In the given figure, AB = BC and AC = CD. Show that: ∠BAD : ∠ADB = 3 : 1

In △ABC,

AB = BC

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

In △ACD,

AC = CD

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

From figure,

⇒ ∠ACB + ∠ACD = 180° (Linear pair)

⇒ x + ∠ACD = 180°

⇒ ∠ACD = 180° - x

In △ACD,

By angle sum property of triangle,

⇒ ∠CAD + ∠ADC + ∠ACD = 180°

⇒ y + y + (180° - x) = 180°

⇒ 2y + 180° - x = 180°

⇒ 2y - x = 0

⇒ 2y = x

From figure,

∠BAD = ∠BAC + ∠CAD

⇒ ∠BAD = x + y

⇒ ∠BAD = 2y + y

⇒ ∠BAD = 3y

⇒ ∠BAD = 3∠ADC

From figure,

⇒ ∠ADC = ∠ADB

Thus,

⇒ ∠BAD = 3∠ADB

⇒ $\dfrac{\text{∠BAD}}{\text{∠ADB}} = \dfrac{3}{1}$

Hence, proved that ∠BAD : ∠ADB = 3 : 1.