Question

show that the bisectors of the base angles of a triangle can never enclose a right angle.​

Answer 1

Step-by-step explanation:

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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Answer 2

Step-by-step explanation:

Let BP and CP are bisector of angles B and C respectively

we need to prove that bisectors of the base angles of a triangle can never enclose a right angle.

In ∆ABC, ∠A+∠B+∠C = 180°

Now, ∠A+ 2∠1 + 2∠C = 180°

2(∠1+∠2) = 180°-∠A

⇒∠1+∠2= 90° − $\frac{∠A}{2}$

In ∆PBC,∠P+∠1+∠2=180°

∠P+90°- $\frac{∠A}{2}$= 180°

∠P = 90°+ $\frac{∠A}{2}$

Hence angle P is always greater then 90°

Thus, PBC can never be a right angled triangle