Question

The bisectors of angle $A,B$ and $C$ of a $\Delta ABC$ intersect its circumcricle at $D,E$ and $F$ respectively, Prove that angles of $\Delta DEF$ arc ${90}^0–\frac{1}{2}A,{90}^0–\frac{1}{2}B$ and ${90}^0–\frac{1}{2}C$.

Answer 1

According to question $AD$ is bisector of $\angle A$
$\therefore \angle 1=\angle 2=\frac{A}{2}$
$BE$, is bisector of $\angle B$
$\therefore \angle 3=\angle 4=\frac{B}{2}$
$CF$, is bisector of $\angle C$
$\therefore \angle 5=\angle 6=\frac{C}{2}$
We know that angles in same segment are equal
$\angle 9=\angle 3$ (By $\widehat{AE}$ angle subtented by)
$\angle 8=\angle 5$ (By $\widehat{FA}$ angle subtended by)
$\angle 9+\angle 8=\angle 3+\angle 5$
$\angle D=\frac{B}{2}+\frac{C}{2}$
Similarly, $\angle E=\frac{A}{2}+\frac{C}{2}$
and $\angle F=\frac{A}{2}+\frac{B}{2}$
In $\Delta DEF$
$\angle D+\angle E+\angle F={180}^0$
$\angle D={180}^0-\angle E-\angle F$
$\angle D={180}^0-(\angle E+\angle F)$
$\angle D={180}^0-(\frac{A}{2}+\frac{C}{2}+\frac{A}{2}+\frac{B}{2})$
$\angle D={180}^0-(\frac{A}{2}+\frac{B}{2}+\frac{C}{2})-\frac{A}{2}$
$\angle D={180}^0-{90}^0-\frac{A}{2}$
$\because$ In $\Delta ABC,\angle A+\angle B+\angle C={180}^0$
$\Rightarrow \frac{A}{2}+\frac{B}{2}+\frac{C}{2}={90}^0$
$\angle D={90}^0-\frac{A}{2}$
Similarly, we can prove that
$\angle E={90}^0-\frac{B}{2}$
$\angle F={90}^0-\frac{F}{2}$