【正文】 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(jié)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(xí)：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(diǎn)D，DM⊥AB于M，DN⊥AC延長線于N求證：BM = CN規(guī)律37. 有垂直時(shí)常構(gòu)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(jié)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(jié)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(jié)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(jié)CE，則BE = CE∵∠A = 90o ∴AE2＋AC2 = EC2∴AE2＋AC2= BE2∴BE2－AE2 = AC2練習(xí)：已知，如圖，在△ABC中，∠BAC = 90o，AB = AC，P為BC上一點(diǎn)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(duì)角線AC、BD相交于點(diǎn)O，△AOB的周長比△BOC的周長多8cm，求這個(gè)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(gè)四邊形各邊長分別為19cm、11cm、19cm、11cm.，相鄰兩個(gè)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(xí)：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(diǎn)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(diǎn)與對(duì)邊的兩個(gè)端點(diǎn)的連線所構(gòu)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(gè)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(guī)律45～規(guī)律49請(qǐng)同學(xué)們自己證明）.例：已知，如圖，E為矩形ABCD的邊AD上一點(diǎn)，且BE = ED，P為對(duì)角線BD上一點(diǎn)，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(diǎn)C作對(duì)角線BD的垂線與∠BAD的平分線交于點(diǎn)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(dāng)有下列情況時(shí)常作斜邊中線：①有斜邊中點(diǎn)時(shí)例：已知，如圖，AD、BE是△ABC的高， F是DE的中點(diǎn)，G是AB的中點(diǎn)求證：GF⊥DE證明：連結(jié)GE、GD∵AD、BE是△ABC的高，G是AB的中點(diǎn)∴GE = AB，GD = AB∴GE = GD∵F是DE的中點(diǎn)∴GF⊥DE②有和斜邊倍分關(guān)系的線段時(shí)例：已知，如圖，在△ABC中，D是BC延長線上一點(diǎn)，且DA⊥BA于A，AC = BD求證：∠ACB = 2∠B證明：取BD中點(diǎn)E，連結(jié)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(duì)角線BD上一點(diǎn)P，作PE⊥BC于E，作PF⊥CD于F 求證：AP = EF 證明:連結(jié)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(diǎn)，MN⊥MD，BN平分∠CBE并交MN于N求證：MD = MN證明：取AD的中點(diǎn)P，連結(jié)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(jìn)為AB中點(diǎn)∴AM = MB = AB∴DP = MB AP = AM∴∠APM =∠AMP = 45o∴∠DPM =135o∵BN平分∠CBE∴∠CBN = 45o∴∠MBN =∠MBC＋∠CBN = 90o＋45o= 135o即∠DPM =∠MBN∴△DPM≌△MBN∴DM = MN注意：把M改為AB上任一點(diǎn)，其它條件不變，結(jié)論仍然成立。練習(xí)：已知，如圖，在正方形ABCD中，E為AD上一點(diǎn)，BF平分∠CBE交CD于F求證：BE = CF＋AE，常把這條線段延長，構(gòu)造全等三角形.例：如圖，在正方形ABCD中，E、F分別是CD、DA的中點(diǎn)，BE與CF交于P點(diǎn)求證：AP = AB 證明：延長CF交BA的延長線于K∵四邊形ABCD為正方形∴BC = AB = CD = DA ∠BCD =∠D =∠BAD = 90o ∵E、F分別是CD、DA的中點(diǎn)∴CE = CD DF = AF