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Question

ΔABC is a right angled at A such that AB = AC and bisector of C intersects the side AB at D. Prove that AC + AD = BC.

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Solution

Let AB = AC = a and AD = b

In a right angled triangle ABC , BC2 = AB2 + AC2

BC2 = a2 + a2

BC = a√2

Given AD = b, we get

DB = AB – AD or DB = a – b

We have to prove that AC + AD = BC or (a + b) = a√2.

By the angle bisector theorem, we get

ADDB=ACBCb(ab)=aa2b(ab)=12b=(ab)2b2=abb(1+2)=ab=a(1+2)

Rationalizing the denominator with (1 - √2)

b=a(12)(1+2)×(12)b=a(12)(1)

b = a(√2 - 1)

b = a√2 – a

b + a = a√2

or AD + AC = BC [we know that AC = a, AD = b and BC = a√2]

Hence it is proved.


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