In △ ABC, CD and CF are bisectors of internal and external angles of triangle ABC respectively, DF is parallel, BC intersects AC at e, then de = DF? Please explain the reason

In △ ABC, CD and CF are bisectors of internal and external angles of triangle ABC respectively, DF is parallel, BC intersects AC at e, then de = DF? Please explain the reason

Maybe it's a mistake. It should be de = EF,
Analysis: ∵ DF ∥ BC,
∴∠EDC=∠DCB,∠DFC=∠FCM,
Note: m is on the extension line of BC
∵ CD bisection ∠ ACB, CF bisection ∠ ACM,
∴∠BCD=∠DCE,∠ECF=∠FCM,
∴∠EDC=∠ECD,∠EFC=∠FCE,
∴ED=EC,EF=EC,
Then de = EF

It is known that CD and CF are bisectors of inner and outer angles of triangle ABC respectively. DF is parallel to BC and intersects AC at E. DF = 2DE

Let G be a point on the extension line of edge BC.. because DF is parallel to BC, ∠ BDC = ∠ EDC, ∠ EFC = ∠ FCG, because CD and CF are the bisectors of the inner and outer angles of triangle ABC, so ∠ BDC = ∠ DCE, ∠ FCG = ∠ ECF, so ∠ EDC = ∠ DCE, ∠ EFC = ∠ ECF, so de = EC, EC = EF, so de =

It is known that the area of triangle ABC is 84 square centimeters, which is twice the area of parallelogram efcd. Find the area of triangle FBE

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