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Question
ABC is a right triangle with AB = AC. Bisector of ∠A meets BC at D. Prove that BC = 2AD.
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Solution
We have given, ΔABC which is an isosceles right triangle with AB = AC and AD is the bisector of ∠A.
Now in ΔABC,
AB = AC ...[Given]
⇒ ∠C = ∠B ...(1) [Angles opposite to equal sides are equal]
Now, in ΔABC, ∠A = 90°
∠A + ∠B + ∠C = 180° ...[Angle sum property of Δ]
⇒ 90° + ∠B + ∠B = 180° ...[From (1)]
⇒ 2∠B = 90°
⇒ ∠B = 45°
⇒ ∠B = ∠C = 45° or ∠3 = ∠4 = 45°
Also, ∠1 = ∠2 = 45° ...[∵ AD is bisector of ∠A]
Also, ∠1 = ∠3, ∠2 = ∠4 = 45°
⇒ BD = AD, DC = AD ...(2) [Sides opposite to equal angles are equal]
Thus, BC = BD + DC = AD + AD ...[From (2)]
⇒ BC = 2AD
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