8. 如图，已知$AB = 12$，$AB\perp BC$于点$B$，$AB\perp AD$于点$A$，$AD = 5$，$BC = 10$。点$E$是$CD$的中点，求$AE$的长。
答案：
解：如答图，延长 AE 交 BC 于点 F.
$\because AB \perp BC, AB \perp AD, \therefore AD // BC,$
$\therefore \angle D = \angle C, \angle DAE = \angle CFE.$
又∵点 E 是 CD 的中点，
$\therefore DE = CE.$
在$\triangle AED$与$\triangle FEC$中，
$\left\{\begin{array}{l} \angle D = \angle C, \\ \angle DAE = \angle CFE, \\ DE = CE, \end{array}\right.$
$\therefore \triangle AED \cong \triangle FEC(AAS),$
$\therefore AE = FE, AD = FC.$
$\because AD = 5, BC = 10, \therefore BF = 5.$
在$Rt\triangle ABF$中，$AB^2 + BF^2 = AF^2, \therefore AF = 13,$
$\therefore AE = \frac{1}{2}AF = 6.5.$
解：如答图，延长 AE 交 BC 于点 F.
$\because AB \perp BC, AB \perp AD, \therefore AD // BC,$
$\therefore \angle D = \angle C, \angle DAE = \angle CFE.$
又∵点 E 是 CD 的中点，
$\therefore DE = CE.$
在$\triangle AED$与$\triangle FEC$中，
$\left\{\begin{array}{l} \angle D = \angle C, \\ \angle DAE = \angle CFE, \\ DE = CE, \end{array}\right.$
$\therefore \triangle AED \cong \triangle FEC(AAS),$
$\therefore AE = FE, AD = FC.$
$\because AD = 5, BC = 10, \therefore BF = 5.$
在$Rt\triangle ABF$中，$AB^2 + BF^2 = AF^2, \therefore AF = 13,$
$\therefore AE = \frac{1}{2}AF = 6.5.$
9. （2023·苏州期中）如图，公路上$A$，$B$两点相距50 km，$C$，$D$为两村庄，$DA\perp AB$于点$A$，$CB\perp AB$于点$B$，已知$DA = 30$km，$CB = 20$km。现要在公路$AB$上建一个土特产市场$E$，使得$C$，$D$两村庄到市场$E$的距离相等，则市场$E$应建在距$A$点
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千米处？判断此时$\triangle DEC$的形状，并请说明理由。答案：解：设$AE = x km$，则$BE = (50 - x)km,$
$\because DA \perp AB, CB \perp AB, \therefore \angle A = \angle B = 90^{\circ}.$
在$Rt\triangle ADE$中，$DE^2 = 30^2 + x^2,$
在$Rt\triangle CBE$中，$CE^2 = 20^2 + (50 - x)^2,$
$\because DE = CE, \therefore 30^2 + x^2 = 20^2 + (50 - x)^2,$
解得$x = 20$，即$AE = 20 km.$
答：市场 E 应建在距 A 点 20 km 的位置.
此时$\triangle DEC$是等腰直角三角形.
理由：可知$BE = 50 - 20 = 30(km),$
$\because AE = BC, AD = BE, \angle A = \angle B = 90^{\circ},$
$\therefore \triangle ADE \cong \triangle BEC(SAS),$
$\therefore \angle AED = \angle BCE,$
$\because \angle BCE + \angle BEC = 90^{\circ},$
$\therefore \angle AED + \angle BEC = 90^{\circ},$
$\therefore \angle DEC = 90^{\circ},$
$\because DE = CE, \therefore \triangle DEC$是等腰直角三角形.
$\because DA \perp AB, CB \perp AB, \therefore \angle A = \angle B = 90^{\circ}.$
在$Rt\triangle ADE$中，$DE^2 = 30^2 + x^2,$
在$Rt\triangle CBE$中，$CE^2 = 20^2 + (50 - x)^2,$
$\because DE = CE, \therefore 30^2 + x^2 = 20^2 + (50 - x)^2,$
解得$x = 20$，即$AE = 20 km.$
答：市场 E 应建在距 A 点 20 km 的位置.
