There are three triangles. Two compose the largest one. I'll call the leftmost part $L,$ its counterpart on the right $R,$ and their composite $K.$
Looking at $R,$ we can write down an equation for our goal $x_2$ as follows: $$x_2^2=a_1^2+a_2^2-2a_1a_2\cos E.$$ Thus we only need find $a_1$ and $a_2,$ and we'd be done.
To find $a_1,$ look at $L,$ the we have that $$\frac{\sin B}{\text{Length}}=\frac{\sin A}{a_1}.$$ Similarly, by looking at $K,$ we find that $$\frac{\sin D}{\text{Length}}=\frac{\sin A}{a_2}.$$ This helps us to find $a_2.$
From the last two equations we eliminate $A$ to find an equation in $a_1,\,a_2.$
Now look at the triangle $R,$ from which we may write down $$\frac{\sin E}{x_2}=\frac{\sin D}{a_1}.$$
Together with the first equation, we find a $3×3$ system in $a_1,\,a_2\,x_2,$ which we can now solve for $x_2.$
OK, let's do this together. From the two middle equations we find that $$\sin A=\frac{a_1\sin B}{\text{Length}}=\frac{a_2\sin D}{\text{Length}},$$ so that $$a_1=\frac{\sin D}{\sin B}a_2.$$ From the last equation we also have $$a_1=\frac{\sin D}{\sin E}x_2,$$ so that we have that $$a_2=\frac{\sin D}{\sin E}\frac{\sin B}{\sin D}x_2=\frac{\sin B}{\sin D}x_2.$$
Substituting now into the first equation and solving for $x_2$ is straightforward, and gives $$x_2^2\left(\frac{\sin B+\sin D}{\sin E}+1\right)=\frac{2\sin B\sin D\cot E}{\sin E},$$ amd finally recall that $x_2>0.$
Let $E(x,y),$ $A(0,0)$, $D(0,1),$ $B(1,0)$ and $C(1,1).$
Thus, we obtain the following system.
$$\left(x-\frac{1}{2}\right)^2+(y-1)^2=\left(\frac{1}{2}\right)^2$$ and $$\frac{\sqrt{x^2+(y-1)^2}}{\sqrt{x^2+y^2}}=\frac{4}{5}.$$
Can you end it now?
I got $E\left(\frac{16}{65},\frac{37}{65}\right)$ or $E\left(\frac{16}{17},\frac{13}{17}\right)$ and from here easy to find the area.
Best Answer
Pick $D'$ on $AC$ such that $CD'=CB$. Then triangle $CBD'$ is isosceles with a top angle of $40^\circ$. The remaining two angles are $70^\circ$.
Now take a look at points $D,D'$. By hypothesis $AD=CD'$ so $D,D'$ are symmetric with respect to the midpoint of $AC$.
Moreover, $BD, BD'$ are also symmetric with respect to the bisector of the angle $\angle ABC$ since $\angle ABD = \angle D'BC = 70^\circ$.
This is only possible if the triangle $ABC$ is isosceles and $AB=BC$. This gives $\angle B = 100^\circ$ and $\angle DBC = 30^\circ$.
To conclude that the triangle is isosceles do the following
consider a triangle $AD'B'$, translating the triangle $DCB$ until $AD'$ and $DC$ overlap.
then $\angle ABD'= \angle AB'D'$, showing that $A,D',B,B'$ are on the same circle
by construction $BB'||AC$, thus $A,D',B,B'$ are the vertices of an isosceles trapeze. The non-parallel edges and diagonals are equal, implying that the triangle $ABC$ is isosceles also.