Sides AB and AC of a triangle ABC are equal. BC is produced through C upto point D such that AC = CD. D and A are joined and produced (through vertex A) upto point E. If angle BAE = 108°; find angle ADB.

In △ ABD,

⇒ ∠BAE = ∠3 + ∠ADB (An exterior angle is equal to sum of two opposite interior angles)

⇒ 108° = ∠3 + ∠ADB …..(1)

Given,

⇒ AB = AC

∴ ∠3 = ∠2 (Angles opposite to equal sides are equal)

Substituting value of ∠3 from above equation in (1), we get :

⇒ 108° = ∠2 + ∠ADB ………..(2)

In △ ACD,

⇒ ∠2 = ∠1 + ∠ADC (An exterior angle is equal to sum of two opposite interior angles)

⇒ ∠2 = ∠1 + ∠ADB ……….(3)

Given,

⇒ AC = CD

∴ ∠1 = ∠ADC (Angles opposite to equal sides are equal)

⇒ ∠1 = ∠ADB

Substituting above value of ∠1 in (3), we get :

⇒ ∠2 = ∠ADB + ∠ADB

⇒ ∠2 = 2∠ADB

Substituting above value of ∠2 in (2), we get :

⇒ 108° = 2∠ADB + ∠ADB

⇒ 108° = 3∠ADB

⇒ ∠ADB = $\dfrac{108°}{3}$ = 36°.

Hence, ∠ADB = 36°.