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

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

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 $$\triangle AHE$$,
$$(x-2)^2 + x^2 = 100 \implies x = 8$$
$$\therefore BE = x \sqrt2 = 8 \sqrt2$$