15. 如图，四边形$ABCD$为平行四边形，延长$BC$到点$E$，使$BE = CD$，连结$AE$交$CD$于点$F$.
(1)求证：$AE$平分$∠ BAD$.
(2)连结$BF$，若$BF⊥ AE$，$∠ E = 60°$，$AB = 4$，求▱$ABCD$的面积.

15. (1)证明:$\because$四边形$ABCD$为平行四边形,
$\therefore AB=CD$,$AD// BE$,$\therefore ∠ DAE=∠ E$,
$\because BE=CD$,$\therefore AB=BE$,
$\therefore ∠ BAE=∠ E$,
$\therefore ∠ BAE=∠ DAE$,$\therefore AE$平分$∠ BAD$.
(2)解:由$BE=AB$,$∠ E=60°$,
$\therefore △ ABE$为等边三角形,
$\therefore AB=AE=4$,又$\because BF⊥ AE$,
$\therefore AF=EF=2$,$\therefore BF=\sqrt{4^{2}-2^{2}}=2\sqrt{3}$,
$\because ∠ DAE=∠ E$,$AF=EF$,$∠ AFD=∠ CFE$,
$\therefore △ ADF≌ △ ECF$ (ASA),
$\therefore$平行四边形$ABCD$的面积$=△ ABE$的面积$= \dfrac{1}{2}×4×2\sqrt{3}=4\sqrt{3}$.

16. 如图，在平行四边形$ABCD$中，$∠ BAD$的平分线交$BC$于点$E$，过点$D$作$AE$的垂线交$AE$于点$G$，交$AB$延长线于点$F$，连接$EF$，$ED$.
(1)求证：$EF = ED$.
(2)若$∠ ABC = 60°$，$AD = 6$，$CE = 2$，求$BF$的长.

16. (1)证明:$\because AE$平分$∠ BAD$,
$\therefore ∠ FAG=∠ DAG$,
$\because DG⊥ AE$,
$\therefore ∠ AGF=∠ AGD=90°$,
又$\because AG=AG$,
$\therefore △ FAG≌ △ DAG(\mathrm{ASA})$,
$\therefore GF=GD$.
又$\because DF⊥ AE$,
$\therefore EF=ED$.
(2)解:$\because △ FAG≌ △ DAG$,
$\therefore AF=AD=6$,
$\because$四边形$ABCD$是平行四边形,
$\therefore AD// BC$,$BC=AD=6$,
$\therefore ∠ BAD=180°-∠ ABC=180°-60°=120°$,
$\therefore ∠ FAE=\dfrac{1}{2}∠ BAD=60°$,
$\therefore ∠ FAE=∠ B=60°$,$\therefore △ ABE$为等边三角形,
$\therefore AB=AE=BE=BC-CE=6-2=4$,
$BF=AF-AB=6-4=2$.