Let $BX \cap AY=W$. Since $MA=MC$, $\angle WAT=\angle ACM=\angle MAT$, so $AT$ bisects $\angle WAB$; similarly, $BT$ bisects $\angle WBA$. Thus, $T$ is the incenter of $\triangle WAB$ and, moreover, $\triangle MXY$ is the medial triangle of $\triangle WAB$ since $XY=\frac{AB} {2}$ (from now on, delete $C, D$ and the semi-circle and focus on $\triangle WAB$). Thus, due to homothety at the centroid of $\triangle WAB$, if $N$ is the midpoint of $XY$, then $IN \parallel MT$, which means that $E$ is the $M$-excircle touchpoint with $XY$ in $\triangle MXY$. If $MH$ is the altitude of $\triangle MXY$, then $IE$ bisects $MH$, so $HEMP$ is a parallelogram and $HEQM$ is a rectangle. This means that $MP=EH=MQ$, as desired.

Let $U,V$ be the intersections of $AD,BC$ with $XY$. $MA=MD$ implies $DY=YU$, similarly $XV=VC$. Then the perpendicular bisectors of $DU,VC,DC$ all pass through $I$ and hence $DUVC$ is cyclic with center $I$. Let this circle intersect $AC$ at $K$ and $BD$ at $L$. A quick angle chasing shows $UK, VL \perp XY$. Then $UVLK$ is a rectangle. Also $EM$ passes through the midpoint of $KL$, say $R$, and let the midpoint of $UV$ be $S$. Projecting $-1=(R,S,I,RS_{\infty})$ from $T$ on $AB$ gives the desired.

We will start with well known lemma. Denote by $I$ and $E$ the incenter of $ABC$ and tangency point of $A$-excircle with $BC$, respectively.Let $D$ be point on $BC$ such that $AD \perp BC$. Then the line $EI$ passes through the midpoint of $AD$. From the following lemma we want to prove that $E$ is the tangency point of triangle $MXY$. We can change the problem Let $ABC$ be a triangle with incenter $I$. Let points P and Q on the ray $AB$ and $AC$ respectively such that $AP=AQ=BC$. The lines through $P$ and $Q$ parallel to $BI$ and $CI$ intersect at point $T$. And $E$ is the tangency point of $A$-excircle with $BC$. Then we will prove that $A$, $T$, $E$ collinear. Let $D$ and $F$ are other tangency point, then $ED \parallel PT$ and $EF \parallel QT$ and from $AD=AF$ and $AP=AQ$ we get that triangles $EDF$ and $TPQ$ are homothetic so $A$, $T$, $E$ are collinear. Done!

Returning back to this problem after some time, and solved it in 20 mins. Kinda sad I didn't do it in contest, and thus sweep, but oh well. Claim: $T$ is the $M$-excircle touchpoint of $\triangle MXY$. Proof. By ratio lemma spamming, we have: \[ \frac{YT}{TX} = \frac{MY}{MX} \cdot \frac{\sin{\angle DME}}{\sin{\angle XME}} = \frac{\sin{\angle YXM}}{\sin{\angle XYM}} \cdot \frac{\frac{\sin{\angle DME}}{DE}}{\frac{\sin{\angle XME}}{EC}} \cdot \frac{DE}{EC} = \]\[ = \frac{\sin{\angle YXM}}{\sin{\angle XYM}} \cdot \frac{\frac{\sin{\angle BEM}}{DM}}{\frac{\sin{\angle AEM}}{MC}} \cdot \frac{DE}{EC} = \frac{\sin{\angle YXM}}{\sin{\angle XYM}} \cdot \frac{DE}{EC} \cdot \frac{AE}{EB} = \frac{\sin{\angle YXM}}{\sin{\angle XYM}} \cdot \left(\frac{\sin{\angle \frac{XYM}{2}}}{\sin{\angle \frac{YXM}{2}}}\right)^2 = \frac{\sin{\angle \frac{XYM}{2}}}{\sin{\angle \frac{YXM}{2}}} \cdot \frac{\cos{\angle \frac{YXM}{2}}}{\cos{\angle \frac{XYM}{2}}}. \] If $T'$ is the desired tangency point, and $p$ is the semi-perimeter of $\triangle MXY$, then: \[ \frac{YT'}{T'X} = \frac{p - YM}{p - MX} = \frac{\cos{\angle \frac{YXM}{2}}}{\cos{\angle \frac{XYM}{2}}} \cdot \frac{\sin{\angle \frac{XYM}{2}}}{\sin{\angle \frac{YXM}{2}}} = \frac{YT}{TX}. \quad \square \] Now, let $S$ be the $M$-excircle excenter of $\triangle MXY$, and let $N = SM \cap XY$. Claim: $MQ = MP$. Proof. From the first claim, we have that $S, T, Q$ are collinear. Now, from angle bisector harmonic construction: \[ (S, I; M, N) = -1 \overset{(T)}{\Rightarrow} (Q, P; M, P_{\infty AB}) = -1 \Rightarrow MQ = MP. \quad \square \]

Let $A' = AT \cap XY$ and $B' = BT \cap XY$ Angle chasing gives $YD=YB'$ and $XC=XC'$ Also by homothety through $E$ we have $TA'=TB'$ and thus also $QA'=QB'$ let $I_M$ be the $M-$excenter of triangle $MXY$, we claim that $I_M$ lies on the perpendicular bisector of $A'B'$ This follows from $I_M$ lying on the perpendicular bisectors of $B'D$, $CA'$ and $CD$ Finally let $J$ be the intersection of $TQ$ with the circle $XIYI_M$ and $G = MI \cap XY$ We now project: $-1 = (MG;II_M) \stackrel{P_{\infty}}{=} (QT;JI_M) \stackrel{I}{=} (QP;MP_{\infty})$ which is what we wanted to prove

whoever came up with this qn: salute to you