Ankoganit wrote:

Triangle $ABC$ has orthocenter $H$ . Let $D$ be a point distinct from the vertices on the circumcircle of $ABC$ . Suppose that circle $BHD$ meets $AB$ at $P\ne B$ , and circle $CHD$ meets $AC$ at $Q\ne C$ . Prove that as $D$ moves on the circumcircle, the reflection of $D$ across line $PQ$ also moves on a fixed circle.

Michael Ren

I prove a more general result.

Now we prove that the following result.

$\textbf{Lemma}:$ $T$ is a random point inside triangle $\triangle ABC$ and the circumcircle of $\triangle BTC$ intersects line $AB$ at a second point $D$ . $C'$ is the reflection of $C$ wrt line $DT$ . now as $C$ traverses the circumcircle of $\triangle ABC$ then $C'$ is moving on a circle.

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(5.533356307592783cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(7); defaultpen(dps); /* default pen style */ real xmin = -2.5967225715269686, xmax = 2.9366337360658137, ymin = -1.4500901983026113, ymax = 1.497749970650213; /* image dimensions */ pen qqffff = rgb(0.,1.,1.); pair A = (-0.38182256002963183,0.9242356477935798), B = (-0.8712073830836633,-0.49091516136753727), T = (-0.4884619389510197,0.0716836776184096), C = (0.8243241283679402,-0.5661181249442873), D = (-0.7910498768987462,-0.25912423857179945), G = (0.1367007089328688,0.9906123945203044), O = (0.,0.); filldraw((-1.2404832561853285,1.3225483131102684)--A--C--cycle, qqffff + opacity(0.10000000149011612), linewidth(0.) + qqffff); filldraw(C--O--T--cycle, qqffff + opacity(0.10000000149011612), linewidth(0.) + qqffff); /* draw figures */ draw(circle(O, 1.), linewidth(0.4)); draw(B--T, linewidth(0.4)); draw(A--B, linewidth(0.4)); draw(circle((-0.02893771651931373,-0.6524318709577974), 0.8576163703494034), linewidth(0.4)); draw(A--C, linewidth(0.4)); draw(B--C, linewidth(0.4)); draw(T--D, linewidth(0.4)); draw(C--T, linewidth(0.4)); draw(T--(-1.2404832561853285,1.3225483131102684), linewidth(0.4)); draw(T--G, linewidth(0.4)); draw(D--(-1.2404832561853285,1.3225483131102684), linewidth(0.4)); draw(A--G, linewidth(0.4)); draw(C--D, linewidth(0.4)); draw(C--(-1.2404832561853285,1.3225483131102684), linewidth(0.4)); draw(A--O, linewidth(0.4)); draw(C--O, linewidth(0.4)); draw(A--(-1.2404832561853285,1.3225483131102684), linewidth(0.4)); draw(O--T, linewidth(0.4)); draw(C--G, linewidth(0.4)); /* dots and labels */ dot(A,linewidth(4.pt)); label("$A$", (-0.37001275416763546,1.0052707061473545), NE * labelscalefactor); dot(B,linewidth(4.pt)); label("$B$", (-1.0454128883429783,-0.6023223215512621), NE * labelscalefactor); dot(T,linewidth(4.pt)); label("$T$", (-0.4438846438430636,-0.13094931124138326), NE * labelscalefactor); dot(C,linewidth(4.pt)); label("$C$", (0.9033979154754588,-0.6374994118728948), NE * labelscalefactor); dot(D,linewidth(4.pt)); label("$D$", (-0.9258107812494281,-0.2611045454314244), NE * labelscalefactor); dot((-1.2404832561853285,1.3225483131102684),linewidth(4.pt)); label("$C'$", (-1.3690421193019968,1.2128155390449877), NE * labelscalefactor); dot(G,linewidth(4.pt)); label("$G$", (0.19282069097848353,1.0650717596941301), NE * labelscalefactor); dot(O,linewidth(4.pt)); label("$O$", (0.013417530338158087,0.02734759520596411), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]$

$\textbf{proof:}$ note that $\triangle COT \sim \triangle CAC'$ ,on the other hand we have:
$\frac{OT}{R}=\frac{AC'}{AC}$ now since the ratio between $AC'$ and $AC$ is constant then $C'$ acts like $C$ and moves on a circle. $\blacksquare$

back to the problem,
Consider that $H$ is a random point in the problem. redefine the problem resp.
note that $P,H,Q$ are collinear now guess what! we throw away the point $C$ and $Q$ . now apply the lemma on triangle $\triangle ADC$ and we are done.