It is known that when △ ABC, ∠ ACB = 90 °, CE ⊥ AB is on e, D is a point on AB, and ad = AC, AF square ∠ CAE intersects CE on F, FD ∥ BC is proved

It is known that when △ ABC, ∠ ACB = 90 °, CE ⊥ AB is on e, D is a point on AB, and ad = AC, AF square ∠ CAE intersects CE on F, FD ∥ BC is proved

AF is equal to CAE, so CAF = DAF
Ad = AC, AF = AF
So triangle AFD is congruent with triangle AFC,
Therefore, ADF = ACF
In the right triangle ABC, CE ⊥ AB,
Therefore, ACF = ABC,
Therefore, ADF = ABC
So FD / / CB
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