$EQ \cap BC = T$. $AE \cap BC = P'$ The condition that $EQ \parallel PX <=> \angle QTP = \angle QPT$, The condition that $EQ$ bisects $\angle AEP <=> $ $\angle BEP' = \angle CEP$. The tangent at the point $E$ to the circle $\odot(BEC)$ intersects the line $BC$ at the point $X$, $AX \cap \odot(ABC) = L$. $=> XE = XT => \angle XET = \angle XTE = \angle XPQ$ and $\angle XLQ = \angle ACQ = \angle ACB + \angle BCQ = \angle DQC + \angle BCQ = \angle XPQ => X,P,L,Q,E - $ they lie on the same circle. $=>$ After Inversion with the center at the point $X$ and the radius $EX$, $P$ and $P'$ are swapped. What was required.

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Let point $R$ on $\odot\Omega$ satisfy $AQ=AR$. Easy to see that $DR\parallel PX\parallel EQ$. Let’s be on the coordinate plane that the unit circle is $\odot\Omega$ and y-axis is parallel to $EQ$. Let $A(\cos\alpha,\sin\alpha)$ and $Q(\cos(\alpha+2\beta),\sin(\alpha+2\beta))$. Then $R(\cos(\alpha-2\beta),\sin(\alpha-2\beta))$, so $D(\cos(2\beta-\alpha),\sin(2\beta-\alpha))$. Since $AD\parallel BC$, let $B(\cos(\beta+\gamma),\sin(\beta+\gamma))$ and $C(\cos(\beta-\gamma),\sin(\beta-\gamma))$, and $t=\cos\gamma$. Then $BC:x\cos\beta+y\sin\beta=\cos\gamma=t$. Since $DQ:x\cos2\beta+y\sin2\beta=\cos\alpha$, we have that $P(2t\cos\beta-\cos\alpha,\frac{\cos\beta\cos\alpha-t\cos2\beta}{\sin\beta})$. Let $E(\cos(\alpha+2\beta),s)$, then $0=k_{EB}+k_{EC}=\frac{s-\sin(\beta+\gamma)}{\cos(\alpha+2\beta)-\cos(\beta+\gamma)}+\frac{s-\sin(\beta-\gamma)}{\cos(\alpha+2\beta)-\cos(\beta-\gamma)}$, that is $(s-\sin(\beta+\gamma))(\cos(\alpha+2\beta)-\cos(\beta-\gamma))+(s-\sin(\beta-\gamma))(\cos(\alpha+2\beta)-\cos(\beta+\gamma))=0$, that is $2s(\cos(\alpha+2\beta)-\cos\beta\cos\gamma)=2\cos(\alpha+2\beta)\sin\beta\cos\gamma-\sin2\beta$, so $s=\frac{\sin\beta(t\cos(\alpha+2\beta)-\cos\beta)}{\cos(\alpha+2\beta)-t\cos\beta}$. Then $y_E-y_P=s-\frac{\cos\beta\cos\alpha-t\cos2\beta}{\sin\beta}=\frac M{\sin\beta(\cos(\alpha+2\beta)-t\cos\beta)}$, here $M=\sin^2\beta(t\cos(\alpha+2\beta)-\cos\beta)-(\cos\beta\cos\alpha-t\cos2\beta)(\cos(\alpha+2\beta)-t\cos\beta)$ $=\cos\beta(\cos(\alpha+\beta)-t)(t\cos2\beta-\cos(\alpha+\beta))$. Since $x_E-x_P=\cos(\alpha+2\beta)-2t\cos\beta+\cos\alpha=2\cos\beta(\cos(\alpha+\beta)-t)$, so $k_{EP}=\frac{y_E-y_P}{x_E-x_P}=\frac{t\cos2\beta-\cos(\alpha+\beta)}{2\sin\beta(\cos(\alpha+2\beta)-t\cos\beta)}$. And $k_{EA}=\frac{s-\sin\alpha}{\cos(\alpha+2\beta)-\cos\alpha}=\frac{\frac{\sin\beta(t\cos(\alpha+2\beta)-\cos\beta)}{\cos(\alpha+2\beta)-t\cos\beta}-\sin\alpha}{-2\sin\beta\sin(\alpha+\beta)}$ $=\frac{\sin\beta(t\cos(\alpha+2\beta)-\cos\beta)-\sin\alpha(\cos(\alpha+2\beta)-t\cos\beta)}{-2\sin\beta\sin(\alpha+\beta)(\cos(\alpha+2\beta)-t\cos\beta)}=-\frac{t\cos2\beta-\cos(\alpha+\beta)}{2\sin\beta(\cos(\alpha+2\beta)-t\cos\beta)}$. So $k_{EP}+ k_{EA}=0$, which means that $EQ$ bisects $\angle AEP$. $\Box$

