equal.

⇒ AD = AB …(1)

Similarly, we have AB = BC …(2)

BC = CD …(3)

CD = DA …(4)

From (1), (2), (3) and (4) we have: AB = BC = CD = DA

∴ Quadrilateral ABCD is having all sides equal.

In ΔAOD and ΔCOB, we have AO = CO [Given]

OD = OB [Given]

∠AOD = ∠COB [Vertically opposite angles]

∴ ΔAOD ≌ ΔCOB

⇒ Their corresponding parts are equal. ⇒ ∠1= ∠2 But, they form a pair of interior alternate angles.

∴ AD || BC

Similarly, AB || DC

∴ ABCD is a parallelogram.

∵ Parallelogram having all of its sides equal is a rhombus.

∴ ABCD is a rhombus.

Now, in ΔABC and ΔBAD, we have AC = BD [Given]

BC = AD [Proved]

AB = BA [Common]

∴ ΔABC ≌ ΔBAD [SSS criteria]

⇒ Their corresponding angles are equal.

∴ ∠ABC = ∠BAD

Since, AD || BC and AB is a transversal.

∴ ∠ABC + ∠BAD = 180º [Interior opposite angles are supplementary]

i.e. The rhombus ABCD is having one angle equal to 90º.

Thus, ABCD is a square.

**Question 6. Diagonal AC of a parallelogram ABCD bisects ∠ A (see figure). Show that**

**(i) it bisects ∠ C also, (ii) ABCD is a rhombus. Solution: **We have a parallelogram ABCD in which diagonal AC bisects ∠A.

⇒ ∠DAC = ∠BAC

(i) To prove that AC bisects ∠C.

∵ ABCD is a parallelogram.

∴ AB || DC and AC is a transversal.

∴ ∠1 = ∠3[Alternate interior angles] …(1)

Also, BC || AD and AC is a transversal.

∴ ∠2 = ∠4[Alternate interior angles] …(2)

But AC bisects ∠A. [Given]

∴ ∠1 = ∠2 …(3)

From (1), (2) and (3), we have ∠3= ∠4

⇒ AC bisects ∠C.

(ii) To prove ABCD is a rhombus.

In DABC, we have ∠1= ∠4[∵ ∠1 = ∠2 = ∠4]

⇒ BC = AB [Sides opposite to equal angles are equal] …(4)

Similarly, AD = DC …(5)

But ABCD is a parallelogram [Given]

∴ AB = DC [Opposite sides of a parallelogram] …(6)

From (4), (5) and (6), we have AB = BC = CD = DA

Thus, ABCD is a rhombus.

**Question 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.**

∴ AB = BC = CD = AD

Also, AB || CD an AD || BC

Now, AD = CD ⇒ ∠1 = ∠2 …(1)

[Angles opposite to equal sides are equal]

Also, CD || AB [Opposite sides of the parallelogram] and AC is AC is transversal.

∴ ∠1= ∠3 [Alternate interior angles] …(2)

From (1) and (2), we have ∠2= ∠3 and ∠1 = ∠4 ⇒ AC bisects ∠C as well as ∠A.

Similarly, we prove that BD bisects ∠B as well as ∠D.

**Question 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: **We have a rectangle ABCD such that AC bisects ∠ A as well as ∠C.

i.e. ∠ 1= ∠ 4 and ∠ 2 = ∠ 3 ...(1)

(i) Since, rectangle is a parallelogram.

∴ ABCD is a parallelogram.

⇒ AB || CD and AC is a transversal.

∴ ∠2= ∠4 [Alternate interior angles] . ..(2)

From (1) and (2), we have ∠3= ∠4

⇒ AB = BC [∵ Sides opposite to equal angles in D ABC are equal.]

∴ AB = BC = CD = AD

⇒ ABCD is a rectangle having all of its sides equal.

∴ ABCD is a square.

(ii) Since, ABCD is a square, and diagonals of a square bisect the opposite angles.

∴ BD bisects ∠B as well as ∠D.

**Question 9. In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see figure). Show that:**

**(i) **Δ**APD ≌ **Δ**CQB (ii) AP = CQ (iii) **Δ**AQB ≌ **Δ**CPD (iv) AQ = CP (v) APCQ is a parallelogram. Solution:** We have parallelogram ABCD. BD is a diagonal and ‘P’ and ‘Q’ are such that

PD = QB [Given]

(i) To prove that ΔAPD ≌ ΔCQB

∵ AD || BC and BD is a transversal. [∵ ABCD is a parallelogram.]

∴ ∠ADB = ∠CBD [Interior alternate angles]

⇒ ∠ADP = ∠CBQ

Now, in ΔAPD and ΔCQB, we have

AD = CB [Opposite side of the parallelogram]

P D = QB [Given]

∠CBQ = ∠ADP [Proved]

∴ Using SAS criteria, we have ΔAPD ≌ ΔCQB

(ii) To prove that AP = CQ Since ΔAPD ≌ ΔCQB [Proved]

∴ Their corresponding parts are equal. ⇒ AP = CQ

(iii) To prove that ΔAQB ≌ ΔCPD.

∵ AB || CD and BD is a transversal. [∵ ABCD is a parallelogram.]

