Question

ABC is a right angled triangle such that AB is equal to AC and the bisector of angle C intersects the side AB at D.Prove that AC➕ AD is equal to BC

Answer 1

ANSWER:-

Given:

ABC is a right angled ∆ such that AB=AC and the bisector of angle C intersects the side AB at D.

To prove:

Prove that AC+AD = BC.

Solution:

Let AB=AC = a & AD =b

Therefore,

In ∆ABC,

Using Pythagoras Theorem:

BC² = AB² + AC²

BC² = a² + a²

BC² = 2a²

BC= a√2

Since,

AD=b,we get;

DB= AB - AD

DB= a - b

We have to prove that AC+AD=BC

=) (a+b) = a√2.

Using the angle bisector theorem, we get;

$= > \frac{AD}{DB} = \frac{AC}{BC} \\ \\ = > \frac{b}{(a - b)} = \frac{a}{a \sqrt{2} } \\ \\ = > \frac{b}{(a - b)} = \frac{1}{ \sqrt{2} } \\ \\ = > b = \frac{(a - b)}{ \sqrt{2} } \\ \\ = > b \sqrt{2} = a - b \\ \\ = > b(1 + \sqrt{2} ) = a \\ \\ = > b = \frac{a}{(1 + \sqrt{2}) }$

Rationalizing the denominator,we get;

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

b + a= a√2

=) AD + AC= BC

Hence,

proved.

Hope it helps ☺️