# Measure of the Angles of the Cyclic Quadrilateral

We will prove that, in the figure ABCD is a cyclic quadrilateral and the tangent to the circle at A is the line XY. If ∠CAY : ∠CAX = 2 : 1 and AD bisects the angle CAX while AB bisects ∠CAY then find the measure of the angles of the cyclic quadrilateral. Also, prove that DB is a diameter of the circle.

Solution:

∠CAY + ∠CAX = 180° and ∠CAY : ∠CAX = 2 : 1.

Therefore, ∠CAY = $$\frac{2}{3}$$ × 180° = 120° and ∠CAX = $$\frac{1}{3}$$ × 180° = 60°.

As AD bisects ∠CAX, ∠DAX = ∠CAD = $$\frac{1}{2}$$ × 60° = 30°

As AB bisects ∠CAY, ∠YAB = ∠CAB = $$\frac{1}{2}$$ × 120° = 60°.

Now, ∠CAY = ∠ADC = 120° (Since, angle between tangent and chord is equal to the angle in the alternate segment).

Therefore, ∠CBA = 180° - ∠ADC = 180° - 120° = 60° (Since opposite angles of a cyclic quadrilateral are supplementary).

Again, ∠DAB = ∠DAC + ∠CAB = 30° + 60° = 90°.

Therefore, ∠BCD = 180° - ∠DAB = 180° - 90° = 90°.

We can see that the chord DB subtends a right angle at A.

Therefore, DB is a diameter of the circle (as an angle in a semicircle is a right-angle).

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