Oh wow, what a problem, its just angle chase but unless u have a fixed plan to solve this its hard to find motivation for the following solution: Let $AC \cap (AHT)=J$, first note that $M$ is center of $(CDH)$ so we have $\angle CPQ=\angle MDC=\angle MCD$ so $PQ=QC$, now notice that $\angle AJP=180-\angle AHP=180-\angle ABC=\angle TAC$ so we get $PJ \parallel TA$ and also since $\angle PQC=180-2\angle HCD=2\angle ABC$ we get in fact that $Q$ is the center of $(JPC)$ so $QJ=QP$ but since $AT \parallel PJ$ we get $ATPJ$ Cyclic Isosceles Trapezoid, so by symetry we also get $QA=QT$ as desired thus we are done .