IMO 1994- Problem 2

Problem: $ABC$ is an isosceles triangle with $AB=AC$. Suppose that:
(i) $M$ is the midpoint of $BC$ and $O$ is the point on the line $AM$ such that $OB$ is perpendicular to $AB$.
(ii) $Q$ is an arbitrary point on the segment $BC$ different from $B$ and $C$.
(iii) $E$ lies on the line $AB$ and $F$ lies on the line $AC$ such that $E,Q$ and $F$ are distinct and collinear.
Prove that $OQ$ is perpendicular to $EF$ if and only if $QE=QF$.
Solution: First assume that $OQ$ is perpendicular to $EF$. Now $OEBQ$ and $OCFQ$ are cyclic.
Hence $\angle{OEQ}=\angle{OBQ}=\angle{OCQ}=\angle{OFQ}$. It follows that $QE=QF$.
Suppose now that $QE=QF$ and that the perpendicular through $O$ to $EF$ meet $BC$ at $Q'\ne Q$. Draw the line through $Q'$ parallel to $EF$, meeting the lines $AB$ and $AC$ at $E'$ and $F'$,respectively. Then $Q'E'=Q'F'$ as before. Let $AQ'$ meet $EF$ at $N$. Then $N\ne Q$ and $NE=NF$, so that $QE\ne QF$, a contradiction. So $Q'=Q$.
Source: http://sms.math.nus.edu.sg/simo/IMO_Problems/94.pdf