From Kiselev / Givental:
From the intersection point of the diagonals of a rhombus perpendiculars are dropped to the sides of the rhombus. Prove that the feet of these perpendiculars are vertices of a rectangle.
My solution is below, to which I request verification, improvements, or feedback. The proof is surprisingly (at least to me) simple.
Let $PQRS$ be a rhombus with center $O$. From $O$, drop perpendiculars to $A,B,C,D$ respectively.
Since $PQRS$ is a rhombus, then $\triangle OPQ, \triangle OQR, \triangle ORS, \triangle OSP$ are congruent right triangles, whose altitudes are congruent. But $OA,OB,OC,OD$ are simply the altitudes of those triangles and therefore congruent. [1] Thus, $AC$ and $BD$ are congruent and mutually bisecting and therefore the diagonals of rectangle $ ABCD$.
Update [1] (As suggested by Calvin Lin): Furthermore, note that $OA$ and $OC$ are colinear, since they share point $O$ and are perpendicular to parallel lines $PQ$ and $RS$. Likewise, $OB$ and $OD$ are colinear.