# 1987 IberoAmerican Mathematical Olympiad, Problem 6

Let ABCD be a plain convex quadrilateral. P, Q are points of AD and BC respectively such that

	 \frac{AP}{PD} = \frac{AB}{DC} = \frac{BQ}{QC}.


Show that the angles that are formed by the lines PQ with AB and CD are equal.

Solution by Steve Dinh, a.k.a. Vo Duc Dien (dedicated to the lovely Quynh-Diep Katrina Ngo)

From P draw a line || to DC and intercept AC at I. Link IQ. We have IQ || AB. We then have ∠QEC = ∠QPI and ∠QFB = ∠PQI.

To prove that the angles that are formed by the lines PQ with AB and CD are equal, we then need to prove ∠QPI = ∠PQI,

or IP = IQ.

From I draw a line parallel to AD and intercept DC at J. We have

\frac{IQ}{AB} = \frac{IC}{AC} = \frac{JC}{DC},

or

\frac{AB}{DC} = \frac{IQ}{JC}

We also have \frac{AB}{DC} = \frac{AP}{PD} = \frac{IP}{JC};

Therefore, IP = IQ.