Then, triangle ABD is a right triangle with angle ADB= angle BDC = 90 degrees.

But from the angle ABC is a right angle, BC is tangent to the circle O and tangent to the point B.

Also, DE is the tangent of circle o,

Therefore, angle BDE = angle DBE, DE = EB...(*).

Because triangle BDC is a right triangle,

90 degrees = angle BDE ++ angle EDC = angle DBE+ angle c,

So, angle EDC = angle c,

Triangle EDC is an isosceles triangle. CE = DE。

Reorganization (*), yes,

BE = DE = CE。

The original proposition was proved. .