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 =∞.