Question

∆ABC is an isosceles triangle in which AB = AC, side BA is produced to D such that AD = AB. Show that ∠BCD is a right angle.

Answer 1

Given: In ∆ABC, AB = AC, side BA is produced to D such that AB = AD. 

To prove: ∠BCD = 90° 

Construction: Join CD. 

Proof: In ∆ABC given that 

AB = AC 

=> ∠ACB = ∠ABC 

Now AB = AD 

AD = AC 

=> ∠ACD = ∠ADC …(2) [Opposite angles of equal sides] 

Adding Eqn. (1) & Eqn. (2) 

∠ACB + ∠ACD = ∠ABC + ∠ADC 

=> ∠BCD = ∠ABC + ∠BDC [ v ∠ADC = ∠BDC] 

=> ∠BCD + ∠BCD = ∠ABC + ∠BCD + ∠BDC 

=> 2∠BCD = 180° [Adding ∠BCD on both sides] 

[By angle sum property of ∆] 

∠BCD = 180/2 [∠ABC + ∠BCD + ∠BDC = 180°] 

∠BCD = 90° 

So ∠BCD is a right angle.