AABC is a right triangle right angled at A such that AB- AC andbisecto

1.	AABC is a right triangle right angled at A such that AB- AC andbisector of 'C intersects the side AB at D. Prove that AC+ AD- BC
Answer» Let AB = AC = a and AD = b In a right angled triangle ABC , BC²= AB²+ AC²BC²= a²+ a² BC = a√2 Given AD = b, we getDB = 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√2b/(a - b) = 1/√2b = (a – b)/ √2b√2 = a – bb(1 + √2) = ab = 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 – ab + a = a√2or AD + AC = BC [we know that AC = a, AD = b and BC = a√2] Hence it is proved.

AABC is a right triangle right angled at A such that AB- AC andbisector of 'C intersects the side AB at D. Prove that AC+ AD- BC

Answer»

Let AB = AC = a and AD = b

In a right angled triangle ABC ,

BC²= AB²+ AC²BC²= a²+ a²

BC = a√2

Given AD = b,

we getDB = 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√2b/(a - b)

= 1/√2b = (a – b)/ √2b√2

= a – bb(1 + √2) = ab

= 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 – ab + a

= a√2or AD + AC = BC

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

Hence it is proved.