In the given figure, AB = AD; CB = CD; ∠A = 42° and ∠C = 108°, find ∠ABC.

Join BD.

In △ABD,

AB = AD

⇒ ∠ABD = ∠ADB = x (let) (Angles opposite to equal sides in a triangle are equal)

By angle sum property of triangle,

⇒ ∠BAD + ∠ABD + ∠ADB = 180°

⇒ 42° + x + x = 180°

⇒ 2x = 180° - 42°

⇒ 2x = 138°

⇒ x = $\dfrac{138°}{2}$

⇒ x = 69°

⇒ ∠ABD = ∠ADB = 69°.

In △BCD,

CB = CD

⇒ ∠CBD = ∠CDB = y (let) (Angles opposite to equal sides in a triangle are equal)

By angle sum property of triangle,

⇒ ∠BCD + ∠CBD + ∠CDB = 180°

⇒ 108° + y + y = 180°

⇒ 2y = 180° - 108°

⇒ 2y = 72°

⇒ y = $\dfrac{72°}{2}$

⇒ y = 36°

⇒ ∠CBD = ∠CDB = 36°

From figure,

∠ABC = ∠ABD + ∠CBD = 69° + 36° = 105°.

Hence, ∠ABC = 105°.