In the given figure, BD=DC and ∠CBD = 40°. Find ∠BAC.

In ∆BDC, BD = DC [given]

∴ ∠DCB =∠CBD = 40…(i)

[∵ angles opposite to equal sides are equal]

Also, ∠CBD +∠BDC + ∠DCB = 180°

[∵sum of angles of a triangle is 180°]

∴ 40° +∠BDC + 40° = 180° [from Eq. (i)]

=> ∠BDC = 180°- 80° = 100°

As A, B, D and C are four points on a circle.

Therefore, ABDC forms a cyclic quadrilateral.

Hence, ∠A +∠D = 180°

[∵ in a cyclic quadrilateral, sum of two opposite angles is equal to two right angles, i.e. 180°]

=> ∠A + 100° = 180° [ ∠D = ∠BDC = 100°]

∴ ∠A = 80°