In Δ ABC, D is a point on BC such that AB = AD = BD = DC. Show that :

∠ADC : ∠C = 4 : 1.

Given,

AB = AD = BD

∴ Δ ABD is an equilateral triangle.

∴ ∠ABD = ∠ADB = ∠BAD = 60°.

Since, BDC is a straight line.

∴ ∠ADB + ∠ADC = 180°

⇒ 60° + ∠ADC = 180°

⇒ ∠ADC = 180° - 60° = 120°.

In Δ ADC,

⇒ AD = DC (Given)

∴ ∠DAC = ∠DCA = x (let) [Angles opposite to equal sides are equal]

By angle sum property of triangle,

⇒ ∠DAC + ∠DCA + ∠ADC = 180°

⇒ x + x + 120° = 180°

⇒ 2x = 180° - 120°

⇒ 2x = 60°

⇒ x = $\dfrac{60°}{2}$ = 30°

⇒ ∠DCA = ∠C = 30°.

⇒ ∠ADC : ∠C = 120° : 30° = 4 : 1.

Hence, proved that ∠ADC : ∠C = 4 : 1.