nAalniaOMliO wrote: Point $E$ lies on the arc $AC$ of circle $\Omega$ such that $\angle ABD=\angle CBE$. Arc $AC$ that contains $B$ or that does not contain $B$?

Hopefully not a fakesolve The main idea is reflecting the orthocenter lemma. Let $BD$ intersect $\Omega$ at $Z\ne B$. Since $E$ lies on arc $AC$ that does not contain $B$, we have $ZE\parallel AC$. Let the line perpendicular to $PQ$ through $E$ intersect $AC$ at $X$ and $\Omega$ at $Y\ne E$. Hence, $BZ\parallel EY$. Since $DXEZ$ is parallelogram, observe that $DX=ZE=BY$, So $BDXY$ is an isoceles trapezoid. Hence, $X$ and $Y$ are symmetric w.r.t $PQ$. Hence, $\measuredangle PXQ = - \measuredangle PYQ = -\measuredangle PEQ = \measuredangle QEP$. Since $EX\perp PQ$ and $\measuredangle PXQ = \measuredangle QEP$, thus $X$ is the orthocenter of $\Delta PQE$. Hence, the orthocenter of $\Delta PQE$ lies on $AC$, proven $\blacksquare$

GreenTea2593 wrote: nAalniaOMliO wrote: Point $E$ lies on the arc $AC$ of circle $\Omega$ such that $\angle ABD=\angle CBE$. Arc $AC$ that contains $B$ or that does not contain $B$? Thank you! Edited the problem statement. GreenTea2593 wrote: Hopefully not a fakesolve I think your solution is correct

For any point $ P $ on $ \odot (ABC) $, let $ \mathcal{S}_{P} $ denote the Steiner line of $ P $ WRT $ \triangle EPQ $. Let $ H $ be the orthocenter of $ \triangle EPQ $ and let $ AH, BD$ intersect again with $ \odot (ABC) $ at $ C', X$, respectively. It is sufficient to show that $ \measuredangle ABD = \measuredangle EBC'$. Note that $ \mathcal{S}_{A} \parallel BC', \mathcal{S}_{X} \parallel BE$, so we conclude that $ \measuredangle ABD = \measuredangle (\mathcal{S}_{X}, \mathcal{S}_{A}) = \measuredangle EBC'$. $\blacksquare$