4. 已知在$△ ABC$中,$AB=AC$,$∠ B=α$.
(1) 如图①,当$α=30°$时,过点$A$作$AD⊥ AB$交$BC$于点$D$,若$AD=4\ \mathrm{cm}$,则$BC$的长为$\mathrm{cm}$;
(2) 如图②,当$α=45°$时,过点$B$作$BD$平分$∠ ABC$交$AC$于点$D$,过点$C$作$CE⊥ BD$交$BD$的延长线于点$E$,求证:$BD=2CE$;
(3) 当$0°＜α＜90°$时,$AB=4$,$BC=5$,$BE$为$∠ ABC$的平分线,$CE⊥ BE$于点$E$,连接$AE$,若$S_{△ ABC}=m$,则$△ ACE$的面积为.(用含$m$的式子表示)

(1)12 解析:$\because AB=AC,\therefore ∠ B=∠ C=30°.\because AB⊥ AD,AD=4\ \mathrm{cm},\therefore ∠ BAD=90°,BD=2AD=2×4=8(\mathrm{cm}),\therefore ∠ B+∠ ADB=90°,\therefore ∠ ADB=60°.\because ∠ ADB=∠ DAC+∠ C=60°,\therefore ∠ DAC=30°,\therefore ∠ DAC=∠ C,\therefore DC=AD=4\ \mathrm{cm},$
$\therefore BC=BD+DC=8+4=12(\mathrm{cm}).$
(2)延长 CE 与 BA 的延长线交于点 F,如图①.$\because ∠ ABC=45°,AB=AC,\therefore ∠ BAC=90°.\because CE⊥ BD,\therefore ∠ BAC=∠ DEC.$$\because ∠ ADB=∠ CDE,\therefore ∠ ABD=∠ DCE$.在$△ BAD$和$△ CAF$
$\mathrm{中,}\begin{cases} ∠ BAD=∠ CAF, \\ AB=AC, \\ ∠ ABD=∠ ACF, \end{cases}\therefore △ BAD≌△ CAF(\mathrm{ASA}),\therefore BD=CF.$
$\because BD$平分$∠ ABC,CE⊥ DB,\therefore ∠ FBE=∠ CBE$.在$△ BEF$和
$△ BEC\mathrm{中,}\begin{cases} ∠ FBE=∠ CBE, \\ BE=BE, \\ ∠ BEF=∠ BEC, \end{cases}\therefore △ BEF≌△ BEC(\mathrm{ASA}),$
$\therefore CE=EF,\therefore BD=2CE.$



(3)$\frac{1}{8}m$ 解析:延长 CE 与 BA 的延长线交于点 F,作$CH⊥ AB$于点 H,如图②,由(2)可知$△ BEF≌△ BEC,$$\therefore CE=FE,BC=BF=5,\therefore S_{△ ACE}=S_{△ AFE}=\frac{1}{2}S_{△ ACF}$.又$\because AB=4,\therefore AF=1.\because S_{△ ABC}=m$,即$\frac{1}{2}AB· CH=m,\therefore CH=\frac{1}{2}m,$$\therefore S_{△ ACF}=\frac{1}{2}AF· CH=\frac{1}{2}×1×\frac{1}{2}m=\frac{1}{4}m,\therefore S_{△ ACE}=\frac{1}{2}S_{△ ACF}=\frac{1}{8}m.$

5. 如图,在$△ ABC$中,$∠ BAC=120°,AD ⊥ BC$于点$D$,且$AB+BD=DC$,求$∠ C$的度数.(用截长法和补短法两种方法解答)

方法1:(截长法)在 CD 上取点 E,使$DE=BD$,连接 AE,则$CE=AB=AE,\therefore ∠ B=∠ AED=∠ C+∠ CAE=2∠ C.$$\because ∠ BAC=120°,\therefore ∠ B+∠ C=2∠ C+∠ C=60°,\therefore ∠ C=20°.$
方法2:(补短法)延长 DB 至点 F,使$BF=AB$,连接 AF,则$AB+BD=DF=CD,\therefore AF=AC,∠ C=∠ F=\frac{1}{2}∠ ABC.$$\because ∠ BAC=120°,\therefore ∠ ABC+∠ C=∠ ABC+\frac{1}{2}∠ ABC=60°,$$\therefore ∠ ABC=40°,\therefore ∠ C=20°.$
归纳总结 利用“截长补短法”构造等腰三角形:基本图形1:如图①,在$△ ABC$中,$∠ C=36°,CA=CB,∠1=∠2$,则$CD=AD=AB$.基本图形 2: 如图②,在$△ ABC$中,$∠ C=90°,AC=BC,∠1=∠2,DE⊥ AB$于点 E,则$AC=BC=AE.$

6. 在 $△ ABC$ 中, $∠ ACB=2∠ B, BC=2AC$, 求证:
$∠ A=90°.$

如图①,在 BC 的延长线上截取$CH=AC$,在 BC 上截取$CE=CA.\because BC=2AC,\therefore BE=CE=AC.\because AC=CH,\therefore ∠ H=∠ CAH,\therefore ∠ ACB=∠ H+∠ CAH=2∠ H$,且$∠ ACB=2∠ B,$$\therefore ∠ H=∠ B,\therefore AH=AB$.又$HC=BE,\therefore △ AHC≌△ ABE$$(\mathrm{SAS}),\therefore AE=AC,\therefore AE=AC=CE,\therefore △ ACE$是等边三角形,$\therefore ∠ ACB=60°,\therefore ∠ B=30°,\therefore ∠ BAC=90°.$



一题多解 如图②,作$∠ ACB$的平分线 CD 交 AB 于点 D,过点 D 作$DE⊥ BC$于点 E.$\because ∠ ACB=2∠ B,\therefore ∠ B=∠ BCD=\frac{1}{2}∠ ACB,\therefore BD=CD,\therefore BE=CE=\frac{1}{2}BC.\because BC=2AC,\therefore AC=CE.$
$\mathrm{在}△ ACD\mathrm{和}△ ECD\mathrm{中,}\begin{cases} AC=EC, \\ ∠ ACD=∠ ECD, \\ CD=CD, \end{cases}\therefore △ ACD≌$
$△ ECD(\mathrm{SAS}),\therefore ∠ A=∠ CED=90°.$