Note that if EQ bisects <BEC then EQ passes through the midpoint M of arc BC of (BEC). Let Q’ be on (ABCD) such that QQ’ || BC, and M’ be the reflection of M over QQ’; then D, M’, Q and A, M’, Q’ are collinear. Now let AP and EQ intersect at R, and A’, P’ be the projections of A, P onto EQ; then EQ is the external bisector of <AEP if and only if RA’ / RP’ = EA’ / EP’. The former equals RA / RP = QM’ / QP = QM / QS if S is the intersection of BC and EQ. Thus, it suffices to show that EA’ / EP’ = QM / QS. Note that if EQ intersects (ABCD) again at T, then by power of a point at S, ET / ES = QM / QS, so it remains to show that A’T / P’S = QM / QS, which is true since AA’T ~ PP’S and AT / PS = QQ’ / PS = QM / QS, done.

Let $\measuredangle$ denote directed angles modulo $180^\circ$. After some angle chasing, we can reduce the problem to the following. Restated problem wrote: In $\triangle QBC$, let $P$ be a point on $\overline{BC}$, and let the isogonal of $\overline{QP}$ in $\angle BQC$ intersect the circumcircle $\Omega$ of $QBC$ again at $A$. Prove that if $E$ lies on the reflection of $\overline{QP}$ over the $Q$-altitude of $QBC$, then $\measuredangle BEQ=\measuredangle QEC$ implies $\measuredangle PEQ=\measuredangle QEA$. Let $F$ be the image of $E$ under force-overlaid inversion at $Q$ with radius $\sqrt{QB \cdot QC}$. Let $M$ and $N$ be the midpoints of $\overline{BC}$ and $\overline{PA}$, respectively. By the first isogonality lemma, $\measuredangle BEQ=\measuredangle QEC$ if and only if the reflection of $F$ over $M$ lies on $\overline{QE}$, and $\measuredangle PEQ=\measuredangle QEA$ if and only if the reflection of $F$ over $N$ lies on $\overline{QE}$. Thus, it suffices to prove that $\overline{MN} \parallel \overline{QE}$. Let $P'$ and $A'$ be the reflections of $P$ and $A$ over the perpendicular bisector of $\overline{BC}$. Notice that $Q$, $P$, and $A'$ are collinear. We know that $\overline{MN} \parallel \overline{P'A}$ and both $\overline{QE}$ and $\overline{P'A}$ are reflections of $\overline{QA'}$ over a line perpendicular to $\overline{BC}$, so $\overline{MN} \parallel \overline{QE}$, as desired.

Transform the problem using pole-polar duality at $Q$ and let $Z_m$ denote the pole of $m$. Details are omitted but the diagram looks like: where $Z_{EP}Z_{BCP}$ is tangent to $\omega = (Z_{BCP}Z_{AB}QZ_{AC})$, and $\ell = Z_{EP}Z_{BE}Z_{CE}Z_{AE}$ is parallel to the tangent to $\omega$ at $Q$. We want to show that if $\angle Z_{BE}Z_{CE}Q = \angle Z_{CE}Z_{BE}Q$ then $\angle Z_{EP}Z_{AE}Q = \angle Z_{AE}Z_{EP}Q$ as well. This is immediate by DIT on points $Z_{BCP},Z_{BCP},Z_{AB},Z_{AC}$ and line $\ell$.