In the given figure AB = AC and ∠ACD = 105°, then ∠BAC is equal to

This uses two facts together: angles on a straight line form a linear pair and sum to 180°, and in an isosceles triangle the angles opposite the two equal sides are equal, with all three interior angles of any triangle summing to 180°. Equivalently, the exterior angle theorem says an exterior angle equals the sum of the two non-adjacent interior angles.

1. ∠ACD and ∠ACB lie on the straight line BD, so they form a linear pair: ∠ACB = 180° − 105° = 75°.

2. Since AB = AC, triangle ABC is isosceles, and the angles opposite these equal sides are equal: ∠ABC = ∠ACB = 75°.

3. The angles of triangle ABC sum to 180°, so ∠BAC = 180° − (∠ABC + ∠ACB) = 180° − 150° = 30°.

By the exterior angle theorem, ∠ACD = ∠ABC + ∠BAC, i.e., 105° = 75° + ∠BAC, which again gives ∠BAC = 30° — confirming the result.

So ∠BAC = 30°.