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The circle which passes through the vertices a and c of a parallelogram OABC has a centre at O if BA produced meets the circle at D then prove that ABC=2 times of angle OCD
Question is to prove that ∠APD = ∠ADPSee the attached picture to represent the problem. ABCD is a parallelogram. =>Opposite angles are same. =>∠ABC = ∠ADC=>∠BAD = ∠BCDA circle is drawn in such a way that it passes through A, B and C. P is a point on circle intersecting with line CD.ABCP is cyclic quadrilateral. (Circle is passing through all four vertices. A, B, C and P).From the property of Cyclic Quadrilateral, ∠ABC + ∠APC = 180°.(Opposite angles are supplementary)∠APC + ∠APD = 180°. (Angle on straight line)From above 2, we can derive ∠ABC = ∠APD. ∠ABC = ∠ADP = ∠APD. ∠ADP = ∠APD.
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