$(1)$证明：如图$1$，

$\because AB=AC$，
$\therefore \angle ABC=\angle ACB$，
$\because \angle BAC+\angle ABC+\angle ACB=180^{\circ}$，
$\therefore \angle BAC+2\angle ACB=180^{\circ}$，
$\because \angle BFE=\angle ABH+\angle BAF$，$\angle ABH=\angle CAE$，
$\therefore \angle BFE=\angle CAE+\angle BAF=\angle BAC$，
$\therefore \angle BFE+2\angle ACB=180^{\circ}$，
$\because \angle BFE+\angle AFB=180^{\circ}$，
$\therefore \angle AFB=2\angle ACB$；
$(2)$证明：过点$A$作$AK\bot BF$于$K$，过点$C$作$CG\bot FH$交$FH$的延长线于$G$，作$CT\bot AE$交$AE$的延长线于$T$，

则$\angle AKB=\angle ATC=\angle CGH=\angle AKH=90^{\circ}$，
$\because FC$平分$\angle EFH$，$CG\bot FH$，$CT\bot FE$，
$\therefore CG=CT$，
在$\triangle ABK$和$\triangle CAT$中，
$\left\{\begin{array}{l}{∠AKB=∠ATC}\\{∠ABH=∠CAE}\\{AB=AC}\end{array}\right.$，
$\therefore \triangle ABK$≌$\triangle CAT\left(ASA\right)$，
$\therefore AK=CT$，
$\therefore AK=CG$，
在$\triangle AHK$和$\triangle CHG$中，
$\left\{\begin{array}{l}{∠AHK=∠CHG}\\{∠AKH=∠CGH}\\{AK=CG}\end{array}\right.$，
$\therefore \triangle AHK$≌$\triangle CHG\left(AAS\right)$，
$\therefore AH=CH$；
$(3)$如图$3$，过点$A$作$AK$∥$CD,$交$BH$于$K$，过点$C$作$CG$∥$AE$交$BD$于$G$，作$CM$∥$BD$交$AE$的延长线于点$M$，

则$\angle AKH=\angle CDH$，$\angle CFG=\angle FCM=\angle CFM=\angle FCG$，$\angle CGF=\angle CMF$，
$\therefore FG=FM=CG=CM$，
在$\triangle AKH$和$\triangle CDH$中，
$\left\{\begin{array}{l}{∠AKH=∠CDH}\\{∠AHK=∠CHD}\\{AH=CH}\end{array}\right.$，
$\therefore \triangle AKH$≌$\triangle CDH\left(AAS\right)$，
$\therefore KH=DH$，
$\because CG$∥$AE$，
$\therefore \angle FAH=\angle GCH$，
在$\triangle AFH$和$\triangle CGH$中，
$\left\{\begin{array}{l}{∠FAH=∠GCH}\\{AH=CH}\\{∠AHF=∠CHG}\end{array}\right.$，
$\therefore \triangle AFH$≌$\triangle CGH\left(ASA\right)$，
$\therefore AF=CG$，$FH=GH$，
$\therefore KH-FH=DH-GH$，
即$KF=DG$，
$\because CM$∥$BD,AK$∥$CD$，
$\therefore \angle ECM=\angle CBH$，$\angle KAH=\angle ACD$，
$\because \angle ACB=\angle ABC=\angle ABH+\angle CBH=\angle ACG+\angle ECM$，
$\therefore \angle GCM=2\angle ACB=2\angle EFC$，
$\therefore \angle ACB=\angle EFC$，
$\because \angle ABH+\angle AFB+\angle BAF=180^{\circ}$，
$\therefore \angle ABH+2\angle EFC+\angle BAK+\angle ACD-\angle CAE=180^{\circ}$，
即$2\angle EFC+\angle BAK+\angle ACD=180^{\circ}$，
$\because \angle ACD+3\angle EFC=180^{\circ}$，
$\therefore \angle BAK=\angle EFC$，
$\therefore \angle BAK=\angle ACB$，即$\angle BAK=\angle ACE$，
在$\triangle BAK$和$\triangle ACE$中，
$\left\{\begin{array}{l}{∠ABH=∠CAE}\\{AB=AC}\\{∠BAK=∠ACE}\end{array}\right.$，
$\therefore \triangle BAK$≌$\triangle ACE\left(ASA\right)$，
$\therefore BK=AE$，
$\because AE+DF=14$，$BH+AF=16$，
$\therefore AE+DG+2FH=14$，$AE+DG+3FH=16$，
$\therefore FH=2$.

