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Show that the bisectors of the base angles of a triangle cannot enclose a right angle in any case.
To do:
We have to show that the bisectors of the base angles of a triangle cannot enclose a right angle in any case.
Solution:
Let in a right-angled triangle $ABC$, $\angle A$ be the vertex angle and $OB$ and $OC$ be the bisectors of $\angle B$ and $\angle C$ respectively.
$\angle A+\angle B+\angle C=180^{\circ}$
Dividing both sides by 2, we get,
$\frac{1}{2} \angle A+\frac{1}{2} \angle B+\frac{1}{2} \angle C=180^{\circ}$
$\frac{1}{2} \angle A+\angle O B C+\angle O B C=90^{\circ}$
$\angle O B C+\angle O C B=90^{\circ}-\frac{1}{2}A$
In $\triangle B O C$,
$\angle B O C+\angle O B C+\angle O C B=180^{\circ}$
$\angle B O C+90^{\circ}-\frac{1}{2} \angle A=180^{\circ}$
$\angle B O C=90^{\circ}+\frac{1}{2} \angle A$
This implies,
$\angle BOC > 90^{\circ}$
Hence, the bisectors of the base angles of a triangle cannot enclose a right angle in any case.
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