Question

Bisectors of vertex angles $A,B$ and $C$ of a triangle $ABC$ intersect its circumcircle at the points $D,E$ and $F$ respectively. Prove that angle $EDF={90}^o-\frac{1}{2}\angle A$

Answer 1

We have,

Given that, $BE$ is the bisector of $B$.

$\therefore \angle ABE=\frac{\angle B}{2}$

$\angle ADE=\angle ABE$ (Angles in the same segment for chord $AF$)

Similarly, $\angle ACF=\angle ADF=\frac{\angle C}{2}$ (Angle in the same segment for chord $AF$ )

$\angle D=\angle ADE+\angle ADF$

$\angle D=\frac{\angle B}{2}+\frac{\angle C}{2}$

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

$=\frac{1}{2}({180}^o-\angle A)(\therefore \angle A+\angle B+\angle C={180}^o)$

$\angle D={90}^o-\frac{\angle A}{2}$

Hence, this is the answer.