Question

$\Delta ABC\text{is a right angled at A such that AB = AC and bisector of}\angle C\text{intersects the side AB at D. Prove that AC + AD = BC.}$

Answer 1

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

$\frac{AD}{DB}=\frac{AC}{BC}\frac{b}{(a-b)}=\frac{a}{a\surd 2}\frac{b}{(a-b)}=\frac{1}{\surd 2}b=\frac{(a–b)}{\surd 2}b\surd 2=a–bb(1+\surd 2)=ab=\frac{a}{(1+\surd 2)}$

Rationalizing the denominator with (1 - √2)

$b=\frac{a(1-\surd 2)}{(1+\surd 2)\times (1-\surd 2)}b=\frac{a(1-\surd 2)}{(-1)}$

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.