It is known that in the isosceles RT triangle ABC, ∠ ACB = 90 °, D is the midpoint of BC, e is the point on AB, and AE = 2EB, the proof of ad ⊥ CE is obtained

Connect de and make ef perpendicular to BC through E
The similarity between △ CAE and △ BDE is obtained according to the edge ∠ edge
So, ACE = EDB
Because EF, BC is vertical
So △ bef is similar to △ ABC
So EF = 1 / 3aC
BF=1/3BC
Because BD = 1 / 2BC BF = 1 / 3bC
So DF = 1 / 6BC
So DF = 1 / 3CD
So △ ACD is similar to △ def
Therefore, ADC = EDF
Because the previous results of △ CAE and △ BDE are similar
Therefore, ACE = ADC, ad ⊥ CE

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