As shown in the figure, in the triangle ABC, D is a point on the edge AB, and ad = AC, De is parallel to BC, CD bisection angle EDF

As shown in the figure, in the triangle ABC, D is a point on the edge AB, and ad = AC, De is parallel to BC, CD bisection angle EDF

It is proved that CDE = DCF = FDC
Thus, FD = FC can be determined
From AF = AF, ad = AC
From this, we can determine △ ADF ≌ △ ACF
That is, DAF = caf
According to the principle of symmetry
It can be proved
AF vertical split CD
prove:
∵ CD bisection ∠ EDF
∴∠EDC=∠FDC
∵DE//BC
∴∠EDC=∠DCF
∴∠DCF=∠FDC
∴CF=DF
And ∵ AC = ad, AF = AF
∴⊿ACF≌⊿ADF（SSS）
∴∠CAF=∠DAF
That is, AF is the vertex bisector of the isosceles triangle ACD. According to the three lines in one, the angle bisector is also the middle perpendicular of the bottom
The AF splits the CD vertically