滬科版初中九年級(jí)數(shù)學(xué)上冊(cè)專(zhuān)項(xiàng)素養(yǎng)鞏固訓(xùn)練卷(五)證比例式或等積式的技巧練課件

專(zhuān)項(xiàng)素養(yǎng)鞏固訓(xùn)練卷(五)證比例式或等積式的技巧(練方法)類(lèi)型一構(gòu)造平行線法1.(★☆☆)如圖,在△ABC中,AD為BC邊上的中線,過(guò)C任意作一條直線交AD于E,

交AB于F,求證:AE∶ED=2AF∶FB.

證明:如圖,過(guò)點(diǎn)D作DN∥CF,交AB于點(diǎn)N.∵DC=DB,∴FN=NB=

FB,∵DN∥CF,∴AE∶ED=AF∶FN,即AE∶ED=AF∶

FB,∴AE∶ED=2AF∶FB.

2.(2024安徽宣城期末,20,★★☆)如圖,在△ABC中,點(diǎn)D,E分別在AB,AC上,DE∥

BC,點(diǎn)F在邊AB上,BC2=BF·BA,CF與DE相交于點(diǎn)G.

(1)求證:△BAC∽△DGF;(2)當(dāng)點(diǎn)E為AC的中點(diǎn)時(shí),求證:

=

.

證明:(1)∵BC2=BF·BA,∴BC∶BF=BA∶BC,∵∠ABC=∠CBF,∴△BAC∽△BCF,∵DE∥BC,∴△BCF∽△DGF,∴△BAC∽△DGF.對(duì)應(yīng)目標(biāo)編號(hào)M9122005(2)過(guò)點(diǎn)A作AH∥BC交CF的延長(zhǎng)線于H,如圖,∵DE∥BC,∴AH∥DE,∵點(diǎn)E為AC的中點(diǎn),∴AH=2EG,∵AH∥DG,∴△AHF∽△DGF,∴

=

,∴

=

.

類(lèi)型二三點(diǎn)定型法3.[一題多解](★☆☆)如圖,四邊形ABCD的對(duì)角線AC,BD交于點(diǎn)F,點(diǎn)E是BD上一

點(diǎn),且∠BAC=∠BDC=∠DAE.求證:

=

.

證明:證法一:∵∠BAC=∠DAE,∴∠BAC+∠CAE=∠DAE+∠CAE,即∠BAE=∠CAD.∵∠BAC=∠BDC,∠BFA=∠CFD,∴180°-∠BAC-∠BFA=180°-∠BDC-∠CFD,即∠ABE=∠ACD,∴△ABE∽△ACD,∴

=

.證法二:∵∠BAC=∠DAE,∴∠BAC+∠CAE=∠DAE+∠CAE,即∠BAE=∠CAD.又∵∠BEA=∠DAE+∠ADE,∠ADC=∠BDC+∠ADE,∠DAE=∠BDC,∴∠AEB=∠ADC.∴△ABE∽△ACD,∴

=

.4.(★☆☆)如圖,△ABC中,∠BAC=90°.M為BC的中點(diǎn),DM⊥BC交CA的延長(zhǎng)線于

D,交AB于E.求證:AM2=MD·ME.

證明:∵∠BAC=90°,M為BC的中點(diǎn),∴AM=BM=CM,∴∠B=∠BAM,∵∠B+∠C=90°,∴∠BAM+∠C=90°,∵∠C+∠D=90°,∴∠BAM=∠D,∵∠AME=∠DMA,∴△AME∽△DMA,∴

=

,∴AM2=MD·ME.類(lèi)型三構(gòu)造相似三角形法5.(★☆☆)如圖,在等邊三角形ABC中,點(diǎn)P是BC邊上任意一點(diǎn),AP的垂直平分線

分別交AB,AC于點(diǎn)M,N.求證:BP·CP=BM·CN.

證明:如圖,連接PM,PN,∵M(jìn)N垂直平分AP,∴AM=MP,AN=PN,又∵M(jìn)N為公共邊,∴△AMN≌△PMN(SSS),∴∠MPN=∠BAC=60°,∵∠BPM+∠CPN=120°,∠BPM+∠BMP=120°,∴∠BMP=∠CPN,又∵∠B=∠C=60°,∴△MPB∽△PNC,∴

=

,即BP·CP=BM·NC.

6.(新獨(dú)家原創(chuàng),★★☆)如圖,F為正方形ABCD的邊AB的中點(diǎn),E是AD上的一點(diǎn),

AE=

AD,FG⊥CE于G.求證:FG2=EG·CG.

