Singapore Mathematical Olympiad

Let the lines $BC$ and $AD$ intersect at $M$. The conditions $\angle A < 90^\circ$, $\angle B < 90^\circ$ and $AB > CD$ imply that $C$ lies between $MB$ and $D$ lies between $MA$. Given the triangles $APD$ and $BQC$ are similar, we have $\angle BCQ = \angle ADP$ so that $\angle MCQ = \angle MDP$. From this, we see that the triangles $MCQ$ and $MDP$ are similar. Thus $MC/MD = MQ/MP$. Using the given condition that the triangles $APD$ and $BQC$ are similar, we have $\angle PAD = \angle QBC$ so that $\angle PAM = \angle QBM$. It follows that the triangles $PAM$ and $QBM$ are similar. Thus $MQ/MP = MB/MA$. Therefore, $MB/MA = MC/MD$. This implies that $AB$ is parallel to $CD$.