Boxes Out, Segments In

Place the diagram on the coordinate plane so that $B$ is at $B(0, 0)$ and $C$ is at $C(14, 0)$ .

The $13$ - $14$ - $15$ triangle is a $5$ - $12$ - $13$ right triangle combined with a $9$ - $12$ - $15$ right triangle, so the coordinates of $A$ are $A(5, 12)$ .

To go from $B$ to $A$ you go $5$ right and then $12$ up, so to go from $A$ to $A'$ you go $12$ left and $5$ up (since $\angle BAA'$ is a right angle), so the coordinates of $A'$ are $A'(5 - 12, 12 + 5) = A'(-7, 17)$ .

In a similar manner, you can calculate the coordinates $B'(-12, 5)$ , $B''(0, -14)$ , $C'(14, -14)$ , $A''(17, 21)$ , and $C''(26, 9)$ .

Then the line through $B'(-12, 5)$ and $C''(26, 9)$ follows $y = \frac{2}{19}(x - 26) + 9$ , the line through $A'(-7, 17)$ and $C'(14, -14)$ follows $y = -\frac{31}{21}(x + 7) + 17$ , and the line through $B''(0, -14)$ and $A''(17, 21)$ follows $y = \frac{35}{17}x - 14$ .

The intersection of $y = \frac{2}{19}(x - 26) + 9$ and $y = -\frac{31}{21}(x + 7) + 17$ is $D(\frac{161}{631},\frac{3969}{631})$ , the intersection of $y = \frac{2}{19}(x - 26) + 9$ and $y = \frac{35}{17}x - 14$ is $E(\frac{6545}{631},\frac{4641}{631})$ , and the intersection of $y = -\frac{31}{21}(x + 7) + 17$ and $y = \frac{35}{17}x - 14$ is $F(\frac{3689}{631},-\frac{1239}{631})$ .

Then $\overrightarrow{FE} = (\frac{2856}{631}, \frac{5880}{631})$ and $\overrightarrow{FD} = (-\frac{3528}{631}, \frac{5208}{631})$ , so the area of $\triangle DEF = \frac{1}{2}\begin{vmatrix}\frac{2856}{631} & \frac{5880}{631}\\-\frac{3528}{631} & \frac{5208}{631}\end{vmatrix} = \frac{28224}{631}$ , so $a = 28224$ , $b = 631$ , and $a + b = \boxed{28855}$ .