Solution:

Place $A = (0, 0)$ and $B = (9, 0).$ From $CA = 17$ and $BC = 10,$ solving $x^2 + y^2 = 289$ and $(x - 9)^2 + y^2 = 100$ gives $C = (15, 8).$ The congruence $\triangle ABC \cong \triangle BAD$ swaps $A$ and $B,$ so $D$ is the reflection of $C$ across the line $x = \frac{9}{2},$ namely $D = (-6, 8).$

A point lies in triangle $ABC$ exactly when it is on or above $\overline{AB},$ on $B$'s side of line $AC,$ and on $A$'s side of line $BC;$ similarly for triangle $BAD.$ In the overlap the binding constraints are $y \ge 0,$ line $AC,$ and line $BD,$ so the intersection is the triangle with base $\overline{AB}$ and apex $E = AC \cap BD.$ Line $AC$ is $y = \frac{8x}{15}$ and line $BD$ is $y = -\frac{8(x - 9)}{15},$ which meet at $E = \left(\frac{9}{2}, \frac{12}{5}\right).$

The area is $$\frac{1}{2} \cdot 9 \cdot \frac{12}{5} = \frac{54}{5},$$ so $m + n = 54 + 5 = 59.$