∆ ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see Fig. 7.34). Show that ∠ BCD is a right angle.

 

In isosceles triangle ABC,

AB = AC (Given)

∴ ∠ACB = ∠ABC (Angles opposite to equal sides of a triangle are equal)

Let ∠ACB = ∠ABC be x. ----------- (1)

In ΔACD,

AC = AD (Since, AB = AD)

∴ ∠ADC = ∠ACD (Angles opposite to equal sides of a triangle are equal)

Let ∠ADC = ∠ACD be y. ----------- (2)

Thus, ∠BCD = ∠ACB + ∠ACD = x + y ----------- (3)

In ΔBCD,

∠ABC + ∠BCD + ∠ADC = 180° (Angle sum property of a triangle)

Substituting the values we get,

x + (x + y) + y = 180° [From equation (1), (2) and (3)]

2 (x + y) = 180°

2(∠BCD) = 180° [From equation(3)]

∴ ∠BCD = 90°