1987 OIM Problems/Problem 6

Problem

Let $ABCD$ be a planar convex quadrilateral, $P$ and $QQ$ are points of $AD$ and $BC$ respectively such that: \[\frac{AP}{PD}=\frac{AB}{DC}=\frac{BQ}{QC}\] Prove that the angles that are formed between line $PQ$ and lines $AB$ and $DC$ are equal.

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

This problem needs a solution. If you have a solution for it, please help us out by adding it.

See also

https://www.oma.org.ar/enunciados/ibe2.htm