Difficult triangle/circle geometry problem Let $AB$ be a segment of length $12$. $ω1$ is a circle centered at $A$ with
radius $13$, $ω2$ is a circle centered at $B$ with radius $7$. Let $l$ be a common
tangent of the circles and $l$ intersects $ω1$ and $ω2$ at $C$ and $D$, respectively.
ω1 and ω2 intersect each other at $M$ and $N$, where $N$ is closer to $l$. If
lines $CN$ and $DN$ intersect $MD$ and $MC$ at $X$ and $Y$ respectively, then
find $XY$.
It is very easy to find the coordinates of $M$ and $N$ by Pythagorean theorem and using similar triangles you can also find the coordinates of $C$ and $D$. We can then finish the problem by finding the appropriate lines and the two intersection points $X$ and $Y$. But this is tedious and time consuming. 
Is there are purely synthetic solution?
 A: A possible synthetic solution would be to calculate various segments involving triangle $MCD$. 
Let $MN$ intersect $CD$ at $T$. We have three cevians $MT$, $CX$, and $DY$. $MT$ is a median.
Using Ceva's theorem or otherwise, show that $XY$ is parallel to $CD$.
Next similar triangles can be used to find $XY$.
Here with the details:
$\text{CD}^2+6^2=12^2$  gives $\text{CD} = 6\sqrt{3}$
$$\text{TC}^2=\text{TD}^2=\text{TN}\cdot\text{TM}$$
$$\text{TC}=\text{TD} = \text{CD}/2 = 3\sqrt{3}$$
Let $MN$ intersect $AB$ at $H$
$$\text{TN}\cdot\text{TM} = \text{TH}^2-\text{NH}^2=27$$
To find $NH$, calculate the area of triangle $ANB$ in two ways, once using the height and base and again using Heron's formula
We find $\text{NH}=4\sqrt{3}$ and from this $\text{TH}=5\sqrt{3}$ , $\text{TN}=\sqrt{3}$ , $\text{TM}=9\sqrt{3}$
Finally notice that angle $MTD$ = 180 degrees - angle $HTC$ = angle $BAC$ = 60 degrees
The final action takes place in triangle $MCD$
Let $MT$ intersect $XY$ at $G$. Let $x=YG$
$$\frac{x}{\text{CT}}=\frac{\text{MG}}{\text{MT}}\text{ using }MGY \sim MTC$$
$$\frac{x}{\text{TD}}=\frac{\text{NG}}{\text{NT}}\text{ using }NGY\sim NTD$$
this takes us to $$x=\frac{12}{5}\sqrt{3}$$ $$\text{XY}=\frac{24}{5}\sqrt{3}$$
