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中考数学二轮复习:几何辅助线、二次函数压轴题全解__参考答案 |
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如图2,由(1)知AP=AD,AB=AC,∠BAP=∠CAD=β, ∴△ABP≌△ACD,∴∠ABE=∠ACD.∵AC⊥BD,∴∠GDN=90°-β, ∵GN=GD,∴∠GND=∠GDN=90°-β, ∴∠NGD=180°-∠GND-∠GDN=2β.∴∠AGF=∠NGD=2β. ∴∠AFG=∠BAD-∠AGF=3β-2β=β. ∵FN平分∠BFM,∴∠NFM=∠AFG=β,∴FM∥AE,∴∠FMN=90°. ∵H为BF的中点,∴BF=2MH. 在FB上截取FR=FM,连接RM, ∴∠FRM=∠FMR=90°-β.∵∠ABC=90°-β,∴∠FRM=∠ABC,∴RM∥BC, ∴∠CBD=∠RMB.∵∠CAD=∠CBD=β,∴∠RMB=∠CAD. BRBMBM33 ∵∠RBM=∠ACD,∴△RMB∽△DAC,∴===,∴BR=CD. CDACAB44 333 ∵BR=FB-FM,∴FB-FM=BR=CD,FB=FM+CD.∴2MH=FM+CD. 444 【方法二】 如图3,由(1)得AP=AD,AB=AC,∠BAP=∠CAD=β, ∴△ABP≌△ACD,∴∠ABE=∠ACD.∵AC⊥BD,∴∠GDN=90°-β, ∵GN=GD,∴∠GND=∠GDN=90°-β, ∴∠NGD=180°-∠GND-∠GDN=2β.∴∠AGF=∠NGD=2β. ∴∠AFG=∠BAD-∠AGF=3β-2β=β. ∵FN平分∠BFM,∴∠NFM=∠AFG=β,∴FM∥AE,∴△ABE∽△FMB, ABAEBE ∴==,∠FMN=90°,∵H为BF的中点,∴BF=2MH. FBFMBM 过点M作MK⊥BA于点K,则∠MKB=∠AEB=90°, 设AB=4x,AE=4y,则AC=4x,CE=4(x-y), 2222 ∴在在Rt△ABE中,BE=AB-AE=4x-y, ∵BM:AB=3:4,∴BM=3x,∵∠MBK=∠ABE,∴△MKB∽△AEB, 2 ABAEBE3xy3x 22 ∴==,∴MK=3y,MK=3x-y,∴FM=,BF=, 2222 MBMKBK x-yx-y 2 BABE4(x-xy) 又∵∠AEB=∠DEC,∴△AEB∽△DEC,∴=,∴CD=, 22 CDCE x-y 22 3x3xy34(x-xy)3 ∴2MH=BF==+2=FM+CD. 222222 44 x-yx-yx-y FF A AA GG HH K
R D B E P BDBD EMEM PNPN C CC 100/100
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