Irish Mathematical Olympiad

Let $O$ be the centre of the circle $T$. Extend $PB$ to meet the circle at $C$. Let $\angle QPA=\alpha$ and let $\angle BPO=\beta$. Since $|AO|=|OB|$ the join of the two centres is perpendicular to $AB$. The join of the two centres is also perpendicular to $PQ$. Thus $PQ\parallel AB$. Hence $\angle BAP=\alpha$. Hence $|PB|=|QA|$.

Since $|AO|=|OB|$, $\angle APO=\angle BPO=\beta$. Hence $|PC|=|PD|$. $\angle PCD=\pi /2-\beta$, $\angle DQP=\pi /2+\beta$. Also $\angle PDC=\angle PCD=\pi /2-\beta$.

As $QDCP$ is a cyclic quadrilateral $$\angle QDP=\pi =(\alpha +2\beta +\pi /2-\beta )=\pi /2-(\alpha +\beta )$$ Also $\angle AQP=\angle BPQ=\alpha +2\beta$. Then since $\angle DQP+\angle DCP=\pi$ we get $$\angle DQA=\pi -(\pi /2-\beta +\alpha +2\beta =\pi /2-(\alpha +\beta )=\angle QDP)$$ Hence $|AD|=|QA|=|PB|$.