Question

In the given figure, $△$ABC is inscribed in a circle. The bisector of $\angle$BAC meets BC at D and the circle at E. If EC is joined then $\angle$ECD = 30${}^o$. Yhe value of $\angle$BAC is

• A. 30${}^o$
• B. 40${}^o$
• C. 50${}^o$
• D. 60${}^o$

Answer 1

The correct option is D 60${}^o$

Given- $\Delta ABC$ has been inscribed in a circle. AE, the bisector of $\angle BAC$, meets BC at D and the arc BEC at E. $\angle ECD={30}^o$ when EC is joined.
To find out- $\angle BAC=?$ Solution- We join BE. Now, BE, the chord of the given circle, subtends $\angle BAE\&\angle BCE$ to the circumference of the given circle at A & C respectively. So $\angle BAE=\angle BCE.......\left(i\right)$ (since angles, subtended by a chord of a circle to the circumference of the same circle at different points, are equal.)
But $\angle BAE=\angle EAC........\left(ii\right)$ since AE is the bisector of \angle BAC.
$\therefore$ From (i) & (ii) $\angle BCE=\angle BAE=\angle EAC={30}^o$.
$\therefore \angle BAC=\angle BAE+\angle EAC={30}^o+{30}^o={60}^o$.
Ans- Option D.