证明:(1) $\because FD$ 平分$∠ AFE,$
$\therefore ∠ AFD=∠ DFE。$
$\because FD// BC,$
$\therefore ∠ AFD=∠ B,$$∠ DFE=∠ FEB。$
$\therefore ∠ B=∠ FEB。$
$\therefore FB=FE。$
$\therefore △ FBE$ 是等腰三角形。
(2) 解:$\because ∠ B=∠ FEB,$$∠ B=50°,$
$\therefore ∠ FEB=50°。$
$\because ∠ FEB+∠ FEC=180°,$
$\therefore ∠ FEC=130°。$
$\because EH$ 平分$∠ FEC,$
$\therefore ∠ CEH=\dfrac{1}{2}∠ FEC=65°。$
$\because FD// BC,$
$\therefore ∠ FHE=∠ CEH=65°。$