Question

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

Answer 1

Answer:

Step-by-step explanation:

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

AD/ DB = AC / BC

b/(a - b) = a/ a√2

b/(a - b) = 1/√2

b = (a – b)/ √2

b√2 = a – b

b(1 + √2) = a

b = a/ (1 + √2)

Rationalizing the denominator with (1 - √2)

 

b = a(1 - √2) / (1 + √2) × (1 - √2)

b = a(1 - √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.

Answer 2

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