北師大版初中九年級(jí)數(shù)學(xué)上冊(cè)專項(xiàng)素養(yǎng)鞏固訓(xùn)練卷(四)相似三角形的五種基本類型(練模型)課件

相似三角形的五種基本類型(練模型)專項(xiàng)素養(yǎng)鞏固訓(xùn)練卷(四)類型一“A”字型模型解讀(1)“A”字型如圖1,已知:DE∥BC.結(jié)論:△ADE∽△ABC?

=

=

.(2)反“A”字型如圖2,已知:∠AED=∠C.結(jié)論:△ADE∽△ABC?

=

=

.(3)反“A”字型(共邊共角)如圖,已知:∠ABD=∠C.結(jié)論:①△ABD∽△ACB;②

=

=

;③AB2=AD·AC.第1題圖1.(2022四川涼山州中考,11,★☆☆)如圖,在△ABC中,點(diǎn)D、E分別在邊AB、AC

上,若DE∥BC,

=

,DE=6cm,則BC的長(zhǎng)為()CA.9cm

B.12cmC.15cm

D.18cm解析

C∵

=

,∴

=

,∵DE∥BC,∴∠ADE=∠B,∠AED=∠C,∴△ADE∽△ABC,∴

=

,∴

=

,∴BC=15cm,故選C.2.(★★☆)如圖,在△ACB中,AC=30cm,BC=25cm.動(dòng)點(diǎn)

P從點(diǎn)C出發(fā),沿CA向終點(diǎn)A勻速運(yùn)動(dòng),速度是2cm/s,同時(shí),動(dòng)點(diǎn)Q從點(diǎn)B出發(fā),沿BC

向終點(diǎn)C勻速運(yùn)動(dòng),速度是1cm/s.當(dāng)△CPQ與△CAB相似時(shí),求運(yùn)動(dòng)的時(shí)間.考向動(dòng)態(tài)變化試題解析設(shè)運(yùn)動(dòng)的時(shí)間為ts,由題意可知,CP=2tcm,BQ=tcm,則CQ=(25-t)cm,①當(dāng)△CPQ∽△CAB時(shí),

=

,即

=

,解得t=

;②當(dāng)△CPQ∽△CBA時(shí),

=

,即

=

,解得t=

.綜上所述,運(yùn)動(dòng)時(shí)間為

s或

s時(shí),△CPQ與△CAB相似.類型二“X”字型模型解讀特點(diǎn):有一組對(duì)頂角.(1)正“X”字型:如圖1,AB∥DE?△ABC∽△EDC.(2)斜“X”字型:如圖2,另有一組角相等,如

?△ABC∽△DEC.3.(★☆☆)如圖,點(diǎn)P是?ABCD邊AB上的一點(diǎn),射線CP交DA的延長(zhǎng)線于點(diǎn)E,則

圖中相似的三角形有

()

A.0對(duì)

B.1對(duì)

C.2對(duì)

D.3對(duì)D解析

D∵四邊形ABCD是平行四邊形,∴AB∥DC,AD∥BC,∴△EAP∽△EDC,△EAP∽△CBP,∴△EDC∽△CBP,故有3對(duì)相似三角形.故選D.4.(2023陜西西安鐵一中學(xué)月考,7,★★☆)如圖,在菱形ABCD中,EF⊥AC于點(diǎn)H,

交AD于點(diǎn)E,交CB的延長(zhǎng)線于點(diǎn)F,交AB于G,且AE∶FB=1∶3,則GB∶CD的值為

()

A.

B.

C.

D.

