It is known that in the triangle ABC, ab = AC, D point is on AB, e point is on the extension line of AC, and BD = CE, connecting de with BC at f point No picture
prove:
Crossing point D as DH ‖ AE to BC at h
∴∠DHB=∠ACB,∠HDF=∠E
∵AB=AC
∴∠B=∠ACB
∴∠B=∠DHB
∴BD=HD
∵BD=CE
∴HD=CE
In △ DHF and △ ECF
∠HDF=∠E
∠DFH=∠EFC
HD=CE
∴△DHF≌△ECF
∴DF=EF
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