證明:如圖,連接EF,CF.∵AE=

AD,AF=BF=

AB,四邊形ABCD為正方形,∴

=

=

,∵∠A=∠B=90°,∴△EFA∽△FCB,∴∠AFE=∠BCF.∵∠BFC+∠BCF=90°,∴∠AFE+∠BFC=90°,∴∠EFC=90°,∴∠EFG+∠CFG=90°.對(duì)應(yīng)目標(biāo)編號(hào)M9122005又∵FG⊥CE,∴∠EFG+∠FEG=90°,∴∠CFG=∠FEG.∵∠EGF=∠CGF=90°,∴△EFG∽△FCG,∴

=

,∴FG2=EG·CG.

類(lèi)型四等比或等積代換法7.(2024安徽滁州定遠(yuǎn)期末,19,★☆☆)如圖,點(diǎn)E為?ABCD的邊CD延長(zhǎng)線上的

一點(diǎn),連接BE交AC于點(diǎn)O,交AD于點(diǎn)F.(1)求證:

=

;(2)求證:BO2=EO·FO.

證明:(1)∵四邊形ABCD是平行四邊形,∴AB∥CD,AB=CD,∴△AOB∽△COE,∴

=

,∴

=

.(2)∵△COE∽△AOB,∴

=

,∵AD∥BC,∴△COB∽△AOF,∴

=

,∴

=

,即OB2=OF·OE.8.(★☆☆)如圖,在△ABC中,AB=AC,DE∥BC,點(diǎn)F在邊AC上,DF與BE相交于點(diǎn)G,

且∠EDF=∠ABE.求證:

(1)△DEF∽△BDE;(2)△GDE∽△EDF;(3)DG·DF=DB·EF.

證明:(1)∵AB=AC,∴∠ABC=∠ACB,∵DE∥BC,∴∠ABC+∠BDE=180°,∠ACB+∠CED=180°.∴∠BDE=∠CED,對(duì)應(yīng)目標(biāo)編號(hào)M9122005∵∠EDF=∠ABE,∴△DEF∽△BDE.(2)∵△DEF∽△BDE,∴∠BED=∠DFE.∵∠GDE=∠EDF,∴△GDE∽△EDF.(3)由(2)知,△GDE∽△EDF,∴

=

,∴DE2=DG·DF,∵△DEF∽△BDE,∴

=

,∴DE2=DB·EF,∴DG·DF=DB·EF.9.(★☆☆)如圖,D是△ABC的邊AB上一點(diǎn),DE∥BC,交邊AC于點(diǎn)E,延長(zhǎng)DE至點(diǎn)

F,使EF=DE,連接BF,交邊AC于點(diǎn)G,連接CF.(1)求證:

=

;(2)如果CF2=FG·FB,求證:CG·CE=BC·DE.

證明:(1)∵DE∥BC,∴△ADE∽△ABC,△EFG∽△CBG,∴

=

,

=

,又∵DE=EF,∴

=

,∴

=

.(2)∵CF2=FG·FB,∴

=

,又∵∠CFG=∠CFB,∴△CFG∽△BFC,∴

=

,∠FCE=∠CBF,又∵DF∥BC,∴∠EFG=∠CBF,∴∠FCE=∠EFG,又∵∠FEG=∠CEF,∴△EFG∽△ECF,∴

=

=

,∴

=

,即CG·CE=BC·DE.10.(2024安徽合肥廬陽(yáng)期末,21,★★☆)如圖,在△ABC中,點(diǎn)D,E分別在邊AB,AC

上,ED、CB的延長(zhǎng)線相交于點(diǎn)F.(1)如圖①,若∠FBD=∠FEC,BF=4,FD=5,FE=8,求FC的長(zhǎng);(2)如圖②,若BD=CE,求證:

=

.

解析

(1)∵∠FBD=∠FEC,∠BFD=∠EFC,∴△FBD∽△FEC,∴FB∶FE=FD∶

FC,即4∶8=5∶FC,解得FC=10.(2)證明:過(guò)點(diǎn)D作DM∥AC交FC于點(diǎn)M,如圖,∵DM∥AC,∴△BDM∽△BAC,∴

=

,∴

=

,∵BD=CE,∴

=

.∵DM∥CE,∴△FCE∽△FMD,∴

=

,∴

=

.

類(lèi)型五等線段代換法11.(2024安徽六安裕安中學(xué)月考,19,★☆☆)如圖,直線DN平行于△ABC的中線

AF交AB于點(diǎn)D,交CA的延長(zhǎng)線于點(diǎn)E,交BC于點(diǎn)N,求證:

=

.

證明:由DN∥AF易得

=

,

=

,∵在△ABC中,AF是BC邊上的中線,∴FB=FC,∴

=

.12.(★☆☆)如圖,在等腰三角形ABC中,AB=AC,AD⊥BC于點(diǎn)D,P是AD上一點(diǎn),CF∥AB,延長(zhǎng)BP交AC于點(diǎn)E,交CF于點(diǎn)F.求證:BP2=PE·PF.