In the figure (2) given below, prove that ∠BAD : ∠ADB = 3 : 1.

From figure,

∠ADC = ∠CAD (As angles opposite to equal sides of triangle are equal) ……(i)

From figure,

∠BAD = ∠BAC + ∠CAD = ∠BAC + ∠ADC = ∠BAC + ∠ADB …….(ii)

From △ABC,

∠BAC = ∠ACB (As angles opposite to equal sides of triangle are equal) ……(iii)

∠ACB = 180 - (∠ACD) = 180 - [180 - (∠CAD + ∠ADC)] = ∠CAD + ∠ADC ……..(iv)

From figure,

∠ADC = ∠ADB

∴ ∠CAD = ∠ADB

From (iv)

∠ACB = ∠CAD + ∠ADC = ∠ADB + ∠ADB = 2∠ADB.

From (iii) we get,

∠BAC = ∠ACB = 2∠ADB.

Substituting value of ∠BAC in (ii) we get,

∠BAD = ∠BAC + ∠ADB = 2∠ADB + ∠ADB = 3∠ADB.

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

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