Two circles $G_1$ and $G_2$ intersect at two points $M$ and $N$. Let $AB$ be the line tangent to these circles at $A$ and $B$, respectively, so that $M$ lies closer to $AB$ than $N$. Let $CD$ be the line parallel to $AB$ and passing through the point $M$, with $C$ on $G_1$ and $D$ on $G_2$. Lines $AC$ and $BD$ meet at $E$; lines $AN$ and $CD$ meet at $P$; lines $BN$ and $CD$ meet at $Q$. Show that $EP=EQ$.
Attached above is my diagram.
I've proved that $NX$ bisects $AB$, as $MN$ is the radical axis of the two circles (power of $X$ with respect to both circles is equal). So, $XA = XB$. Now, $AB$ and $PQ$ are parallel, $M$ bisects $PQ$. Hence, $MP = MQ$.
I'm stuck here. One of my friends suggested me to prove that triangles $MAB$ and $EAB$ are congruent, which would require some angle chasing and using (asa) congruence criterion.
Could someone please help me understand the angle chasing part? I'm not able to figure out why the triangles should be congruent, and I'm not particularly great at angle chasing. You could use the attached diagram for explanation, if needed. Thanks a lot.
We have\begin{eqnarray*}\angle MAB &=& \angle MCA \;\;\;{\rm tangent-chord}\\ &=& \angle BAE\;\;\; {\rm parallel \;line} \end{eqnarray*} and the same way we have $\angle MBA = ABE$, so mentioned triangles are congruent by $(asa)$.