Problem 2086

Let $M G$ meet $\Gamma$ at $P$. Since $\angle M C D = \angle C A E$ and $\angle M D C = \angle C A E$, we have $M C = M D$. Thus
$$M D^{2} = M C^{2} = M G \cdot M P$$
and hence $M D$ is tangent to the circumcircle of $\triangle D G P$. Therefore $\angle D G P = \angle E D P$.
Let $\Gamma'$ be the circumcircle of $\triangle B D E$. If $B = P$, then, since $\angle B G D = \angle B D E$, the tangent lines of $\Gamma'$ and $\Gamma$ at $B$ should coincide, that is $\Gamma'$ is tangent to $\Gamma$ from inside. Let $B \neq P$. If $P$ lies in the same side of the line $B C$ as $G$, then we have
$$\angle E D P + \angle A B P = 180^{\circ}$$
because $\angle D G P + \angle A B P = 180^{\circ}$. That is, the quadrilateral $B P D E$ is cyclic, and hence $P$ is on the intersection of $\Gamma'$ with $\Gamma$.

Otherwise,
$$\angle E D P = \angle D G P = \angle A G P = \angle A B P = \angle E B P .$$
Therefore the quadrilateral $P B D E$ is cyclic, and hence $P$ again is on the intersection of $\Gamma'$ with $\Gamma$.
Similarly, if $L H$ meets $\Gamma$ at $Q$, we either have $Q = B$, in which case $\Gamma'$ is tangent to $\Gamma$ from inside, or $Q \neq B$. In the latter case, $Q$ is on the intersection of $\Gamma'$ with $\Gamma$. In either case, we have $P = Q$.