Question

ABC is a right triangle such that AB = AC and bisector of angle C intersects the side AB at D. Prove that AC + AD = BC.

Answer 1

Given: In right angled ΔABC, AB = AC and CD is the bisector of ∠C.

Construction: Draw DE ⊥ BC.

To prove: AC + AD = BC

Proof: In right angled ΔABC, AB = AC and BC is a hypotensue  ...[Given]

∴ ∠A = 90°

In ΔDAC and ΔDEC, ∠A = ∠3 = 90°

∠1 = ∠2   ...[Given, CD is the bisector of ∠C]

DC = DC   ...[Common sides]

∴ ΔDAC ≅ ΔDEC   ...[By AAS congruence rule]

⇒ DA = DE   [By CPCT]  ...(i)

And AC = EC   ...(ii)

In ΔABC, AB = AC

∠C = ∠B  [Angles opposite to equal sides are equal] ...(iii) 

Again, in ΔABC, ∠A + ∠B + ∠C = 180°   ...[By angles sum property of a triangle]

⇒ 90° + ∠B + ∠B = 180°   ...[From equation (iii)]

⇒ 2∠B = 180° – 90°

⇒ 2∠B = 90°

⇒ ∠B = 45°

In ΔBED, ∠5 = 180° – (∠B + ∠4)  ...[By angle sum property of a triangle]

= 180° – (45° + 90°)

= 180° – 135°

= 45°

∴ ∠B = ∠5

⇒ DE = BE  [∵ Sides opposite to equal angles are equal] ...(iv)

From equations (i) and (iv),

DA = DE = BE   ...(v)

∵ BC = CE + EB

= CA + DA   ...[From equations (ii) and (v)]

∴ AD + AC = BC  

Hence proved.