$ABCD$ is a quadrilateral inscribed in a circle with centre O. If $\angle COD = 120$ degrees and $\angle BAC = 30$ degrees, then the value of $\angle BCD$ (in degrees) is

Given Information:

ABCD is a quadrilateral inscribed in a circle with center O

$\angle COD = 120°$

$\angle BAC = 30°$

Find: $\angle BCD$

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Since ABCD is inscribed in a circle, all four vertices lie on the circle. This makes ABCD a cyclic quadrilateral.

The key insight is that $\angle COD$ is a central angle (formed by two radii from center O), while $\angle BAC$ is an inscribed angle (formed by two chords meeting at a point on the circle).

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The Inscribed Angle Theorem states: An inscribed angle is half the central angle that subtends the same arc.

Since $\angle COD = 120°$ is the central angle subtending arc CD, the inscribed angle $\angle CAD$ (which also subtends arc CD) is:

$\angle CAD = \tfrac{1}{2} \times \angle COD = \tfrac{1}{2} \times 120° = 60°$

Note: We write $\angle CAD$ instead of $\angle DAC$ to follow the standard notation where the vertex of the angle is in the middle.

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We know:

$\angle CAD = 60°$ (from above)

$\angle BAC = 30°$ (given)

Therefore: $\angle BAD = \angle BAC + \angle CAD = 30° + 60° = 90°$

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In a cyclic quadrilateral, opposite angles are supplementary (they add up to 180°).

Since $\angle BAD$ and $\angle BCD$ are opposite angles in cyclic quadrilateral ABCD:

$\angle BAD + \angle BCD = 180°$

$90° + \angle BCD = 180°$

$\angle BCD = 180° - 90° = 90°$

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The beauty of this solution lies in two fundamental circle theorems:

The Inscribed Angle Theorem: Connects central and inscribed angles

The Cyclic Quadrilateral Property: Opposite angles are supplementary

These theorems work together to give us the answer efficiently.

Therefore, $\angle BCD = 90°$