Solution 1: Let $I_{1}, I_{2}$ be the centers of $\omega_{1}, \omega_{2}$, respectively, $X_{1}, X_{2}$ be the tangency points of $\omega_{1}, \omega_{2}$ with $B C$, respectively, and $Y_{1}, Y_{2}$ be the tangency points of $\omega_{1}, \omega_{2}$ with $A D$, respectively. Let the two common external tangents of $\omega_{1}, \omega_{2}$ meet at $P$. Note that line $I_{1} I_{2}$ also passes through $P$. By Heron's formula, the area of triangle $A B C$ is 84. Thus, $\frac{1}{2} A D \cdot B C=84$, and so $A D=12$. By the Pythagorean Theorem on right triangles $A B D$ and $A C D, B D=5$ and $C D=9$. The inradius of $A B D, r_{A B D}$, is $\frac{[A B D]}{s_{A B D}}$, where $[A B D]$ is the area of $A B D$ and $s_{A B D}$ is its semiperimeter. $A B D$ is a $5-12-13$ right triangle, so $[A B D]=30$ and $s_{A B D}=15$. Thus, $r_{A B D}=2$. Similarly, we get that $A C D$'s inradius is $r_{A C D}=3. I_{1} Y_{1} D X_{1}$ is a square, so $X_{1} D=I_{1} X_{1}=r_{A B D}=2$, and similarly $X_{2} D=3. X_{1}$ and $X_{2}$ are on opposite sides of $D$, so $X_{1} X_{2}=X_{1} D+X_{2} D=5$. Since $P$ lies on lines $I_{1} I_{2}, X_{1} X_{2}$, and $I_{1} X_{1}, I_{2} X_{2}$ are parallel, triangles $P X_{1} I_{1}$ and $P X_{2} I_{2}$ are similar. Thus, $\frac{X_{1} I_{1}}{X_{2} I_{2}}=\frac{2}{3}=\frac{P X_{1}}{P X_{2}}=\frac{P X_{1}}{P X_{1}+X_{1} X_{2}}=\frac{P X_{1}}{P X_{1}+5}$. Solving gives $P X_{1}=10$. Letting $\angle I_{1} P X_{1}=\theta$, since $I_{1} X_{1} P$ is a right angle, we have $\tan \theta=\frac{X_{1} I_{1}}{X_{1} P_{1}}=\frac{1}{5}$. $D$ and $E$ lie on different common external tangents, so $P I_{1}$ bisects $\angle E P D$, and thus $\angle E P D=2 \theta$. Thus, $\tan \angle E P D=\tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^{2} \theta}=\frac{5}{12}$. $E D$ is perpendicular to $B C$, so triangle $E D P$ is a right triangle with right angle at $D$. Thus, $\frac{5}{12}=$ $\tan \angle E P D=\frac{E D}{P D}. P D=P X_{1}+X_{1} D=12$, so $E D=5. A D=12$, so it follows that $A E=7$. Solution 2: Lemma: Let $\Omega_{1}, \Omega_{2}$ be two non-intersecting circles. Let $\ell_{A}, \ell_{B}$ be their common external tangents, and $\ell_{C}$ be one of their common internal tangents. $\Omega_{1}$ intersects $\ell_{A}, \ell_{B}$ at points $A_{1}, B_{1}$, respectively, and $\Omega_{2}$ intersects $\ell_{A}, \ell_{B}$ at points $A_{2}, B_{2}$. If $\ell_{C}$ intersects $\ell_{A}$ at $X$, and $\ell_{B}$ at $Y$, then $X Y=A_{1} A_{2}=B_{1} B_{2}$. Proof: Let $\ell_{C}$ intersect $\Omega_{1}, \Omega_{2}$ at $C_{1}, C_{2}$, respectively. Then, examining the tangents to $\Omega_{1}, \Omega_{2}$ from $X, Y$, we have $A_{1} X=C_{1} X, A_{2} X=C_{2} X, B_{1} Y=C_{1} Y, B_{2} Y=C_{2} Y$. Thus, $2 A_{1} A_{2}=2 B_{1} B_{2}=$ $A_{1} A_{2}+B_{1} B_{2}=A_{1} X+A_{2} X+B_{1} Y+B_{2} Y=C_{1} X+C_{2} X+C_{1} Y+C_{2} Y=2 X Y$, and the conclusion follows. Using the notation from above, we apply the lemma to circles $\omega_{1}, \omega_{2}$, and conclude that $E D=X_{1} X_{2}$. Then, we proceed as above to compute $X_{1} X_{2}=5=E D$. Thus, $A D=7$.
Let $A B C$ be a triangle with $A B=13, B C=14, C A=15$. The altitude from $A$ intersects $B C$ at $D$. Let $\omega_{1}$ and $\omega_{2}$ be the incircles of $A B D$ and $A C D$, and let the common external tangent of $\omega_{1}$ and $\omega_{2}$ (other than $B C$) intersect $A D$ at $E$. Compute the length of $A E$.