Question

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

Find: 

1. ∠DCB 
2. ∠CBD

Answer 1

Given:

Triangle ABC is isosceles with AC = BC.

CD bisects AB at D

∠CAB = 55°

Find the remaining angles of triangle ABC

Since AC = BC

∠CAB = ∠CBA = 55°

∠A + ∠B + ∠C = 180°

55° + 55° + ∠C = 180°

∠C = 180° − 110° = 70°

So, ∠ACB = 70°

Since CD bisects AB in an isosceles triangle, it also acts as an angle bisector and median from vertex C.

`∠DCB = 1/2 ∠ACB = 70^\circ/2`

∠DCB = 35°

In triangle CDB:

∠DCB + ∠CBD + ∠CDB = 180°

But since CD is the bisector and △CDB is isosceles (as CD bisects AB),

∠CDB = ∠CBD

35° + 2∠CBD = 180°

2∠CBD = 145°

∠CBD = 72.5°