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

(i) ∠DCB

(ii) ∠CBD

Isosceles triangle ABC is shown in the figure below:

(i) In △ ACD and △ BCD,

⇒ ∠CAD = ∠CBD (Since, AC = BC and angles opposite to equal sides are equal.)

⇒ AD = BD (CD bisects AB)

⇒ AC = BC (Given)

∴ △ ACD ≅ △ BCD (By S.A.S. axiom)

We know that,

Corresponding sides of congruent triangle are equal.

∴ ∠ADC = ∠BDC = x (let)

Since, ADB is a straight line.

∴ ∠ADC + ∠BDC = 180°

⇒ x + x = 180°

⇒ 2x = 180°

⇒ x = $\dfrac{180°}{2}$ = 90°.

∴ ∠ADC = ∠BDC = 90°.

In △ BDC,

⇒ ∠BDC + ∠DCB + ∠CBD = 180°

⇒ 90° + ∠DCB + 55° = 180°

⇒ 145° + ∠DCB = 180°

⇒ ∠DCB = 180° - 145° = 35°.

Hence, ∠DCB = 35°.

(ii) From part (i), we get :

⇒ ∠CBD = 55°.

Hence, ∠CBD = 55°.