In the diagram above. O is the centre of the circle and |BD| = |DC|. If < DCB = 35º, find < BAO
∠DCB = ∠CBD = 35º (base angles of an isosceles triangle are equal)
∠ADB = 35º + 35º = 70° (ext. angle of a triangle equals two opp. interior angles)
∠ABD = 90º (angle in a semicircle is a right angle)
∠BAO + ∠ADB + ∠ADB = 180º (sum of angles in a triangle)
∠BAO + 90º + 70º = 180º
∠BAO + 160º = 180º
∠BAO = 180º - 160º
∠BAO = 20º
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