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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