初中幾何輔助線做法大全(編輯修改稿)

　

【文章內容簡介】 B =∠PCB∴△ABE≌△PBC∴AB = BP∴∠BAP =∠BPA∵∠ABP =∠ABC－∠PBC = 50o－10o = 40o∴∠PAB = (180o－∠ABP) = (180o－40o)= 70o⑴構造等腰三角形使二倍角是等腰三角形的頂角的外角例：已知，如圖，在△ABC中，∠1 = ∠2，∠ABC = 2∠C，求證：AB＋BD = AC證明：延長AB到E，使BE = BD，連結DE則∠BED = ∠BDE∵∠ABD =∠E＋∠BDE∴∠ABC =2∠E∵∠ABC = 2∠C∴∠E = ∠C 在△AED和△ACD中∠E = ∠C∠1 = ∠2AD = AD∴△AED≌△ACD∴AC = AE∵AE = AB＋BE∴AC = AB＋BE即AB＋BD = AC⑵平分二倍角例：已知，如圖，在△ABC中，BD⊥AC于D，∠BAC = 2∠DBC求證：∠ABC = ∠ACB證明：作∠BAC的平分線AE交BC于E，則∠BAE = ∠CAE = ∠DBC∵BD⊥AC∴∠CBD ＋∠C = 90o∴∠CAE＋∠C= 90o ∵∠AEC= 180o－∠CAE－∠C= 90o∴AE⊥BC∴∠ABC＋∠BAE = 90o∵∠CAE＋∠C= 90o∠BAE = ∠CAE∴∠ABC = ∠ACB⑶加倍小角例：已知，如圖，在△ABC中，BD⊥AC于D，∠BAC = 2∠DBC求證：∠ABC = ∠ACB證明：作∠FBD =∠DBC,BF交AC于F（過程略）.例：已知，如圖，△ABC中，AB = AC，∠BAC = 120o，EF為AB的垂直平分線，EF交BC于F，交AB于E求證：BF =FC證明：連結AF，則AF = BF∴∠B =∠FAB∵AB = AC∴∠B =∠C∵∠BAC = 120o∴∠B =∠C∠BAC =(180o－∠BAC) = 30o∴∠FAB = 30o∴∠FAC =∠BAC－∠FAB = 120o－30o =90o又∵∠C = 30o∴AF = FC∴BF =FC練習：已知，如圖，在△ABC中，∠CAB的平分線AD與BC的垂直平分線DE交于點D，DM⊥AB于M，DN⊥AC延長線于N求證：BM = CN規律37. 有垂直時常構造垂直平分線.例：已知，如圖，在△ABC中，∠B =2∠C，AD⊥BC于D求證：CD = AB＋BD證明：（一）在CD上截取DE = DB，連結AE，則AB = AE∴∠B =∠AEB∵∠B = 2∠C∴∠AEB = 2∠C又∵∠AEB = ∠C＋∠EAC∴∠C =∠EAC∴AE = CE又∵CD = DE＋CE∴CD = BD＋AB（二）延長CB到F，使DF = DC，連結AF則AF =AC（過程略）.例：已知，如圖，在△ABC中，BC = 2AB, ∠ABC = 2∠C,BD = CD求證：△ABC為直角三角形證明：過D作DE⊥BC，交AC于E，連結BE，則BE = CE，∴∠C =∠EBC∵∠ABC = 2∠C∴∠ABE =∠EBC∵BC = 2AB，BD = CD∴BD = AB在△ABE和△DBE中AB = BD∠ABE =∠EBCBE = BE∴△ABE≌△DBE∴∠BAE = ∠BDE∵∠BDE = 90o∴∠BAE = 90o即△ABC為直角三角形，利用勾股定理證題.例：已知，如圖，在△ABC中，∠A = 90o，DE為BC的垂直平分線求證：BE2－AE2 = AC2證明：連結CE，則BE = CE∵∠A = 90o ∴AE2＋AC2 = EC2∴AE2＋AC2= BE2∴BE2－AE2 = AC2練習：已知，如圖，在△ABC中，∠BAC = 90o，AB = AC，P為BC上一點求證：PB2＋PC2= 2PA2.例：已知，如圖，在△ABC中，∠B = 45o，∠C = 30o，AB =，求AC的長. 解：過A作AD⊥BC于D∴∠B＋∠BAD = 90o，∵∠B = 45o，∠B = ∠BAD = 45o，∴AD = BD∵AB2 = AD2＋BD2，AB =∴AD = 1∵∠C = 30o，AD⊥BC∴AC = 2AD = 2四邊形部分.例：已知,□ABCD的周長為60cm，對角線AC、BD相交于點O，△AOB的周長比△BOC的周長多8cm，求這個四邊形各邊長.解：∵四邊形ABCD為平行四邊形∴AB = CD，AD = CB，AO = CO∵AB＋CD＋DA＋CB = 60AO＋AB＋OB－(OB＋BC＋OC) = 8∴AB＋BC = 30，AB－BC =8∴AB = CD = 19，BC = AD = 11答：這個四邊形各邊長分別為19cm、11cm、19cm、11cm.