∴ ∠ABD = ∠CDB

⇒ ∠ABQ = ∠CDP Now, in ΔAQB and ΔCPD,

we have QB = PD [Given]

∠ABQ = ∠CDP [Proved]

AB = CD [Opposite sides of parallelogram ABCD]

∴ ΔAQB ≌ ΔCPD [SAS criteria]

(iv) To prove that AQ = CP.

Since ΔAQB ≌ ΔCPD [Proved]

∴ Their corresponding parts are equal.

⇒ AQ = CP.

(v) To prove that APCQ is a parallelogram.

Let us join AC.

Since, the diagonals of a || gm bisect each other

∴ AO = CO …(1)

and BO = DO

⇒ (BO ∠ BQ) = (DO ∠ DP) [∵ BQ = DP (Given)]

⇒ QO = PO …(2)

Now, in quadrilateral APCQ, we have AO = CO and QO = PO

[from (1) and (2)]

i.e. AC and QP bisect each other at O.

⇒ APCQ is a parallelogram.

**Question 10. ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see figure). Show that (i) **Δ

**Solution:** (i) In ΔAPB and ΔCQD, we have

∠APB = ∠CQD [90º each] AB = CD

[Opposite sides of parallelogram ABCD]

∠ABP = ∠CDQ [AB || CD and AB is a transversal]

⇒ Using AAS criteria, we have ΔAPB ≌ ΔCQD

(ii) Since, ΔAPB ≌ ΔCQD [Proved]

∴ Their corresponding parts are equal.

⇒ AP = CQ

**Example 11. In D ABC and D 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 figure). 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) D ABC ≌ D DEF. Solution:** (i) To prove that ABED is a parallelogram.

Since “A quadrilateral is a parallelogram if a pair of opposite sides is parallel and is of equal length.” Now AB = DE [Given]

AB | | DE [Given]

i.e. ABED is a quadrilateral in which a pair of opposite sides (AB and DE) is parallel and of equal length.

∴ ABED is a parallelogram.

(ii) To prove that BECF is a parallelogram.

∵ BC = EF [Given]

and BC || EF [Given]

i.e. BECF is a quadrilateral in which a pair of opposite sides (BC and EF) is parallel and of equal length.

∴ BECF is a parallelogram.

(iii) To prove that AD || CF and AD = CF ∵ ABED is a parallelogram. [Proved]

∴ Its opposite sides are parallel and equal.

⇒ AD || BE and AD = BE …(1)

Also BEFC is a parallelogram. [Proved]

∴ BE || CF and BE = CF [∵ Opposite sides of a parallelogram are parallel and equal] …(2)

From (1) and (2), we have AD || CF and AD = CF

(iv) To prove that ACFD is a parallelogram.

∵ AD || CF [Proved] and AD = CF [Proved]

i.e. In quadrilateral ACFD, one pair of opposite sides (AD and CF) is parallel and of equal length.

∴ Quadrilateral ACFD is a parallelogram.

(v) To prove that AC = DE.

∵ ACFD is a parallelogram. [Proved]

∴ Its opposite sides are parallel and of equal length. i.e. AC = DF

(vi) To prove that ΔABC ≌ ΔDEF In D ABC and DDEF, we have:

AB = DE [Opposite sides of a parallelogram]

BC = EF [Opposite sides of a parallelogram]

AC = DF [Proved]

∴ Using SSS criteria, we have ΔABC ≌ ΔDEF.

**Question 12. ABCD is a trapezium in which AB || CD and AD = BC (see figure). Show that**

**(i) ∠ A = ∠ B (ii) ∠ C = ∠ D (iii) ∠ ABC ≌ ∠ BAD (iv) Diagonal AC = Diagonal BD Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E. Solution:** We have AB || CD and AD = BC (i) To prove that ∠A = ∠B.

Produce AB to E and draw CE || AD.

∴ AB || DC

⇒ AE || DC [Given]

Also AD || CE [Construction]

∴ AECD is a parallelogram.

⇒ AD = CE [opposite sides of the parallelogram AECD]

But AD = BC [Given]

∴ BC = CE

Now, in ΔBCE, we have BC = CE

⇒ ∠CBE = ∠CEB …(1)

[∵ Angles opposite to equal sides of a triangle are equal]

Also, ∠ABC + ∠CBE = 180° [Linear pair] ...(2)

and ∠A + ∠CEB = 180° [∵ Adjacent angles of a parallelogram are supplementary] …(3)

From (2) and (3), we get ∠ABC + ∠CBE = ∠A + ∠CEB

But ∠CBE = ∠CEB [Using (1)]

∴ ∠ABC = ∠A

or ∠B= ∠A

or ∠A= ∠B

(ii) To prove that ∠C = ∠D.

AB || CD and AD is a transversal.

∴ ∠A + ∠D = 180º [Sum of interior opposite angles]

Similarly, ∠B + ∠C = 180º

⇒ ∠A + ∠D= ∠B + ∠C

But ∠A= ∠B [Proved]

∴ ∠C= ∠D

(iii) To prove ΔABC ≌ ΔBAD In ΔABC and ΔBAD, we have

AB = BA [Common]

BC = AD [Given]

∠ABC = ∠BAD [Proved]

∴ ΔABC ≌ ΔBAD [Using SAS criteria]

(iv) To prove that diagonal AC = diagonal BD ∵

ΔABC ≌ ΔBAD [Proved]

∴ Their corresponding parts are equal.

⇒ the diagonal AC = the diagonal BD.

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