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

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

 

Solution:

In ΔXYZ,

Sum of all angles of a triangle is 180°

i.e., ∠X + ∠Y + ∠Z = 180°

Dividing both sides by ‘2’

$\Rightarrow \frac{1}{2} \angle \mathrm{X}+\frac{1}{2} \angle \mathrm{Y}+\frac{1}{2} \angle \mathrm{Z}=180^{\circ}$

$\Rightarrow \frac{1}{2} \angle \mathrm{X}+\angle \mathrm{OYZ}+\angle \mathrm{OYZ}=90^{\circ}[\because \mathrm{OY}, \mathrm{OZ}, \angle \mathrm{Y}$ and $\angle \mathrm{Z}]$

$\Rightarrow \angle \mathrm{OYZ}+\angle \mathrm{OZY}=90^{\circ}-\frac{1}{2} \angle \mathrm{X}$

Now in ΔYOZ

∴  ∠YOZ + ∠OYZ + ∠OZY = 180°

$\Rightarrow \angle \mathrm{YOZ}+90^{\circ}-\frac{1}{2} \angle \mathrm{X}=180^{\circ}$

$\Rightarrow \angle \mathrm{YOZ}=90^{\circ}-\frac{1}{2} \angle \mathrm{X}$

Therefore, the bisectors of a base angle cannot enclosure right angle.

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