<s>[INST] Geometry problem involving circles and triangles This is the question:
$E$ is a point on the side $AB$ of a rectangle $ABCD$ such that $DE=6$ cm, $DA= 8cm$ and $DC=6cm$. If $CE$ extended meets the circumcircle of the rectangle at $F$, then what is the value of $BF$ ? (use $\sqrt{2}$=$1.414$).

Here is how I tried to solve it

In $\Delta BCD$,    $\frac{CD}{BC} = \frac{DG}{BG}$  (Since CG is the Angle bisector of $\langle C$ )
From this, DG = $\frac{30}{7} cm$;   BG= $\frac{40}{7} cm $
$\langle FCD $ = $\langle FAD$ = $\langle FBD$ = $45°$
Therefore, $\Delta CED$ ~ $\Delta AEF$ ; i.e. $\Delta AEF$ is a right isosceles triangle like $\Delta CED $
Hence, EF = AF = $\sqrt 2 cm $
In $\Delta CED $, CE = 6$\sqrt2$ cm
Now consider $\Delta CGD$ & $\Delta BGF $
$\Delta CGD$ ~ $\Delta BGF $
Assume the value of $GE = x$  $cm$
$\Delta GCD$ ~ $\Delta GBF$   
$\frac {GC}{GB}$ = $\frac {CD}{BF}$  = $\frac {GD}{GF}$ 
$\frac {(6 \sqrt 2 -x)}{\frac{40}{7}}$ = $\frac {6}{BF}$  = $\frac {\frac{30}{7}}{(x+\sqrt 2)}$ 
$(x+\sqrt 2)(6 \sqrt 2 - x)$ = $\frac{30}{7}$ .$\frac{40}{7}$
My plan was to find the value of $x$ so that I could get the ratio and then evaluate $BF$. However, it looks like I am going wrong as I am getting complicated values for $x$. Could anyone please guide?
 [/INST] 
If $DA=8$ and $DE=6$, then $AE=2$.
Since $AC$ and $BD$ are diagonals of a $6 \times 8$ rectangle, $AC = BD = 10.$
$\triangle CDE$ is an isosceles right triangle. So $CE = 6 \sqrt 2.$
$\triangle CDE \sim \triangle AFE$. So $\triangle AFE$ is an isosceles right triangle. Since $AE=2$, then $EF = AF = \sqrt 2.$
$\triangle CAE \sim \triangle DFE \implies
     \dfrac{CA=10}{\color{red}{DF=5\sqrt 2}} 
   = \dfrac{CE=6\sqrt 2}{DE=6} 
   = \dfrac{AE=2}{FE=\sqrt 2}$
Since $BD$ is a diameter of the circle, $\angle BFD$ is a right angle.
Hence $\triangle BFD$ is a right triangle. So
$BF = \sqrt{BD^2 - DF^2} = \sqrt{10^2 - (5\sqrt 2)^2} = 5\sqrt 2$
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