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# Class 9 SELINA Solutions Maths Chapter 9 - Triangles [Congruency in Triangles]

## Triangles [Congruency in Triangles] Exercise Ex. 9(A)

### Solution 1(a)

Correct option: (i) AC = PR

ΔABC ≅ ΔPQR

Then, by c.p.c.t.

AB = PQ

BC = QR

AC = PR

Hence,

AC = PR.

### Solution 1(b)

Correct option: (iii) angle BAC

In ΔABD and ΔACD

AB = AC … given

BD = CD … given

∴ ΔABD ≅ ΔACD … (S.S.S.)

### Solution 1(c)

Correct option: (ii) CD = CB

∠DAC = ∠BAC … given

AC is common

∴ ΔADC ≅ ΔABC … (S.S.S.)

∴ By c.p.c.t.

∠B = ∠D

∠ACD = ∠ACB

CD = CB

### Solution 1(d)

Correct option: (iv) AM bisects ∠BAC

In ΔAMB and ΔAMC

AM is common

∠AMB = ∠AMC … each 90o

MB = MC … AM is the perpendicular bisector of BC

∴ ΔAMB ≅ ΔAMC … (S.A.S.)

∴ By c.p.c.t.

AB = AC

∠BAM = ∠CAM

∴ AM bisects ∠BAC

### Solution 1(e)

Correct option: (iv) ΔABC @ ΔPQR

In ΔABM and ΔPQN

AB = PQ … given

AM = PN … given

BC = QR … given

∴ 2BM = 2QN … M and N are midpoints of BC and QR respectively

∴ .BM = QN

∴ ΔABM ≅ ΔPQN …(S.S.S.)

∴ ∠ABM = ∠PQN … (c.p.c.t.)

Now,

AB = PQ … given

BC = QR … given

∠ABC = ∠PQR

ΔABC ≅ ΔPQR … (S.A.S.)

## Triangles [Congruency in Triangles] Exercise Ex. 9(B)

### Solution 1(a)

Correct option: (iii) AB and CD bisect each other

In, ΔACP and ΔBDP

AC = BD ... given

∠C = ∠D … each 90o

∠APC = ∠BPD … vertically opposite angles

ΔACP ≅ ΔBDP … (A.A.S.)

Then, by c.p.c.t.

CP = DP

AP = BP

Hence, AB and CD bisect each other

### Solution 1(b)

Correct option: (ii) ΔABC ≅ ΔAED

In, ΔABC and ΔAED

∴ ∠BAD + ∠DAC = ∠EAC + ∠DAC

BD = EC

∴ BD + DC = EC + DC

∴ BC = ED

∠B = ∠E … given

∴ ΔABC ≅ ΔAED … (A.A.S.)

### Solution 1(c)

Correct option: (i) ΔABD ≅ ΔACD

AD is ⊥ to both EF and BD.

Hence, EF || BD

Now, if AB is a transversal ∠EAB = ∠ABD…(alternate angles)

Similarly, if AC is a transversal

∠FAC = ∠ACD…(alternate angles)

Now,

∠EAB = ∠FAC … given

Hence,

∠ABD = ∠ACD… (1)

Now,

In, ΔABD and ΔACD

∠ABD= ∠ACD … from (1)

ΔABD ≅ ΔACD … (A.A.S.)

### Solution 1(d)

Correct option: (iii) BP = AR

Join PR

In ΔPOR

PO = RO

∠OPR = ∠ORP … angles opposite to equal sides in a triangle

In, ΔPAR and ΔRBP

∠x = ∠y … given

∠OPR = ∠ORP

PR is common

ΔPAR ≅ ΔRBP … (A.A.S.)

∴ BP = AR … (c.p.c.t.)

### Solution 1(e)

Correct option: (i) AB and CD bisect each other

BC||DA

Now, if AB is a transversal ∠CBO = ∠DAO…(alternate angles)

Similarly, if CD is a transversal

Also, BC = DA … given

Hence,

ΔCOB ≅ ΔDOA … (A.S.A.)

Then, by c.p.c.t.

CO = DO

BO = AO

Hence, AB and CD bisect each other

## Triangles [Congruency in Triangles] Exercise Test Yourself

### Solution 7

In triangles AOE and COD,

A = C  (given)

AOE = COD  (vertically opposite angles)

A + AOE = C + COD

180° - AEO = 180° - CDO

AEO = CDO  ….(i)

Now, AEO + OEB = 180° (linear pair)

And, CDO + ODB = 180° (linear pair)

AEO + OEB = CDO + ODB

OEB = ODB [Using (i)]

Now, in ΔABD and ΔCBE,

A = C  (given)

AB = BC  (given)

ΔABD ΔCBE  (by AAS congruence criterion)

### Solution 8

In ΔAOD and ΔBOC,

AOD = BOC (vertically opposite angles)

DAO = CBO (each 90°)

ΔAOD ΔBOC (by AAS congruence criterion)

AO = BO [cpct]

O is the mid-point of AB.

Hence, CD bisects AB.

### Solution 9

In ΔABC,

AB = AC

B = C (angles opposite to equal sides are equal)

### Solution 12

In ΔABD and ΔBAC,