Let $\angle BAC = \angle BCA = \angle DAC = \angle DCA = a$ Let $\angle PAB = x , \angle QCB = y$ $\angle DPQ = \angle APQ - \angle APD = (180-a-x) - (90-a- \frac{x}{2})=90- \frac{x}{2}$ $\angle DBQ = \angle DBC + \angle CBQ = (90-a) + (90- \frac{y}{2} = 180-a- \frac{y}{2}$ Therefore, $\angle DBQ + \angle DPQ = 270-a - \frac{x}{2} - \frac{y}{2}$ $\angle PAB = a+x , \angle QCA = a+y , APQC$ is a parallelogram $\implies 2a+x+y = 180 \implies a+ \frac{x}{2} + \frac{y}{2} = 90^\circ$ Hence, $\angle DBQ + \angle DPQ = 270-90=180^\circ \implies \angle DBQ = 180- \angle DPQ$ Hence, $B$ lies on the first locus of orthocenter of $\triangle DPQ$........(In $\triangle ABC$, if $H$ is orthocenter, then $\angle BHC = 180-\angle BAC$) Similarly, $\angle PBD = 180 - \angle PQD$ Hence, $B$ lies on the second locus of orthocenter of $\triangle DPQ$ Hence, $B$ is the orthocenter of $\triangle DPQ$