Question

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

Answer 1

In $△ABC,BO$ and $CO$ are bisectors of angle $B$ and $C$ respectively.

Now, in $△ABC$,

$\angle A+\angle B+\angle C=180°$

$\angle B+\angle C=180°-\angle A.....\left(1\right)$

Now, in $△BOC$,

$\angle BOC+\frac{1}{2}\angle B+\frac{1}{2}\angle C=180°$

$\angle BOC+\frac{1}{2}(\angle B+\angle C)=180°$

$\angle BOC+\frac{1}{2}(180°-\angle A)=180°$

$\Rightarrow \angle BOC=180°-90°+\frac{1}{2}\angle A$

$\Rightarrow \angle BOC=90°+\frac{1}{2}\angle A$

$\Rightarrow \angle BOC>90°$

Thus it is proved that the bisector of base angles of a triangle cannot enclose a right angle.