$(1)$证明：如图$1$，

$\because AB=AC$，
$\therefore \angle ABC=\angle ACB$，
$\because \angle BAC+\angle ABC+\angle ACB=180^{\circ}$，
$\therefore \angle BAC+2\angle ACB=180^{\circ}$，
$\because \angle BFE=\angle ABH+\angle BAF$，$\angle ABH=\angle CAE$，
$\therefore \angle BFE=\angle CAE+\angle BAF=\angle BAC$，
$\therefore \angle BFE+2\angle ACB=180^{\circ}$，
$\because \angle BFE+\angle AFB=180^{\circ}$，
$\therefore \angle AFB=2\angle ACB$；
$(2)$证明：过点$A$作$AK\bot BF$于$K$，过点$C$作$CG\bot FH$交$FH$的延长线于$G$，作$CT\bot AE$交$AE$的延长线于$T$，

则$\angle AKB=\angle ATC=\angle CGH=\angle AKH=90^{\circ}$，
$\because FC$平分$\angle EFH$，$CG\bot FH$，$CT\bot FE$，
$\therefore CG=CT$，
在$\triangle ABK$和$\triangle CAT$中，
$\left\{\begin{array}{l}{∠AKB=∠ATC}\\{∠ABH=∠CAE}\\{AB=AC}\end{array}\right.$，
$\therefore \triangle ABK$≌$\triangle CAT\left(ASA\right)$，
$\therefore AK=CT$，
$\therefore AK=CG$，
在$\triangle AHK$和$\triangle CHG$中，
$\left\{\begin{array}{l}{∠AHK=∠CHG}\\{∠AKH=∠CGH}\\{AK=CG}\end{array}\right.$，
$\therefore \triangle AHK$≌$\triangle CHG\left(AAS\right)$，
$\therefore AH=CH$；
$(3)$如图$3$，过点$A$作$AK$∥$CD,$交$BH$于$K$，过点$C$作$CG$∥$AE$交$BD$于$G$，作$CM$∥$BD$交$AE$的延长线于点$M$，

则$\angle AKH=\angle CDH$，$\angle CFG=\angle FCM=\angle CFM=\angle FCG$，$\angle CGF=\angle CMF$，
$\therefore FG=FM=CG=CM$，
在$\triangle AKH$和$\triangle CDH$中，
$\left\{\begin{array}{l}{∠AKH=∠CDH}\\{∠AHK=∠CHD}\\{AH=CH}\end{array}\right.$，
$\therefore \triangle AKH$≌$\triangle CDH\left(AAS\right)$，
$\therefore KH=DH$，
$\because CG$∥$AE$，
$\therefore \angle FAH=\angle GCH$，
在$\triangle AFH$和$\triangle CGH$中，
$\left\{\begin{array}{l}{∠FAH=∠GCH}\\{AH=CH}\\{∠AHF=∠CHG}\end{array}\right.$，
$\therefore \triangle AFH$≌$\triangle CGH\left(ASA\right)$，
$\therefore AF=CG$，$FH=GH$，
$\therefore KH-FH=DH-GH$，
即$KF=DG$，
$\because CM$∥$BD,AK$∥$CD$，
$\therefore \angle ECM=\angle CBH$，$\angle KAH=\angle ACD$，
$\because \angle ACB=\angle ABC=\angle ABH+\angle CBH=\angle ACG+\angle ECM$，
$\therefore \angle GCM=2\angle ACB=2\angle EFC$，
$\therefore \angle ACB=\angle EFC$，
$\because \angle ABH+\angle AFB+\angle BAF=180^{\circ}$，
$\therefore \angle ABH+2\angle EFC+\angle BAK+\angle ACD-\angle CAE=180^{\circ}$，
即$2\angle EFC+\angle BAK+\angle ACD=180^{\circ}$，
$\because \angle ACD+3\angle EFC=180^{\circ}$，
$\therefore \angle BAK=\angle EFC$，
$\therefore \angle BAK=\angle ACB$，即$\angle BAK=\angle ACE$，
在$\triangle BAK$和$\triangle ACE$中，
$\left\{\begin{array}{l}{∠ABH=∠CAE}\\{AB=AC}\\{∠BAK=∠ACE}\end{array}\right.$，
$\therefore \triangle BAK$≌$\triangle ACE\left(ASA\right)$，
$\therefore BK=AE$，
$\because AE+DF=14$，$BH+AF=16$，
$\therefore AE+DG+2FH=14$，$AE+DG+3FH=16$，
$\therefore FH=2$.