# ΔABC 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.

Given: AB = AC and AD = AB
To Prove: ∠ BCD is a right angle

Proof: In

AB = AC (Given)

ACB = ABC (Angles opposite to equal sides are equal)

In

ADC = ACD (Angles opposite to equal sides are equal)

Now,

In

CAB + ACB + ABC = 180o            ( Sum of interior angles of a triangle)

CAB + 2 ACB = 180                   (∠ACB = ∠ABC)

CAB = 180o - 2 ACB (i)

Similarly,

In

CAD = 180o - 2 ACD (ii)

Also,

CAB + CAD = 180o                    (BD is a straight line)

Adding (i) and (ii), we get

CAB + CAD = 180o - 2ACB + 180o - 2ACD

180o = 360o - 2ACB - 2ACD

2 (ACB + ACD) = 180o

BCD = 90o
Hence, Proved.

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