Question

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

Answer 1

In ∆ABC,

AB = AC (Given)

∴ ∠C = ∠B .....(1) (In a triangle, angles opposite to equal sides are equal)

Also, ∠A = 90º (Given)

Now,

∠A + ∠B + ∠C = 180º (Angle sum property of triangle)

⇒ 90º + 2∠B = 180º [Using (1)]

⇒ 2∠B = 180º − 90º = 90º

⇒ ∠B = $\frac{90°}{2}$ = 45º

∴ ∠C = ∠B = 45º

It is given that, AD is the bisector of ∠A.

∴ ∠CAD = ∠BAD$=\frac{\angle \mathrm{A}}{2}=\frac{90°}{2}$ = 45º

In ∆ABD,

∠B = ∠BAD (Each measure 45º)

∴ AD = BD .....(2) (In a triangle, sides opposite to equal angles are equal)

In ∆ACD,

∠C = ∠CAD (Each measure 45º)

∴ AD = CD .....(3) (In a triangle, sides opposite to equal angles are equal)

Adding (2) and (3), we have

AD + AD = BD + CD

⇒ 2AD = BC

Or BC = 2AD

Hence proved.