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 1

Step-by-step explanation:

Question is to prove that ∠APD = ∠ADP

See the attached picture to represent the problem.  

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.