Answer: 129

Solution:

Place line $BC$ on the $x$-axis, so the centers are $P = (0, 1)$ and $Q = (4, 4)$ (their distance is $1 + 4 = 5$), with $B = (0, 0),$ $C = (4, 0),$ and the tangency point $A = P + \frac{1}{5}(Q - P) = \left(\frac{4}{5}, \frac{8}{5}\right).$ The homothety centered at $A$ with ratio $-4$ carries $\mathcal{P}$ to $\mathcal{Q}$ and $D$ to $E,$ so $AE = 4\,AD.$ Since $[DBA] = \frac{1}{2} AD \cdot d(B, \ell)$ and $[ACE] = \frac{1}{2} AE \cdot d(C, \ell),$ the equal-area condition is $d(B, \ell) = 4\,d(C, \ell)$ with $B$ and $C$ on the same side of $\ell.$

Write $\ell$ as $u\left(x - \frac{4}{5}\right) + v\left(y - \frac{8}{5}\right) = 0.$ Its signed values at $B$ and $C$ are $-\frac{4u + 8v}{5}$ and $\frac{16u - 8v}{5},$ so the same-side ratio-$4$ condition reads $-(4u + 8v) = 4(16u - 8v),$ giving $24v = 68u,$ i.e. $v = \frac{17}{6}u.$ Taking $(u, v) = (6, 17),$ the line is $6x + 17y = 32.$

Then $d(B, \ell) = \frac{32}{\sqrt{325}},$ and the center $P = (0, 1)$ is at distance $\frac{15}{\sqrt{325}}$ from $\ell,$ so the chord gives $AD = 2\sqrt{1 - \frac{225}{325}} = \frac{20}{\sqrt{325}}.$ The common area is $$\begin{aligned} &\frac{1}{2} \cdot \frac{20}{\sqrt{325}} \cdot \frac{32}{\sqrt{325}} \\ &= \frac{320}{325} = \frac{64}{65}, \end{aligned}$$ so $m + n = 64 + 65 = 129.$