此时$\triangle DEC$是等腰直角三角形.
理由：可知$BE = 50 - 20 = 30(km),$
$\because AE = BC, AD = BE, \angle A = \angle B = 90^{\circ},$
$\therefore \triangle ADE \cong \triangle BEC(SAS),$
$\therefore \angle AED = \angle BCE,$
$\because \angle BCE + \angle BEC = 90^{\circ},$
$\therefore \angle AED + \angle BEC = 90^{\circ},$
$\therefore \angle DEC = 90^{\circ},$
$\because DE = CE, \therefore \triangle DEC$是等腰直角三角形.
10. 如图，在$\triangle ABC$中，$\angle ABC = 45^{\circ}$，$CD\perp AB$，$BE\perp AC$，垂足分别为$D$，$E$，$F$为$BC$的中点，$BE$与$DF$，$DC$分别交于点$G$，$H$，$\angle ABE = \angle CBE$。
（1）线段$BH$与$AC$相等吗？若相等，请给予证明；若不相等，请说明理由。
解：$BH = AC$. 证明：$\because CD \perp AB, BE \perp AC,$
$\therefore \angle BDH = \angle BEC = \angle CDA = 90^{\circ},$
$\therefore \angle A + \angle ACD = 90^{\circ}, \angle A + \angle HBD = 90^{\circ},$
$\therefore \angle HBD = \angle ACD.$
$\because \angle ABC = 45^{\circ},$
$\therefore \angle BCD = 180^{\circ} - 90^{\circ} - 45^{\circ} = 45^{\circ} = \angle ABC,$
$\therefore DB = DC.$
在$\triangle DBH$和$\triangle DCA$中，$\left\{\begin{array}{l} \angle BDH = \angle CDA, \\ BD = CD, \\ \angle HBD = \angle ACD, \end{array}\right.$
$\therefore \triangle DBH \cong \triangle DCA(ASA), \therefore BH = AC.$
（2）求证：$BG^2 - GE^2 = EA^2$。
证明：连接 CG，由(1)知，$DB = CD,$
$\because F$为 BC 的中点，$\therefore DF$垂直平分 BC，$\therefore BG = CG,$
$\because BE \perp AC, \therefore \angle AEB = \angle CEB = 90^{\circ},$
在$\triangle ABE$和$\triangle CBE$中，$\left\{\begin{array}{l} \angle ABE = \angle CBE, \\ BE = BE, \\ \angle AEB = \angle CEB, \end{array}\right.$
$\therefore \triangle ABE \cong \triangle CBE(ASA),$
$\therefore EC = EA,$
在$Rt\triangle CGE$中，由勾股定理得$CG^2 - GE^2 = CE^2,$
$\because CE = AE, BG = CG,$
$\therefore BG^2 - GE^2 = EA^2.$
（1）线段$BH$与$AC$相等吗？若相等，请给予证明；若不相等，请说明理由。
解：$BH = AC$. 证明：$\because CD \perp AB, BE \perp AC,$
$\therefore \angle BDH = \angle BEC = \angle CDA = 90^{\circ},$
$\therefore \angle A + \angle ACD = 90^{\circ}, \angle A + \angle HBD = 90^{\circ},$
$\therefore \angle HBD = \angle ACD.$
$\because \angle ABC = 45^{\circ},$
$\therefore \angle BCD = 180^{\circ} - 90^{\circ} - 45^{\circ} = 45^{\circ} = \angle ABC,$
$\therefore DB = DC.$
在$\triangle DBH$和$\triangle DCA$中，$\left\{\begin{array}{l} \angle BDH = \angle CDA, \\ BD = CD, \\ \angle HBD = \angle ACD, \end{array}\right.$
$\therefore \triangle DBH \cong \triangle DCA(ASA), \therefore BH = AC.$
（2）求证：$BG^2 - GE^2 = EA^2$。
证明：连接 CG，由(1)知，$DB = CD,$
$\because F$为 BC 的中点，$\therefore DF$垂直平分 BC，$\therefore BG = CG,$
