I would like to prove that triangles $ABD$ and $BCE$ are congruent (I think they are?)
I want the shortest possible reasoning.
Currently, I think I have it by side-side-angle.
Clearly, $AD=EC$ and $BD=BE$ because they are radii of the same circle.
$BDE$ is an isosceles triangle with angles $BDE$ and $BED$ being equal (call this value $x$). But since $ADE$ is a straight line, as is $DEC$, we have $BDA = 180 - x$ and $BEC = 180 - x$.
Hence the triangles are congruent by side-side-angle.
IS this correcT?
Are there any alternatives? Would be interested in as many as possible - in particular, can we argue $BC =AB$ through obvious means or any of the other angles are equal? :)