1. 如图,在$△ ABC$中,$AB=AC$,$AD=DB$,$DE ⊥ AB$于点$E$,若$BC=10$,且$△ BDC$的周长为$24$,求$AE$的长.

1. $\because BC=10$,且$△ BDC$的周长为24,$\therefore BD+CD=24-10=14$.
$\because AD=BD$,$\therefore AD+DC = 14$,$\therefore AC = 14$.$\because AB = AC$,$\therefore AB = 14$.
$\because AD=DB$,$DE ⊥ AB$,$\therefore AE=\frac{1}{2}AB=7$.

2. 如图,在五边形 ABCDE 中,$∠B=∠E,∠BAE=120°$,点 F 为 CD 的中点,$AB=AE,BC=ED$,求$∠BAF$的度数.

2. 连接AC,AD.在$△ ABC$与$△ AED$中,$\begin{cases} AB=AE, \\ ∠B=∠E, \\ BC=ED, \end{cases}$
$\therefore △ ABC ≌ △ AED(\mathrm{SAS})$,$\therefore ∠BAC=∠DAE$,$AC=AD$.又$\because F$是边CD的中点,$\therefore AF ⊥ CD$,$AF$平分$∠CAD$.$\therefore ∠CAF = ∠DAF$,
$\therefore ∠BAC + ∠CAF = ∠EAD + ∠DAF$,即 $∠BAF = ∠EAF$,
$\therefore ∠BAF=\frac{1}{2}∠BAE=60°$.

3. 如图，$AB = AC$，$ST = SB$，$∠ TSB = ∠ SBC = ∠ BAC = 90°$，$M$ 为 $TC$ 的中点，求证：$SM ⊥ AM$，$SM = AM$。

3. 延长SM交BC于点N,连接AS,AN,$\because ∠TSB=∠SBC=90°$,$\therefore TS // BC$,$\therefore ∠STM=∠NCM$,$∠TSM=∠CNM$.$\because TM = MC$,
$\therefore △ TSM ≌ △ CNM (\mathrm{AAS})$,$\therefore SM = MN$,$TS = NC = BS$,$\therefore ∠SBT=45°=∠ABC=∠ACB$.$\because ∠SBC=90°$,$\therefore ∠SBA = 45° = ∠ACN$.在
$△ SBA$ 和 $△ NCA$ 中,$\begin{cases} SB=NC, \\ ∠SBA=∠NCA, \\ AB=AC, \end{cases}$
$\therefore △ SBA ≌ △ NCA (\mathrm{SAS})$,
$\therefore AS=AN$,$∠SAB=∠NAC$,$\therefore ∠SAN=∠BAC=90°$,$\therefore △ ASN$ 为等腰直角三角形.$\because M$ 为 $SN$ 的中点,$\therefore SM ⊥ AM$,$SM=AM$(等腰直角三角形的性质).