Geometry help please?

Question

Triangle ABC is isosceles with AB=AC and D is the midpoint of BC. The point K is on the line through C parallel to AB. K is not equal to C. The line KD intersects the sides AB and AC (or the extensions of the sides) at P and Q respectively. Show that (AP)(QK) = (BP)(QP)

Any help would be great! (Sorry, we weren't given a picture...)

Answer 1

It looks like you want to prove a rewritten form of that equality in which it says that two RATIOS are equal, where each of the ratios is in fact the ratio of lengths of two sides of a triangle, and you are able to prove that the two triangles are similar to each other.

Anyhow, that's the kind of proof I'd look for first.

Answer 2

"ΔABC is isosceles with AB=AC and D is the midpoint of BC"
=> ABD and ACD are congruent right-angled triangles with AB=AC and BD=CD, angle ABD=ACD

"The point K is on the line through C parallel to AB. K is not equal to C"
(Well you could draw a parallelogram ABCZ, and K,C,Z will be collinear - don't think that helps)

Anyway, P,D,K,Q are all collinear. Not sure whether you explicitly bring D in

"Show that (AP)(QK) = (BP)(QP)"
This suggests to me either

* AP/BP = QP/QK => similar triangles are involved. Try to spot which angles are supplementary.

or else :

* (AP)(QK) = (BP)(QP) => might suggest a circle is involved, but there is no simple relation in these four lengths, like intersecting chords or opposite sides of a cyclic quadrilateral.

Can't make much headway on it.
Below is a link to a quick review of all of Euclid's Elements, maybe one will ring a bell? Sorry can't help more.