In the given figure, BD = DC and ∠CBD = 30°, find m(∠BAC).

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Solution

BD = DC
⇒ ∠BCD = ∠CBD = 30°
In ΔBCD, we have:
∠BCD + ∠CBD + ∠CDB = 180° (Angle sum property of a triangle)
⇒ 30° + 30° + ∠CDB = 180°
⇒ ∠CDB = (180° – 60°) = 120°
The opposite angles of a cyclic quadrilateral are supplementary.
Thus, ∠CDB + ∠BAC = 180°
⇒ 120° + ∠BAC = 180°
⇒ ∠BAC = (180° – 120°) = 60°
∴ ∠BAC = 60°

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