$\because BE \perp AC, \therefore \angle AEB = \angle CEB = 90^{\circ},$
在$\triangle ABE$和$\triangle CBE$中，$\left\{\begin{array}{l} \angle ABE = \angle CBE, \\ BE = BE, \\ \angle AEB = \angle CEB, \end{array}\right.$
$\therefore \triangle ABE \cong \triangle CBE(ASA),$
$\therefore EC = EA,$
在$Rt\triangle CGE$中，由勾股定理得$CG^2 - GE^2 = CE^2,$
$\because CE = AE, BG = CG,$
$\therefore BG^2 - GE^2 = EA^2.$
答案：(1) 解：$BH = AC$. 证明：$\because CD \perp AB, BE \perp AC,$
$\therefore \angle BDH = \angle BEC = \angle CDA = 90^{\circ},$
$\therefore \angle A + \angle ACD = 90^{\circ}, \angle A + \angle HBD = 90^{\circ},$
$\therefore \angle HBD = \angle ACD.$
$\because \angle ABC = 45^{\circ},$
$\therefore \angle BCD = 180^{\circ} - 90^{\circ} - 45^{\circ} = 45^{\circ} = \angle ABC,$
$\therefore DB = DC.$
在$\triangle DBH$和$\triangle DCA$中，$\left\{\begin{array}{l} \angle BDH = \angle CDA, \\ BD = CD, \\ \angle HBD = \angle ACD, \end{array}\right.$
$\therefore \triangle DBH \cong \triangle DCA(ASA), \therefore BH = AC.$
(2) 证明：连接 CG，由(1)知，$DB = CD,$
$\because F$为 BC 的中点，$\therefore DF$垂直平分 BC，$\therefore BG = CG,$
$\because BE \perp AC, \therefore \angle AEB = \angle CEB = 90^{\circ},$
在$\triangle ABE$和$\triangle CBE$中，$\left\{\begin{array}{l} \angle ABE = \angle CBE, \\ BE = BE, \\ \angle AEB = \angle CEB, \end{array}\right.$
$\therefore \triangle ABE \cong \triangle CBE(ASA),$
$\therefore EC = EA,$
在$Rt\triangle CGE$中，由勾股定理得$CG^2 - GE^2 = CE^2,$
$\because CE = AE, BG = CG,$
$\therefore BG^2 - GE^2 = EA^2.$
$\therefore \angle BDH = \angle BEC = \angle CDA = 90^{\circ},$
$\therefore \angle A + \angle ACD = 90^{\circ}, \angle A + \angle HBD = 90^{\circ},$
$\therefore \angle HBD = \angle ACD.$
$\because \angle ABC = 45^{\circ},$
$\therefore \angle BCD = 180^{\circ} - 90^{\circ} - 45^{\circ} = 45^{\circ} = \angle ABC,$
$\therefore DB = DC.$
在$\triangle DBH$和$\triangle DCA$中，$\left\{\begin{array}{l} \angle BDH = \angle CDA, \\ BD = CD, \\ \angle HBD = \angle ACD, \end{array}\right.$
$\therefore \triangle DBH \cong \triangle DCA(ASA), \therefore BH = AC.$
(2) 证明：连接 CG，由(1)知，$DB = CD,$
$\because F$为 BC 的中点，$\therefore DF$垂直平分 BC，$\therefore BG = CG,$
$\because BE \perp AC, \therefore \angle AEB = \angle CEB = 90^{\circ},$
在$\triangle ABE$和$\triangle CBE$中，$\left\{\begin{array}{l} \angle ABE = \angle CBE, \\ BE = BE, \\ \angle AEB = \angle CEB, \end{array}\right.$
$\therefore \triangle ABE \cong \triangle CBE(ASA),$
$\therefore EC = EA,$
在$Rt\triangle CGE$中，由勾股定理得$CG^2 - GE^2 = CE^2,$
$\because CE = AE, BG = CG,$
$\therefore BG^2 - GE^2 = EA^2.$