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信息发布者:
B
B
C
$39^{\circ}$
$12$
$4$
证明:$\because D$为$AC$的中点,$\therefore AD = DC,$在$Rt\triangle ADE$与$Rt\triangle CDF$中,$\begin{cases}AD = CD \\ DE = DF\end{cases},$$\therefore Rt\triangle ADE\cong Rt\triangle CDF(HL),$$\therefore \angle A=\angle C.$$\because AB = AC,$$\therefore \angle B=\angle C,$$\therefore \angle A=\angle B=\angle C,$$\therefore \triangle ABC$是等边三角形.
$解:∵△ABC是等边三角形$
$∴AB=BC,∠ABC=∠ACB=60°$
$在△ABD和△BCE中$
$\begin{cases}{AB=BC\ } \\ { ∠DAB=∠EBC} \\{AD=BE } \end{cases}$
$∴△ABD≌△BCE(SAS),∴∠ABD=∠BCE$
$∵∠CFD=∠BCE+∠CBD=∠ABD+∠CBD=∠ABC$
$∴∠CFD=60°$
$7$