，相鄰兩個三角形周長之差等于鄰邊之差.（例題如上）例：已知，如圖，Rt△ABC，∠ACB = 90o，CD⊥AB于D，AE平分∠CAB交CD于F，過F作FH∥AB交BC于H求證：CE = BH證明：過F作FP∥BC交AB于P，則四邊形FPBH為平行四邊形∴∠B =∠FPA，BH = FP∵∠ACB = 90o，CD⊥AB∴∠5＋∠CAB = 45o，∠B＋∠CAB = 90o∴∠5 =∠B∴∠5 =∠FPA又∵∠1 =∠2，AF = AF∴△CAF≌△PAF∴CF = FP∵∠4 =∠1＋∠5，∠3 =∠2＋∠B∴∠3 =∠4∴CF = CE∴CE = BH練習：已知，如圖，AB∥EF∥GH，BE = GC求證：AB = EF＋GH. 例：已知，如圖，在□ABCD中，AB = 2BC，M為AB中點求證：CM⊥DM證明：延長DM、CB交于N∵四邊形ABCD為平行四邊形∴AD = BC，AD∥BC∴∠A = ∠NBA ∠ADN =∠N又∵AM = BM∴△AMD≌△BMN∴AD = BN∴BN = BC∵AB = 2BC，AM = BM∴BM = BC = BN∴∠1 =∠2，∠3 =∠N∵∠1＋∠2＋∠3＋∠N = 180o，∴∠1＋∠3 = 90o∴CM⊥DM.如圖：OE = OF（或這邊所在的直線）上的任意一點與對邊的兩個端點的連線所構成的三角形的面積等于平行四邊形面積的一半.如圖：S△BEC = S□ABCD，不相鄰的兩個三角形的面積之和等于平行四邊形面積的一半.如圖：S△AOB ＋ S△DOC = S△BOC＋S△AOD = S□ABCD，不相鄰的兩條線段的平方和相等.如圖：AO2＋OC2 = BO2 ＋DO2.如圖：四邊形GHMN是矩形（規律45～規律49請同學們自己證明）.例：已知，如圖，E為矩形ABCD的邊AD上一點，且BE = ED，P為對角線BD上一點，PF⊥BE于F，PG⊥AD于G求證：PF＋PG = AB證明：證法一：過P作PH⊥AB于H，則四邊形AHPG為矩形∴AH = GP PH∥AD∴∠ADB =∠HPB∵BE = DE∴∠EBD = ∠ADB∴∠HPB =∠EBD又∵∠PFB =∠BHP = 90o∴△PFB≌△BHP∴HB = FP∴AH＋HB = PG＋PF即AB = PG＋PF證法二：延長GP交BC于N，則四邊形ABNG為矩形，（證明略）：⑴作斜邊上的高例：已知，如圖,若從矩形ABCD的頂點C作對角線BD的垂線與∠BAD的平分線交于點E求證：AC = CE證明：過A作AF⊥BD，垂足為F，則AF∥EG∴∠FAE = ∠AEG∵四邊形ABCD為矩形∴∠BAD = 90o OA = OD∴∠BDA =∠CAD∵AF⊥BD∴∠ABD＋∠ADB = ∠ABD＋∠BAF = 90o∴∠BAF =∠ADB =∠CAD∵AE為∠BAD的平分線∴∠BAE =∠DAE∴∠BAE－∠BAF =∠DAE－∠DAC即∠FAE =∠CAE∴∠CAE =∠AEG∴AC = EC⑵作斜邊中線，當有下列情況時常作斜邊中線：①有斜邊中點時例：已知，如圖，AD、BE是△ABC的高， F是DE的中點，G是AB的中點求證：GF⊥DE證明：連結GE、GD∵AD、BE是△ABC的高，G是AB的中點∴GE = AB，GD = AB∴GE = GD∵F是DE的中點∴GF⊥DE②有和斜邊倍分關系的線段時例：已知，如圖，在△ABC中，D是BC延長線上一點，且DA⊥BA于A，AC = BD求證：∠ACB = 2∠B證明：取BD中點E，連結AE，則AE = BE = BD∴∠1 =∠B∵AC = BD∴AC = AE∴∠ACB =∠2 ∵∠2 =∠1＋∠B∴∠2 = 2∠B∴∠ACB = 2∠B.例：已知，如圖，過正方形ABCD對角線BD上一點P，作PE⊥BC于E，作PF⊥CD于F 求證：AP = EF 證明:連結AC 、PC∵四邊形ABCD為正方形∴BD垂直平分AC，∠BCD = 90o∴AP = CP∵PE⊥BC，PF⊥CD，∠BCD = 90o∴四邊形PECF為矩形∴PC = EF∴AP = EF.例：已知,如圖，正方形ABCD中，M為AB的中點，MN⊥MD，BN平分∠CBE并交MN于N求證：MD = MN證明：取AD的中點P，連結PM，則DP = PA =AD∵四邊形ABCD為正方形∴AD = AB, ∠A =∠ABC = 90o∴∠1＋∠AMD = 90o，又DM⊥MN∴∠2＋∠AMD = 90o∴∠1 =∠2∵M為AB中點∴AM = MB = AB∴DP = MB AP = AM∴∠APM =∠AMP = 45o∴∠DPM =135o∵BN平分∠CBE∴∠CBN = 45o∴∠