Question
The bisectors of base angles of a triangle cannot enclose a right angle in any case.
✓
Answer
In $\triangle\text{XYZ},$ Sum of all angles of a triangle is $180^\circ $
$\text{i.e}.,\angle\text{X}+\angle\text{Y}+\angle\text{Z}=180^\circ$
viding both sides by $'2'$ $\Rightarrow\frac{1}{2}\angle\text{X}+\frac{1}{2}\angle\text{Y}+\frac{1}{2}\angle\text{Z}=180^\circ$
$\Rightarrow\frac{1}{2}\angle\text{X}+\angle\text{OYZ}+\angle\text{OYZ}=90^\circ$
$\big[\therefore\text{OY},\text{OZ},\angle\text{Y}\text{ and }\angle\text{Z}\big]$
$\Rightarrow\angle\text{OYZ}+\angle\text{OZY}=90^\circ-\frac{1}{2}\angle\text{X}$
Now in $\triangle\text{YOZ}$
$\therefore\angle\text{YOZ}+\angle\text{OYZ}+\angle\text{OZY}=180^\circ$
$\Rightarrow\angle\text{YOZ}+90^\circ-\frac{1}{2}\angle\text{X}=180^\circ$
$\Rightarrow\angle\text{YOZ}=90^\circ-\frac{1}{2}\angle\text{X}$
Therefore, the bisectors of a base angle cannot enclosure right angle.
$\text{i.e}.,\angle\text{X}+\angle\text{Y}+\angle\text{Z}=180^\circ$
viding both sides by $'2'$ $\Rightarrow\frac{1}{2}\angle\text{X}+\frac{1}{2}\angle\text{Y}+\frac{1}{2}\angle\text{Z}=180^\circ$
$\Rightarrow\frac{1}{2}\angle\text{X}+\angle\text{OYZ}+\angle\text{OYZ}=90^\circ$
$\big[\therefore\text{OY},\text{OZ},\angle\text{Y}\text{ and }\angle\text{Z}\big]$
$\Rightarrow\angle\text{OYZ}+\angle\text{OZY}=90^\circ-\frac{1}{2}\angle\text{X}$
Now in $\triangle\text{YOZ}$
$\therefore\angle\text{YOZ}+\angle\text{OYZ}+\angle\text{OZY}=180^\circ$
$\Rightarrow\angle\text{YOZ}+90^\circ-\frac{1}{2}\angle\text{X}=180^\circ$
$\Rightarrow\angle\text{YOZ}=90^\circ-\frac{1}{2}\angle\text{X}$
Therefore, the bisectors of a base angle cannot enclosure right angle.
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