In the isosceles right triangle ABC, ∠ BAC = 90 °, the bisector of ∠ CBA intersects AC at D, the perpendicularity of BD is made through C, the perpendicularity is e, and the extension of CE and Ba intersects F If CE = 5, then BD =? Fast

BD=10
It is proved that △ ACF and △ abd are congruent
(Note: △ CBF is isosceles triangle, CE = 1 / 2FC)

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