Question

the circle which passes through the vertices A and C of a parallelogram OABC has center at O. if BA produced meets the circle at D, prove that : angle ABC= 2 times angle OCD

Answer 1

Correct question:-

The circle passing through the vertices A, B and C of a parallelogram ABCD intersects side CD at a point P as shown in the figure. Prove that ∠APD = ∠ADP.

Answer:-

Question is to prove that ∠APD = ∠ADP

ABCD is a parallelogram.

=>Opposite angles are same.

=>∠ABC = ∠ADC

=>∠BAD = ∠BCD

A 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