Let $\angle {DOC}=2x$ Now $\frac {PD}{AD}=\frac {Sin (x+A)}{Cos x}$ $\frac {DQ}{DC}=\frac {Sin (A+x}{Cos (B-x)}$ So finally we've $\frac {DQ}{PD}=\frac {Sin (2x)}{Sin (2B-2x)}$ So now easily we'll get $\angle {DPQ}=2x$

it's easy to prove that $DQ$ and perp from $A$ to $BC$ meet on $(O)$(in $X$) and also that $DP$ and perp from $C$ to $AB$ meet on $(O)$(in $Y$) now by Pascal's theorem $P,Q,H$ are collinear where $H$ is the orthocenter of $\triangle ABC$ now finishing by fact that $H,Y$ are symmetric wrt $AB$ $\angle DPH=2*\angle DYC=\angle DOC$

If you look previous posts, I have done much work . But it depends only angle calculations. Anyway, Let $\angle OCB = \alpha$, $\angle OAB = \beta$, $\angle DBC = \theta$. $\angle ADP = \alpha$, $\angle QDC = \beta$, $\angle DAC=\theta$, $\angle BAC = \angle BDC = 90^\circ - \alpha$, $\angle BCA = \angle ADB = 90-\beta$, $\angle DBA = \alpha + \beta -\theta = \angle ACD$. In $\triangle APD$, $\angle ADP + \angle PAC = 90^\circ$, so $\angle APD = 90^\circ - \angle CAD = 90^\circ - \theta$. Similarly, $\angle DQC = 90^\circ - (\alpha + \beta - \theta)$. Let the perpendicular line to $BC$ at $Q$ meet $BD$ and $AB$ at $X$ and $Y$, respectively. Since $\angle YQD=90^\circ - \angle DQC = \alpha +\beta -\theta = \angle YBD$, then $Y,B,Q,D$ are concyclic. Also, $\angle YXD = \angle BXQ = 90^\circ \angle XBQ = 90^\circ - \theta$ and $\angle YXD = \angle APD = \angle YPD = 90^\circ - \theta$. So $Y,P,X,D$ are concylic. In that case, $\angle YQB = \angle YDV = 90^\circ$, and $\angle YDX = \angle YPX = 90^\circ$. Since $\angle XPB + \angle BQX = 180^\circ$, $BQXP$ is cyclic. So $\angle XPQ = \angle XPQ = \theta$. We have found $\angle APD = 90^\circ - \theta$. So $\angle DPX = 90^\circ - (90^\circ - \theta) = \theta$. And last, $\angle DPQ = 2\theta = \angle DOC$. $\blacksquare$ Note: We can conclude that $X$ is the incenter of $\triangle DPQ$.

Construction of the points, $P, Q$: Take $B'$ the reflection of $B$ in the perpendicular bisector of $AC$, $B'B"$ a diameter of the circle $\odot (ABC)$, $O_a$ the center of the circle passing through $D$ and being tangent to $AB'$ at $A$, $O_c$ the center of the circle passing through $D$ and being tangent to $CB'$ at $C$. Circle $O_a$ intersects $OA$ at $X$, $AB\cap DX=P$, Circle $O_c$ intersects $OC$ at $Y$, $BC\cap DY=Q$. By angle calculation, $\angle XOY+\angle XDY=180^\circ$, so $OYDX$ is cyclic and $\angle DXY=\angle DOY=\angle DOC$, so we now need to prove $XY\parallel PQ$. By angle chasing we get $\Delta CQY\sim\Delta DAX$ and $\Delta PAX\sim\Delta CDY$. From the above triangles similarities we get $\frac{QY}{AX}=\frac{CY}{DX}$ and $\frac{PX}{CY}=\frac{AX}{DY}$. Divide side by side the last two equalities, and get $\frac{QY}{PX}=\frac{DY}{DX}$, showing that, indeed, $XY\parallel PQ$, done. Best regards, sunken rock

My solution is same as leader's solution.