D解析

D∵四邊形ABCD是菱形,∴AB=CD,AE∥BF,∴∠EAB=∠ABF,∠AEF=

∠F,∴△EAG∽△FBG,∴

=

=

,∴

=

,∴

=

,故選D.5.(2023湖北武漢江夏模擬,18,★★☆)如圖,已知平行四邊形ABCD中,∠A=62°,

BE平分∠ABC交AD的延長(zhǎng)線于點(diǎn)E.(1)求∠E的度數(shù).(2)若點(diǎn)D為AE的中點(diǎn),BE交DC于點(diǎn)F,求

的值.解析

(1)∵四邊形ABCD是平行四邊形,∴AD∥BC,CD∥AB,AD=BC,∴∠A+∠ABC=180°,∠CBE=∠E.∵∠A=62°,∴∠ABC=118°.∵BE平分∠ABC,∴∠CBE=

∠ABC=59°,∴∠E=∠CBE=59°.(2)∵點(diǎn)D為AE的中點(diǎn),∴DE=AD.∵BC=AD,∴DE=BC.∵AD∥BC,∴∠E=∠CBF,∠EDF=∠C,∴△DEF≌△CBF(ASA),∴△EAB的面積=四邊形ABCD的面積,∵DF∥AB,∴△EDF∽△EAB,∴

=

=

=

.∴

=

.6.(★★★)如圖,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:△AOB∽△COE.(2)求證:BO2=EO·FO.證明:(1)∵四邊形ABCD是平行四邊形,∴AB∥CD,∴∠BAO=∠ECO,∵∠AOB=∠COE,∴△AOB∽△COE.(2)∵△COE∽△AOB,∴

=

,∵AD∥BC,∴△COB∽△AOF.∴

=

,∴

=

,即OB2=OF·OE.類型三一線三等角模型模型展示(1)如圖1,△CAP∽△PBD(此模型又叫“三垂直模型”);(2)如圖2、圖3,有以下結(jié)論:①△CAP∽△PBD;②連接CD,當(dāng)點(diǎn)P為AB的中點(diǎn)時(shí),△CAP∽△PBD∽△CPD.7.(★☆☆)如圖,在△ABC中,AB=AC,AB=4,BC=3,∠BAC=50°,P為BC上一點(diǎn),且

BP=1,點(diǎn)D為邊AC上一點(diǎn),若∠APD=65°,則CD的長(zhǎng)為

()

A.

B.

C.

D.

C解析

C∵AB=AC,∠BAC=50°,∴∠B=∠C=(180°-50°)÷2=65°,又∵∠APD=65°,∴∠APB+∠DPC=∠APB+∠BAP=180°-65°=115°,∴∠BAP=∠DPC,∴△ABP∽△PCD,∴

=

,∵BC=3,BP=1,∴PC=2,∴

=

,∴CD=

.故選C.8.(★★☆)如圖,AB⊥BC,DC⊥BC,E是BC上一點(diǎn),使得AE⊥DE.(1)求證:△ABE∽△ECD.(2)若AB=4,AE=BC=5,求CD的長(zhǎng).(3)當(dāng)△AED∽△ECD時(shí),請(qǐng)寫出線段AD、AB、CD之間的數(shù)量關(guān)系,并說(shuō)明理由.解析

(1)證明:∵AB⊥BC,DC⊥BC,∴∠B=∠C=90°,∴∠BAE+∠AEB=90°,∵AE⊥DE,∴∠AED=90°,∴∠AEB+∠DEC=90°,∴∠DEC=∠BAE,∴△ABE∽△ECD.(2)Rt△ABE中,∵AB=4,AE=5,∴BE=3,∵BC=5,∴EC=5-3=2,由(1)得,△ABE∽△ECD,∴

=

,∴

=

,∴CD=

.(3)線段AD、AB、CD之間的數(shù)量關(guān)系:AD=AB+CD.理由:如圖,過(guò)E作EF⊥AD于F,

∵△AED∽△ECD,∴∠ADE=∠EDC,∵DC⊥BC,∴EF=EC,∵DE=DE,∴Rt△DFE≌Rt△DCE(HL),∴DF=DC,由(1)得△ABE∽△ECD,又△AED∽△ECD,∴△ABE∽△AED,同理可得,△ABE≌△AFE,∴AF=AB,∴AD=AF+DF=AB+CD.類型四雙垂直模型模型展示如圖,在Rt△ABC中,CD為斜邊AB上的高,則有△ACD∽△ABC∽△CBD,CD2=BD

·AD,BC2=BD·AB,AC2=AD·AB.

