In the triangle ABC, AB > AC, AF is the angle bisector, D is the point on AB and ad = AC, de ‖ BC intersects AC at e to prove the angle EDF of CD bisector

In △ ADF and △ ACF, ad = AC, ∠ DAF = ∠ CAF, AF is the common edge,
So, △ ADF ≌ △ ACF,
It can be concluded that FD = FC,
So, CDF = ∠ DCF
Because, de ‖ BC,
So, DCF = CDE,
It can be concluded that: CDF = ∠ CDE,
That is: CD bisection ∠ EDF

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