What's the measure of the segment $BE$ in the triangle below?

Question

For reference: In the figure: $AC = CD, DE = 2$ and $AE = 10$, calculate $BE$ (Answer: $8\sqrt2$)

My progress:

$ABCM$ is cyclic ($\angle B = \angle M = 90^o )\implies$

$\angle BAC - \theta$

$\triangle BCA \cong \triangle ECD \implies$

$BC=CD, AB=DE=2, AC = AE \therefore \triangle ACE$ is isosceles

$\angle ECD = 90 - \theta = 90 - \angle ACE \implies \angle ACE = \theta$

$\triangle CBE$ (right, isosceles)$\implies \angle EBC = \angle CEB = 45^o\implies\\ BE = CE\sqrt2$

Only CE.... remains to be found

Answer 1

Say $BC = CE = x$.

Drop perp $AH$ from $A$ to $CE$. Then $ABCH$ is a rectangle. $EH = CE - AB = x -2, AH = BC = x$.

Applying Pythagoras in $\triangle AHE$,

$(x-2)^2 + x^2 = 100 \implies x = 8$

$ \therefore BE = x \sqrt2 = 8 \sqrt2$