As shown in the figure, in the isosceles triangle ABC, the angle ACB is equal to 90 degrees, ad is the middle line of the waist BC, CE is perpendicular to ad, AB intersects e, and the angle CDA is equal to the angle EDB

As shown in the figure, in the isosceles triangle ABC, the angle ACB is equal to 90 degrees, ad is the middle line of the waist BC, CE is perpendicular to ad, AB intersects e, and the angle CDA is equal to the angle EDB

It is proved that the extension of the intersection of BF ⊥ BC and CE through point B is at F,
Obviously △ CBF ≌ ACD
∴ ∠F=∠CDA ; CD=BF
∵ CD=BD ∴ BD=BF
∵∠ ABC = ∠ EBF = 45 degrees; be = be
∴ △DBE≌△FBE
∴∠EDB=∠F
∴∠CDA=∠EDB.