Question

Bisetors of vertex angles A, B and C of a triangle ABC intersect its circumcircle at the D, E and F respectively. Then, $\angle EDF=$ ${90}^0-\frac{1}{2}\angle$ A.

• A. True
• B. False

Answer 1

The correct option is A

True

Given - A $\Delta$ ABC whose bisectors of angle A,B and C intersect the circumcircle at D,E, and F respectively, ED, EF and DF are joined.

Construction-Join BF, FA, AE and EC.

$\angle EBF=\angle ECF=\angle EDF$ ....(i) (Angles in the same segment)

In cyclic quad. AFBE,

$\angle EBF+\angle EAF={180}^0$ ....(ii) (Sum of opposite angles) Similarly in cyclic quad. CEAF, $\angle EAF\angle ECF={180}^0$ ....(iii) On adding (ii) and (iii),

$\Rightarrow EBF\Rightarrow \angle ECF+2\angle EAF={360}^0$

$\Rightarrow \angle EDF+\angle EDF+2\angle EAF={360}^0$ [from (i)]

$\Rightarrow 2\angle EDF+2\angle EAF={360}^0$

$\Rightarrow \angle EDF+\angle EAF={180}^0$

$\Rightarrow \angle EDF+\angle 1+\angle BAC+\angle 2={180}^0$

But $\angle 1=\angle 3and\angle 2=\angle 4$ (Angles in the same segment)

$\therefore \angle EDF+\angle 3+\angle BAC+\angle 4={180}^0$

$\Rightarrow$ But $\angle 4=\frac{1}{2}\angle C,\angle 3=\frac{1}{2}\angle B$

$\therefore \angle EDF+\frac{1}{2}\angle C+\angle BAC+\frac{1}{2}\angle B={180}^0$

$\Rightarrow \angle EDF+\frac{1}{2}\angle C+2\times \frac{1}{2}\angle A+\frac{1}{2}\angle B={180}^0$

$\Rightarrow \angle EDF+\frac{1}{2}(\angle A+\angle B+\angle C)+\frac{1}{2}\angle A={180}^0$

$\Rightarrow \angle EDF+\frac{1}{2}\left({180}^0\right)+\frac{1}{2}\angle A={180}^0$

$\Rightarrow \angle EDF+{90}^0+\frac{1}{2}\angle A={180}^0$

$\Rightarrow \angle EDF={180}^0-{90}^0-\frac{1}{2}\angle A={90}^0-\frac{1}{2}\angle A$ Q.E.D