Question

A circle contains vertex A of parallelogram ABCD and intersects AB, AC, and AD at P, Q, and R, respectively. Prove AP•AB + AR•AD = AQ•AC.

Answer 1

I'm supposed to use Ptolemy's law, which I've practiced enough today to recite by memory, but I don't know how to apply it when a corner of the quadrilateral is in the center of the circle and the parallelogram lies outside. Am I supposed to show that APQR is a cyclic quadrilateral and then prove the identity?

Answer 2

**Ptolemy's theorem, sorry.

Answer 3

Can you please draw a diagram?

Answer 4

I'm not at liberty right now to make a diagram, but the general picture that I can see forming from the problem statement is a standard parallelogram ABCD with a circle about A that intersects the three lines that make up the quad.

Answer 5

I'm not sure how else I can interpret the instructions.

Answer 6

I was just wondering if the parallelogram was circumscribed by the circle?

Answer 7

On the problem set, all of the other problems that had a situation like that were specifically stated to mention that they were cyclic quadrilaterals, but this one isn't. What I described was my best interpretation of the problem statement, but I did think the same thing that you did when I first saw the problem. I suppose it MIGHT be possible for the parallelogram to be completely circumscribed by the circle, but then it would have to be a square...

Answer 8

Woops, no, it can't be, under any circumstances -- A is the vertex of the circle! :/ This is probably the trickiest one I've come across today. I've got another one, but I'll post it later..

Answer 9

Oh wow this is more difficult than I though it would be Don't know if I can help. Sorry!

Answer 10

Hm...do you know Ptolemy's theorem? Maybe if you didn't before, you could pick up on something key that I've missed? :P

Answer 11

Ptolemy's theorem: If a quadrilateral ABCD is cyclic (inscribed by all 4 in a circle), then AC•BD = AB•CD + AD•BC.
The basic form of the theorem reflects what I'm supposed to prove, but I can't see how to incorporate the shorter segments...