Question

In the given figure, \[\Delta ABC\] is inscribed in a circle. The bisector of \[\angle BAC\] meets BC at D and the circle at E. If EC is joined then \[\angle ECD={{30}^{\circ }}\]. The value of \[\angle BAC\] is:
A. \[{{30}^{\circ }}\]
B. \[{{40}^{\circ }}\]
C. \[{{50}^{\circ }}\]
D. \[{{60}^{\circ }}\]\[\]

Answer 1

Hint: The angles, subtended by a chord of a circle to the circumference of the same circle, are equal. Thus angle A is bisected so \[\angle BAE\] is equal to \[\angle EAC\]. Thus find \[\angle BAC\].

Complete step-by-step answer:

We have been given a triangle ABC which is inscribed in a circle.AE is the bisector of \[\angle BAC\] and it meets BC at D and the arc BEC at E.

We have been given \[\angle ECD={{30}^{\circ }}\], by joining EC.

Let us join BE.

Now BE is the chord of the given circle.

The angle subtended by a chord in a circle, to the circumference of the same circle at different points are equal.

Thus the chord of the given circle subtends equal angles at \[\angle BAE\] and \[\angle BCE\].

Thus we can say that,

\[\angle BAE=\angle BCE\].

We have been told that AE is the bisector of angle A.

A bisector divides the angle into 2 equal halves.

\[\therefore \angle BAE=\angle EAC\]

Thus we got that \[\angle BAE=\angle BCE\] and \[\angle BAE=\angle EAC\].

Thus we can say that \[\angle BCE=\angle EAC\].

We know the value of \[\angle BCE={{30}^{\circ }}\].

\[\therefore \angle BCE=\angle EAC={{30}^{\circ }}\].

We need to find \[\angle BAC\]. AE is a bisector, so

\[\angle A=\angle BAE+\angle EAC\].

We have got \[\angle BAE={{30}^{\circ }}\] and \[\angle EAC={{30}^{\circ }}\].

\[\therefore \angle A=\angle BAE+\angle EAC={{30}^{\circ }}+{{30}^{\circ }}={{60}^{\circ }}\].

Thus we got \[\angle BAC={{60}^{\circ }}\].

Option D is the correct answer.

Note: We can only solve this problem by using the theorem of the angle subtended by a chord. You should also remember that an angle bisector divides the angle into 2 equal halves.