Point P inside parallelogram to give certain properties

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August 5th 2011, 03:11 PM
pikachu26134

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
August 5th 2011, 05:23 PM
BERMES39

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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
August 6th 2011, 12:52 AM
pikachu26134

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?
August 6th 2011, 01:11 AM
Also sprach Zarathustra

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 $\neq$ <CPD ?
August 6th 2011, 01:20 AM
pikachu26134

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.
August 7th 2011, 12:10 AM
Opalg

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Re: Point P inside parallelogram to give certain properties

Quote:

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.
August 7th 2011, 10:42 PM
pikachu26134

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?
August 7th 2011, 11:26 PM
Opalg

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).