Question

prove that the bisector of base angle of triangle can not enclose a right angle in any case​

Answer 1

Answer:

In ΔABC, BP and CP are bisectors of angles B and C respectively.

Hence ∠A + ∠B + ∠C = 180°

∠A + 2∠1 + 2∠2 = 180°

2(∠1 + ∠2) = 180° – ∠A

(∠1 + ∠2) = 90° – (∠A/2) --- (1)

In ΔPBC, ∠P + ∠1 + ∠2 = 180°

∠P + [90° – (∠A/2)] = 180° [From (1)]

∠P = 180° – [90° – (∠A/2)]

= [90° + (∠A/2)]

Hence angle P is always greater than 90°.

Thus PBC can never be a right angled triangle.

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