As shown in the figure, in the triangle ABC, the angle BAC = 90 ° and BD is the bisector of the angle ABC. The extension line of BD is perpendicular to the line passing through point C and E, and the extension line of CE intersecting Ba is F. the proof is: BD = 2ce Please help me

As shown in the figure, in the triangle ABC, the angle BAC = 90 ° and BD is the bisector of the angle ABC. The extension line of BD is perpendicular to the line passing through point C and E, and the extension line of CE intersecting Ba is F. the proof is: BD = 2ce Please help me

The building owner has one less condition, right? That is, △ ABC should be isosceles right triangle, because I prove that it is not necessarily equal
prove:
∵ be ⊥ FC, be is the angular bisector
The BFC is an isosceles triangle
The original problem is to prove BD = FC
∵ ∠BDA=∠EDC,∠EDC+∠ECD=90º,∠ECD+∠F=90º
∴ ∠BDA=∠F
And ∵ ∠ bad = ∠ CAF = 90 & ordm;
∴ △BAD ∽ △CAF
In my opinion, only when Ba = Ca, can we judge congruence, and then we can get a conclusion