Is this it?

It is known that in trapezoidal ABCD, the extension lines of AD∨BC, BC=DC, CF bisectors ∠BCD, DF∨AB and BF intersect DC at point E. 。

Verification: (1) △ BFC △ DFC; (2)AD=DE。

If that's the case. The answer is as follows:

Proof: (1)∵CF bisection ∠BCD,

∴∠BCF=∠DCF.

In △BFC and △DFC, BC = DC ∠ BCF = ∠ DCF FC = FC ∴△ BFC △ DFC.

(2) connect BD.

∫△BFC?△DFC，

∴BF=DF,∴∠FBD=∠FDB.

∫DF∨AB，

∴∠ABD=∠FDB.∴∠ABD=∠FBD.

∫ AD ∨ BC,

∴∠BDA=∠DBC.

∫ BC =DC,

∴∠DBC=∠BDC.∴∠BDA=∠BDC.

And BD is the common terminal,

∴△BAD≌△BED.

∴AD=DE.

To solve this problem, we should pay attention to (1) dividing ∠BCD by CF ∠BCF=∠DCF, and then prove △ BFC △ DFC by SAS.

(2) Prove AD=DE, connect BD, and prove △ bad △ bed. AB∨DF？ ∠ABD=∠BDF，BF=DF？ ∠DBF=∠BDF, ∴∠ABD=∠EBD, BD=BD, and then prove ∠BDA=∠BDC, it is easy to reason ∠BDA =∠DBC =∞.