In Fig. A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠BEC = 130° and ∠ECD = 20°. Find ∠BAC.

From figure,

BD is a straight line

⇒ ∠BEC + ∠DEC = 180° (Linear pairs)

⇒ 130° + ∠DEC = 180°

⇒ ∠DEC = 180° - 130°

⇒ ∠DEC = 50°.

In ∆ DEC,

⇒ ∠DEC + ∠EDC + ∠ECD = 180° (Angle sum property of a triangle)

⇒ 50° + ∠EDC + 20° = 180°

⇒ ∠EDC + 70° = 180°

⇒ ∠EDC = 180° - 70°

⇒ ∠EDC = 110°.

From figure,

⇒ ∠BDC = ∠EDC = 110°

We know that,

Angles in the same segment of a circle are equal.

⇒ ∠BAC = ∠BDC = 110°.

Hence, ∠BAC = 110°.