Question

An isosceles triangle ABC has AC = BC. CD bisects AB at D and ∠CAB = 55°.

Find: 

1. ∠DCB 
2. ∠CBD

Answer 1

In ΔABC,

AC = BC        ...[Given]

∴ ∠CAB = ∠CBD       ...[angles opp. to equal sides are equal]

⇒ ∠CBD = 55°

In ΔABC,

∠CBA + ∠CAB + ∠ACB = 180°

but, ∠CAB = ∠CBA = 55°

⇒ 55° + 55° + ∠ACB = 180°

⇒ ∠ACB = 180° − 110°

⇒ ∠ACB = 70°

Now,

In ΔACD and ΔBCD,

AC = BC      ...[Given]

CD = CD      ...[Common]

AD = BD     ...[Given: CD bisects AB]

∴ ΔACD ≅ ΔBCD

⇒ ∠DCA = ∠DCB

⇒ ∠DCB = `(∠"ACB")/2`

= `(70°)/2`

⇒ ∠DCB = 35°