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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.
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Solution
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.
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