证明:∵在$\triangle ABC$中,$∠B = 50°,$$∠C = 20°$
∴$∠CAB=180°-∠B-∠C = 110°$
∵$AE\perp BC,$∴$∠AEC = 90°$
∴$∠DAF=∠AEC+∠C = 110°,$∴$∠DAF=∠CAB$
在$\triangle DAF $和$\triangle CAB$中
$\begin {cases}AD = AC\\∠DAF=∠CAB\\AF = AB\end {cases}$
∴$\triangle DAF≌\triangle CAB(S AS),$∴$DF = CB$