$(1)$证明：$\because AB$∥$DF,AC$∥$DE$，
$\therefore$四边形$AEDF$是平行四边形.
$\because AD$是$\triangle ABC$的角平分线，
$\therefore \angle BAD=\angle DAC$.
又$\because AC$∥$DE$，
$\therefore \angle ADE=\angle DAC$.
$\therefore \angle ADE=\angle BAD$.
$\therefore EA=ED$.
$\therefore$四边形$AEDF$是菱形.
$(2)$连接$EF$交$AD$于点$O$.

$\because$四边形$AEDF$是菱形，
$\therefore EF=2FO$.
$\therefore AO=\frac{1}{2}AD=12$.
$\because AD\bot EF$.
在$Rt\triangle AOF$中，由勾股定理得$OF=\sqrt{A{F}^{2}-A{O}^{2}}=\sqrt{1{3}^{2}-1{2}^{2}}=5$.
$\therefore OE=OF=5$.
$\therefore$四边形$AEDF$的面积$=\frac{1}{2}AD×OF+\frac{1}{2}AD×OE=\frac{1}{2}×24×5+\frac{1}{2}×24×5=120$.

$(1)$证明：$\because AB$∥$DF,AC$∥$DE$，
$\therefore$四边形$AEDF$是平行四边形.
$\because AD$是$\triangle ABC$的角平分线，
$\therefore \angle BAD=\angle DAC$.
又$\because AC$∥$DE$，
$\therefore \angle ADE=\angle DAC$.
$\therefore \angle ADE=\angle BAD$.
$\therefore EA=ED$.
$\therefore$四边形$AEDF$是菱形.
$(2)$连接$EF$交$AD$于点$O$.

$\because$四边形$AEDF$是菱形，
$\therefore EF=2FO$.
$\therefore AO=\frac{1}{2}AD=12$.
$\because AD\bot EF$.
在$Rt\triangle AOF$中，由勾股定理得$OF=\sqrt{A{F}^{2}-A{O}^{2}}=\sqrt{1{3}^{2}-1{2}^{2}}=5$.
$\therefore OE=OF=5$.
$\therefore$四边形$AEDF$的面积$=\frac{1}{2}AD×OF+\frac{1}{2}AD×OE=\frac{1}{2}×24×5+\frac{1}{2}×24×5=120$.