In an isosceles triangle ABC, AB = AC and ∠BAC = 90°; the bisector of ∠BAC intersects the side BC at the point D.

Let us prove that,

To Prove:

Proof:

Since, ABC is an isosceles triangle with AB = AC then by the property of isosceles triangle that the angles opposite to equal sides are also equal, we have

∠ABC = ∠ACB

From Triangle ABC,

Sum of angles of triangle = 180°

Therefore,

∠ABC + ∠BAC + ∠ACB = 180°

2 ∠ABC + 90° = 180°

∠ABC = 45°

And ∠ACB = 45°

Sec ∠ACD = sec 45° =

∠CAD = 45° (∠CAD = ∠ACB)

Sin ∠CAD =

Hence,

= 2

Hence, Proved.