In triangle $A B C$, $A B=A C$, angle $A$ is obtuse, $B D$ is the angle bisector, $A M$ is the altitude, $E$ is the foot of the perpendicular dropped from $D$ to side $B C$. A perpendicular is erected from point $D$ to $B D$, intersecting side $B C$ at point $F$. It is known that $M E=F C=a$. Find the area of triangle $A B C$.
Let $\angle D B C=\alpha$ (see the figure). Then $\angle C=\angle B=2 \alpha$. Let $K$ be the midpoint of segment $B F$. Then $D K$ is the median of the right triangle $B D F$. Therefore,

$B K=K D=K F, \angle D K C=2 \angle D B K=2 \alpha=\angle C$.

![](https://cdn.mathpix.com/cropped/2024_05_06_6ed8f1205731b80a0039g-50.jpg?height=375&width=672&top_left_y=1722&top_left_x=699)

Thus, triangle $C D K$ is isosceles. Its height $D E$ is also a median, so $K M+M E=E F+$ $F C$, which means $K M=E F$.

Therefore, $2 K M+a=K F=B K=B M-K M=M C-K M=2 a$, from which $K M=a / 2, K C=3 a, B C=5 a, B E=$ $7 a^{2}, D E^{2}=B E \cdot E F=7 a^{2} / 4$.

Since triangles $A B C$ and $D K C$ are similar, then

$S_{A B C}=\left(\frac{B C}{K C}\right)^{2} S_{D K C}=\left(\frac{B C}{K C}\right)^{2} \cdot \frac{1}{2} K C \cdot D E=\frac{25 a}{3} \cdot \frac{\sqrt{7} a}{4}=\frac{25 \sqrt{7} a}{12}$.

$\frac{25 a^{2} \sqrt{7}}{12}$.