9.(★☆☆)如圖,在Rt△ABC中,∠ACB=90°,CD是邊AB上的高.(1)求證:△ACD∽△ABC.(2)若AC=3,BC=4,求BD的長(zhǎng).解析

(1)證明:∵CD⊥AB,∴∠ADC=90°.∵∠ACB=90°,∴∠ADC=∠ACB.∵∠A=∠A,∴△ACD∽△ABC.(2)∵∠ACB=90°,AC=3,BC=4,∴AB=

=5,∵S△ABC=

AC·BC=

AB·CD,∴AC·BC=AB·CD,∴CD=

=

,∵CD⊥AB,∴BD=

=

=

.類型五手拉手模型模型解讀手拉手模型——相似條件:如圖1,在△AOB中,CD∥AB,將△OCD旋轉(zhuǎn)至圖2所示的位置,旋轉(zhuǎn)角∠AOC=∠BOD,在相應(yīng)的三角形中,旋轉(zhuǎn)角的對(duì)邊AC,BD稱為“拉手線”.結(jié)論:如圖2,△OCD∽△OAB,△AOC∽△BOD,若延長(zhǎng)AC交BD于點(diǎn)E,則必有∠BEC=∠BOA.難點(diǎn):復(fù)雜圖形中尋找“手拉手”模型.突破口:①找旋轉(zhuǎn)角;②找“拉手線”;③“手拉手”構(gòu)造相似三角形.10.(2023甘肅天水八中期末,22,★☆☆)如圖,在△ADE和△ABC中,AB·AE=AD·

AC,且∠EAC=∠DAB.求證:△ABC∽△ADE.證明:∵AB·AE=AD·AC,∴

=

,∵∠EAC=∠DAB,∴∠EAC+∠BAE=∠DAB+∠BAE,即∠BAC=∠DAE,∴△ABC∽△ADE.11.(2023河南新鄉(xiāng)獲嘉模擬,23,★★★)在Rt△ABC中,∠ABC=90°,AB=nBC,P為

AB上的一點(diǎn)(不與端點(diǎn)重合),過(guò)點(diǎn)P作PM⊥AB交AC于點(diǎn)M,得到△APM.(1)【問(wèn)題發(fā)現(xiàn)】如圖①,當(dāng)n=1,P為AB的中點(diǎn)時(shí),CM與BP的數(shù)量關(guān)系為

.(2)【類比探究】如圖②,當(dāng)n=2時(shí),△APM繞點(diǎn)A順時(shí)針旋轉(zhuǎn),連接CM,BP,則在旋

轉(zhuǎn)過(guò)程中CM與BP之間的數(shù)量關(guān)系是否發(fā)生變化?請(qǐng)說(shuō)明理由.(3)【拓展延伸】在(2)的條件下,已知AB=4,AP=2,當(dāng)△APM繞點(diǎn)A順時(shí)針旋轉(zhuǎn)至

B,P,M三點(diǎn)共線時(shí),求出線段BM的長(zhǎng).第11題圖解析

(1)當(dāng)n=1時(shí),AB=BC,∵∠ABC=90°,∴

=

,∵P為AB的中點(diǎn),∴

=

,∴AP=BP=

AB,∵PM⊥AB,∴∠APM=90°,∴∠APM=∠ABC,∴PM∥BC,∴△APM∽△ABC,∴

=

=

,∴AC=

AB,AM=

AP=

AB,∴CM=AC-AM=

AB-

AB=

AB,∴

=

=

,∴CM=

BP,故答案為CM=

BP.(2)CM與BP的數(shù)量關(guān)系不變,理由如下:當(dāng)n=2時(shí),AB=2BC,由勾股定理可得AC=

=

=

AB,∴

=

,由(1)得△APM∽△ABC,∴

=

,由旋轉(zhuǎn)得,∠PAB=∠MAC,∴△ABP∽△ACM,∴

=

=

,∴CM=

BP.