Euclidean geometry : find out the area.

Question

${\angle}ABC=45°$, $AD{\bot}AE$, $AD=AE$, $AC{\bot}BE$, $F$ is the intersection of $AC$ and $BE$, $BF=9$, so what is the area of ${\square}ABCE$?

Answer 1

Let $F(0,0), A(0,a), B(-b,0), C(0,-c), E(e,0)$, then $D=tB+(1-t)C= (-tb,-(1-t)c)$ for some $t$ ($t\in(0;1)$ if we believe the drawing, but I won't use it).
The desired area is then $\frac12(ae+ec+bc+ba)$.
We have:
$AD\perp AE\Leftrightarrow (D-A).(E-A)=0 \Leftrightarrow (-tb,-(1-t)c-a).(e,-a)=0 \Leftrightarrow -tbe+(1-t)ca+a^2=0$
$|AD|=|AE|\Leftrightarrow (D-A)^2=(E-A)^2 \Leftrightarrow t^2b^2+(1-t)^2c^2+a^2+2(1-t)ca=a^2+e^2$
$AC\perp BE$ ve have already used when introducing the coordinate system
$BF=9\Leftrightarrow b=9$
and $\angle ABC=45^\circ \Leftrightarrow (A-B).(C-B)=\frac{\sqrt{2}}{2}|A-B|\cdot|C-B|\Rightarrow ((a,b).(b,-c))^2=\frac{1}{2}(a^2+b^2)(b^2+c^2) \Leftrightarrow 2(ab-bc)^2=(a^2+b^2)(b^2+c^2)$
Then we wolframalpha this thing $\begin{cases} -tbe+(1-t)ca+a^2=0\\ t^2b^2+(1-t)^2c^2+a^2+2(1-t)ca=a^2+e^2\\ b=9\\ 2(ab-bc)^2=(a^2+b^2)(b^2+c^2) \end{cases}$
and see there's one condition missing as the desired area $S=a^2 - \frac{2916}{a - 9} - 243$ can be made arbitrary large if we take $a$ close enough to $9$.

Answer 2

HINT:

We use this fact that if the diameters of a quadrilateral are perpendicular, then the mid points of its sides are cyclic(locate on one circle). Hence GHIJ is a rectangle. The area of ABCE is twice as the area of GHIJ(can you find why?). Now you have to find the measures of sides of this rectangle and its area. Note: there is a typo in figure,O and F must be replaced.