Question

The bisectors of base angles of a triangle cannot enclose a right angle in any case.

Answer 1

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

In a Given triangle $\triangle \mathrm{ABC}$,

CP and BP are bisectors of angles C and B severally.

Hence, $\angle A+\angle B+\angle C=180^{\circ} \rightarrow(1)$

From observing the diagram, we know that,

$\angle B=2 \angle 1$

$\angle C=2 \angle 2$

$\angle B=2 \angle 1$

$\angle C=2 \angle 2$

Substituting, the value of $\angle B\quad and\quad \angle \mathrm{C}\quad in\quad equation \rightarrow(1)$, we get,

$\angle A+2 \angle 1+2 \angle 2=180^{\circ}$

$2(\angle 1+\angle 2)=180^{\circ}-\angle A$

$(\angle 1+\angle 2)=90^{\circ}-\left(\angle \frac{A}{2}\right) \rightarrow(2)$

In $\Delta \mathrm{PBC}, \angle \mathrm{P}+\angle 1+\angle 2=180^{\circ} \rightarrow(3)$

Substituting equation (2) in (3), we get,

In a triangle, $\angle P+\left[90^{\circ}-(\angle A / 2)\right]=180^{\circ}$

$\angle P=180^{\circ}-\left[90^{\circ}-(\angle A / 2)\right]$

$\angle \mathrm{P}=\left[90^{\circ}+(\angle \mathrm{A} / 2)\right]$

Hence, angle P is usually bigger than 90°

Thus, PBC will never be a right-angled triangle.