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Ukrainian National Mathematical Olympiad - Fourth Round

Geometry Difficulty 5.1 AIME, harder Prove it Ukraine

In the trapezoid $ABCD$ with perpendicular diagonals, points $P$ , $N$ , $Q$ , $M$ are the middles of sides $AB$ , $BC$ , $CD$ , $DA$ respectively. On the base $CD$ there is a point $L$ (different from the point $Q$ ) for which the angle $MLN$ is straight. Find the angle $LPA$ .

(Bogdan Rublyov)

Solution

By Varignon's Theorem quadrangle $MPNQ$ is a parallelogram and its sides are parallel to the diagonals of trapezoid $ABCD$ , so it is a rectangle (Fig. 38). Denote $O$ the intersection point of $AC$ and $BD$ . Then from the properties of rectangular $\triangle MNL$ we get that: $OL = OM = ON \Rightarrow OL = OP = OQ$ , so $\triangle PQL$ is rectangular. That means $PL \perp LQ$ , and since $AB \parallel CD$ then $PL \perp AB$ .

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