Point P inside parallelogram to give certain properties

Let a point

P 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

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Re: Point P inside parallelogram to give certain properties

<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

Re: Point P inside parallelogram to give certain properties

So do i just use pythagoras' theorem and prove that square AEBP and PBFC are both say 2x^2, so 4x^2 = ABCD which is 4x^2?

Re: Point P inside parallelogram to give certain properties

Quote:

Originally Posted by

**BERMES39** <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

What about the cases in which <APB $\displaystyle \neq$ <CPD ?

Re: Point P inside parallelogram to give certain properties

I dont know, wouldn't that mean that <APB could be anything so long as its not degenerate? And same with the other? This prospect has confused me, a lot.

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Re: Point P inside parallelogram to give certain properties

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Originally Posted by

**pikachu26134** Let a point P inside a parallelogram ABCD be given such that <APB +<CPD = 180º. 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.

Re: Point P inside parallelogram to give certain properties

Okay, ive got the quadrilateral, PP'xAB = PA x P'B + AP' x PB using ptolemy's theorem. What do i do from there?

Re: Point P inside parallelogram to give certain properties

Quote:

Originally Posted by

**pikachu26134** Okay, ive got the quadrilateral, PP'xAB = PA x P'B + AP' x PB using ptolemy's theorem. What do i do from there?

Well I already mentioned that PP' = AD (that's because DAP'P is a parallelogram). Also, P'B = PC and P'A = PD (that's because the upper parallelogram is an exact copy of the lower one).