hi
$\displaystyle \angle ECA=\angle CBA$ -- 1
$\displaystyle \angle EAC=\angle CBA $(alternate segments)
so $\displaystyle \angle ECA=\angle EAC$
and EA=EC (equal sides of an isosceles triangle)
$\displaystyle \angle OAE=\angle OCE=90^o$ (tangents at A and C)
Hence we can conclude that OAEC is a square
$\displaystyle \angle BCA=90^o$ (angle in a semicircle)
$\displaystyle
\angle DCE+\angle ECA+\angle BCA=180^o
$
$\displaystyle \angle DCE+\angle ECA=90^o$ (angle of a line)
$\displaystyle \angle OAC+\angle BCA+\angle CBA =180^o$
From 1 , we see that $\displaystyle \angle CAB=\angle ECD $
AD=EC (square)
$\displaystyle \angle AOC =\angle DEC =90^o$
so $\displaystyle \triangle ADC \cong \triangle CED $
ED=OC (corresponding angle)
Since OC=EA , ED=EA