<APB = <CPD so parlallelogram ABCD is a square.
AB*AD = area of the square
BP*PD = AP*CD = area of either small square PBFC or AEBP
Let a pointP inside a parallelogram ABCD be given such that <APB +<CPD = 180◦.
Prove that
AB· AD = BP · DP + AP · CP
(here AB, AD etc. refer to the lengths of the corresponding segments).
I think that P could be in the very centre, but am unsure how to go about the question, and initially unsure about a possible position for 'p'.
Help would be great
With the parallelogram ABCD and the point P as in the diagram below, draw an identical parallelogram on top of the given one, with a point P' in the same relative position as P. Then PP'= AD. The angles CPD and AP'B are equal. Thus the angles APB and AP'B are supplementary. Hence PAP'B is a cyclic quadrilateral. The result then follows from Ptolemy's theorem.