Take the right angle side AB of the right triangle ABC as the diameter of the circle O,l and the hypotenuse AC intersect at the point D, over the point D as the tangent to the circle O intersect BC at

Take the right angle side AB of the right triangle ABC as the diameter of the circle O,l and the hypotenuse AC intersect at the point D, over the point D as the tangent to the circle O intersect BC at E, connect OE to prove that EB=.

As shown in the figure above, OA=OB=OD, and ∠OBE=∠ODE=90°, so △OBD is isosceles △, and there is ∠EBD=∠EDB, so △BED is isosceles, so EB=ED.

And because ∠CDE+∠ODA=90°, ∠BCA+∠OAD=90°, ∠ODA=∠OAD, so ∠ODA=∠OAD, so ∠ODA=∠OAD, so ∠ODA=∠OAD. OAD,so ∠CDE=∠BCA,i.e. △ECD is isosceles triangle,so EC=ED

So there?EB=EC=ED