【導(dǎo)讀】2.如圖所示,在△ABC中,AB=AC,∠A=36°,兩條角平分線BE、CD相交于點(diǎn)O,A.90°B.75°C.70°D.60°子表的實(shí)際時(shí)刻是。7.等腰三角形一個(gè)頂角和一個(gè)底角之和是?10,則頂角等于.。,Nba關(guān)于x軸對(duì)稱,則ab??12.如圖,點(diǎn)D,E在△ABC的邊BC上,AB=AC,BD=CE.13.如圖,已知Rt△ABC中,∠ACB=90°,CD⊥AB于D,∠BAC的平分線分別交BC、①依題意將圖2補(bǔ)全;想法1:要證明PA=PM,只需證△APM是等邊三角形;想法3:將線段BP繞點(diǎn)B順時(shí)針旋轉(zhuǎn)60°,得到線段BK,要證PA=PM,只需證PA=CK,11.解:∵A點(diǎn)和E點(diǎn)關(guān)于BD對(duì)稱,又B點(diǎn)、C點(diǎn)關(guān)于DE對(duì)稱,在△ABD與△ACE中,∴∠CAD+∠ACD=90°.∴∠ACD=∠B.∵∠EAB+∠B=∠CEA,∠CAE+∠ACD=∠CFE,∴∠CFE=∠CEF.∴CF=CE.∴∠B=∠C=60°,∴∠BAP=∠CAQ=20°,∴∠PAQ=∠BAC﹣∠BAP﹣∠CAQ=60°﹣20°﹣。20°=20°,∴∠BAQ=∠BAP+∠PAQ=40°;②∵AP=AQ,∴∠APQ=∠AQP,∴∠APB=∠AQC,∵△ABC是等邊三角形,∴∠B=∠C=60°,MAC=∠BAP,∴∠BAP+∠PAC=∠MAC+∠CAP=60°,∴∠PAM=60°,∵AP=AQ,∴AP=AM,