Q10 of 49 Page 151

Suppose ABC is an isosceles triangle with AB = AC. Side BA has produced to D such that BA = AD. Prove that ∠BCD is a right angle.

The figure is as shown


As AB = AC and AB = AD thus AC = AD


Therefore, ΔACD is also isosceles triangle


Let the base angles of ΔABC be x and the base angles of ΔACD be y as shown



In order to prove ∠BCD is a right angle we have to prove that x + y = 90°


Consider ΔBCD


∠DBC = x,


∠DCB = x + y and


∠BDC = y


Sum of angles of a triangle is 180°


⇒ ∠DBC + ∠DCB + ∠BDC = 180°


⇒ x + x + y + y = 180°


⇒ 2x + 2y = 180°


⇒ 2(x + y) = 180°



⇒ x + y = 90°


⇒ ∠DCB = 90°


Hence proved ∠BCD is a right angle.


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