Question 272·Hard·Math

Place $B=(0,0)$ and $C=(14,0)$. Moving or rotating a triangle doesn't change its side lengths, and these coordinates make the distance from $B$ to $C$ equal to the given $14$.

A midpoint divides a segment into equal lengths. Half of $14$ is $7$, so $D=(7,0)$. A midpoint doesn't automatically make a right angle.

Since $AB=13$, $A$ lies on the circle centered at $B$ with radius $13$. Since $AC=15$, it also lies on the circle centered at $C$ with radius $15$. Type both circle equations in Desmos and click their upper intersection. Desmos shows $(5,12)$, so take $A=(5,12)$. The lower intersection, $(5,-12)$, is a reflection across $BC$ and gives the same length for $AD$.

Type the coordinates of $A$ and $D$, then enter $\operatorname{distance}(A,D)^2$. Desmos prints $148$, so $AD^2=148$. This is the Pythagorean theorem at work: the horizontal change is $7-5=2$, and the vertical change is $12-0=12$, giving $2^2+12^2=148$.

Take the positive square root, since a length can't be negative, and pull out the perfect square $4$: $AD=\sqrt{148}=\sqrt{4\cdot37}=2\sqrt{37}$. So the length of $\overline{AD}$ is $2\sqrt{37}$. Choice C.