Question

ABC is a right triangle with AB = AC. Bisector of ∠A meets BC at D. Prove that BC = 2 AD.

Answer 1

Given: A right angles triangle with AB = AC bisector of ∠A meets BC at D.

To prove: BC = 2AD

Proof:

According to the question,

In right Δ ABC,

AB = AC

Since, hypotenuse is the longest side,

BC is hypotenuse

∠BAC = 90^o

Now,

In Δ CAD and Δ BAD,

We have,

AC = AB

Since, AD is the bisector of ∠A,

∠1 = ∠2

AD = AD [Common side]

Now,

By SAS criterion of congruence,

We get,

Δ CAD ≅ Δ BAD

CD = BD [CPCT]

Since, Mid-point of hypotenuse of a right triangle is equidistant from the 3 vertices of a triangle.

AB = BD = CD …(1)

Now, BC = BD + CD

⇒ BC = AD + AD [Using eq.(1)]

⇒ BC = 2AD

Hence, proved.