26. (14分)
【基础巩固】
(1) 如图①,在$\triangle ABC$中$,D$为$BC$上一点,连接$AD,E$为$AD$上一点,连接$CE$.若$\angle BAD=\angle ACE,CD=CE$,求证:$\triangle ABD \backsim \triangle CAE$.
【尝试应用】
(2) 如图②,在$□ ABCD$中,对角线$AC,BD$交于点$O,E$为$OC$上一点,连接$BE,\angle CBE=\angle DCO,BE=DO$.若$BD=12,OE=5$,求$AC$的长.
【拓展提升】
(3) 如图③,在菱形$ABCD$中,对角线$AC,BD$交于点$O,E$为$BC$的中点$,F$为$DC$上一点,连接$OE,AF,AE,\angle AEO=\angle CAF$.若$\frac{DF}{FC}=\frac{5}{3},AC=6$,求菱形$ABCD$的边长.
【基础巩固】
(1) 如图①,在$\triangle ABC$中$,D$为$BC$上一点,连接$AD,E$为$AD$上一点,连接$CE$.若$\angle BAD=\angle ACE,CD=CE$,求证:$\triangle ABD \backsim \triangle CAE$.
【尝试应用】
(2) 如图②,在$□ ABCD$中,对角线$AC,BD$交于点$O,E$为$OC$上一点,连接$BE,\angle CBE=\angle DCO,BE=DO$.若$BD=12,OE=5$,求$AC$的长.
【拓展提升】
(3) 如图③,在菱形$ABCD$中,对角线$AC,BD$交于点$O,E$为$BC$的中点$,F$为$DC$上一点,连接$OE,AF,AE,\angle AEO=\angle CAF$.若$\frac{DF}{FC}=\frac{5}{3},AC=6$,求菱形$ABCD$的边长.
答案：
26.(1)
∵CD = CE,
∴∠CDE = ∠CED.
∴180° - ∠CDE = 180° - ∠CED.
∴∠ADB = ∠CEA.
∵∠BAD = ∠ACE,
∴△ABD∽△CAE (2)
∵四边形ABCD是平行四边形，
∴$BO = DO = \frac{1}{2}BD = \frac{1}{2}×12 = 6.$
∴BE = BO = 6.
∴∠BEO = ∠BOE.
∴180° - ∠BEO = 180° - ∠BOE.
∴∠BEC = ∠COD.
∵∠CBE = ∠DCO,
∴△BEC∽△COD.
∴$\frac{BE}{CO} = \frac{CE}{DO}.$设OC = x,则CE = OC - OE = x - 5.
∴$\frac{6}{x} = \frac{x - 5}{6},$解得x₁ = 9,x₂ = -4(舍去).
∴OC = 9.
∴AC = 2OC = 18 (3)如图，延长AF，BC交于点G.
∵$\frac{DF}{FC} = \frac{5}{3},$设DF = 5t,FC = 3t,则CD = 8t.
∵四边形ABCD是菱形，
∴$AB = AD = BC = CD = 8t,AD//BC,AO = \frac{1}{2}AC = \frac{1}{2}×6 = 3,AC⊥BD.$
∴△CGF∽△DAF.
∴$\frac{CG}{DA} = \frac{CF}{DF},$即$\frac{CG}{8t} = \frac{3}{5}.$
∴$CG = \frac{24}{5}t.$在Rt△BOC中，
∵E为BC的中点，
∴$OE = CE = \frac{1}{2}BC = 4t.$
∴∠COE = ∠ACE.
∴易得∠AOE = ∠GCA.
∵∠AEO = ∠CAF,
∴△AOE∽△GCA.
∴$\frac{OE}{CA} = \frac{OA}{CG}，$即$\frac{4t}{6} = \frac{3}{\frac{24}{5}t}，$解得$t₁ = \frac{√{15}}{4}，$$t₂ = -\frac{√{15}}{4}($舍去).
∴AB = AD = BC = CD = 8t = 2√{15}，即菱形ABCD的边长为2√{15}
26.(1)
∵CD = CE,
∴∠CDE = ∠CED.
∴180° - ∠CDE = 180° - ∠CED.
∴∠ADB = ∠CEA.
∵∠BAD = ∠ACE,
∴△ABD∽△CAE (2)
∵四边形ABCD是平行四边形，
∴$BO = DO = \frac{1}{2}BD = \frac{1}{2}×12 = 6.$
∴BE = BO = 6.
∴∠BEO = ∠BOE.
∴180° - ∠BEO = 180° - ∠BOE.
∴∠BEC = ∠COD.
∵∠CBE = ∠DCO,
∴△BEC∽△COD.
∴$\frac{BE}{CO} = \frac{CE}{DO}.$设OC = x,则CE = OC - OE = x - 5.
∴$\frac{6}{x} = \frac{x - 5}{6},$解得x₁ = 9,x₂ = -4(舍去).
∴OC = 9.
∴AC = 2OC = 18 (3)如图，延长AF，BC交于点G.
∵$\frac{DF}{FC} = \frac{5}{3},$设DF = 5t,FC = 3t,则CD = 8t.
∵四边形ABCD是菱形，
∴$AB = AD = BC = CD = 8t,AD//BC,AO = \frac{1}{2}AC = \frac{1}{2}×6 = 3,AC⊥BD.$
∴△CGF∽△DAF.
∴$\frac{CG}{DA} = \frac{CF}{DF},$即$\frac{CG}{8t} = \frac{3}{5}.$
∴$CG = \frac{24}{5}t.$在Rt△BOC中，
∵E为BC的中点，
∴$OE = CE = \frac{1}{2}BC = 4t.$
∴∠COE = ∠ACE.
∴易得∠AOE = ∠GCA.
∵∠AEO = ∠CAF,
∴△AOE∽△GCA.
∴$\frac{OE}{CA} = \frac{OA}{CG}，$即$\frac{4t}{6} = \frac{3}{\frac{24}{5}t}，$解得$t₁ = \frac{√{15}}{4}，$$t₂ = -\frac{√{15}}{4}($舍去).
∴AB = AD = BC = CD = 8t = 2√{15}，即菱形ABCD的边长为2√{15}