Prove Ab x AD = BP x DP + AP x CP

Let a point P inside a parallelogram ABCD be given such that ∠AP B + ∠CP D = 180◦. Prove that

AB · AD = BP · DP + AP · CP (here AB, AD etc. refer to the lengths of the corresponding segments)

Need a pointe as to what methods i should use, just some help would be very nice, im an 11th grade level mathematician so i would understand some concepts but not so much others - again, help on these would be great.

Cheers

Re: Prove Ab x AD = BP x DP + AP x CP

Quote:

Originally Posted by

**pikachu26134** Let a point P inside a parallelogram ABCD be given such that ∠AP B + ∠CP D = 180◦. Prove that

AB · AD = BP · DP + AP · CP (here AB, AD etc. refer to the lengths of the corresponding segments)

Need a pointe as to what methods i should use, just some help would be very nice, im an 11th grade level mathematician so i would understand some concepts but not so much others - again, help on these would be great.

Cheers

From D draw a line parallel to AP, and from C draw a line parallel to BP. let these two lines intersect at E. then PDEC is cyclic quadrilateral(why?).

PE=AD(why?)

apply ptolemy's theorem to cyclic quad PDEC.