In a trapezium $$PQRS$$, the side $$PQ$$ is perpendicular to the parallel sides $$QR$$, $$PS$$ whose lengths are $$x$$ and $$y$$. If the diagonals $$PR$$ and $$QS$$ intersect in $$M$$, then the perpendicular distance from $$M$$ to the side $$PQ$$ is ________

PLACEHOLDER ANSWER - the answer is an expression in $$x$$ and $$y$$. Triangles $$MQR$$ and $$MSP$$ are similar, so $$\frac{QM}{MS} = \frac{x}{y}$$ and $$\frac{QM}{QS} = \frac{x}{x + y}$$. The distance from a point on $$QS$$ to $$PQ$$ grows linearly from 0 at $$Q$$ to $$y$$ at $$S$$. So the distance from $$M$$ to $$PQ$$ is $$\frac{x}{x + y} \cdot y = \frac{xy}{x + y}$$.