∠CAB=70°.在同一平
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⑴CM+CN+MN=√2CE.在BC上取BG=CN,连接FG,∵ΔABC是等腰直角三角形,E为AB的中点,∴∠MCE=∠B=45°,EC=EB,BC=√2CE∴ΔCEG≌ΔMEG,∴EM=EG、∠ME
∵△ABC绕点A旋转到△AB′C′的位置,∴AC=AC′,∠BAC=∠B′AC′,∵CC′∥AB,∠CAB=75°,∴∠ACC′=∠CAB=75°,∴∠CAC′=180°-2∠ACC′=180°-2×
∵∠C=90°,∴∠ABC+∠BAC=180°-90°=90°,∵∠CAB与∠CBA的平分线相交于O点,∴∠OAB+∠OBA=12(∠ABC+∠BAC)=12×90°=45°,在△AOB中,∠AOB=
CE=CF∵AE平分∠CAB交AD于F∴∠FAD=∠CAE∵在△ADF中,∠FAD∠ADF=∠AFD在△CAE中,∠CAE∠ACE=∠AEC∠ACE=∠ADF=90度∴∠AFD=∠
如图,过点D作DE⊥AB于E,∵∠C=90°,AC=6,BC=8,∴AB=AC2+BC2=62+82=10,∵AD平分∠CAB,∴CD=DE,∴S△ABC=12AC•CD+12AB•DE=12AC•B
如图,延长BA到E,使AE=AC,连接CE,则∠E=∠ECA=45°.∵∠CAD=∠BAD=45°,∴∠E=∠BAD=45°,∴CE∥AD.∴CD:BD=AE:AB,∵AC=AE,∴CD:BD=AC:
∵AB为直径,∴∠ACB=90°,∵∠CAB=30°,∴∠ABC=60°,∴弧BC的度数=12弧AC的度数;∵AD=DC,∴弧AD的度数=弧DC的度数=12弧AC的度数,∴弧BC的度数=弧AD的度数;
(1)CM=DM,且CM⊥DM.证明:∵∠ACE=90°;M为AE的中点.(见原图5.)∴CM=AE/2=AM(直角三角形斜边的中线等斜边的一半)∴∠MCA=∠MAC,则∠CME=∠MCA+∠MAC=
∵CC′∥AB,∠CAB=70°,∴∠C′CA=∠CAB=70°,又∵C、C′为对应点,点A为旋转中心,∴AC=AC′,即△ACC′为等腰三角形,∴∠BAB′=∠CAC′=180°-2∠C′CA=40
在Rt△ABC中,∠CAB=90°,AD是∠CAB的平分线,tanB=1/2,则CD:DB=(1/2)解;过点D作DE⊥AB,垂足为E,过点D作DF⊥AC,垂足为F,∵AD平分∠CAB,∴∠DAC=4
证明:∵AE平分∠CAB,∴∠CAB=2∠BAE,∵∠CAB=2∠B,∴∠EAB=∠B,∴EA=EB,过E作ED⊥AB于D,则AB=2AD,∵AB=2AC,∴AD=AC,在ΔAED与ΔAEC中,AD=
∠D的度数为:70/2=35°.设,∠CAD=∠DAB=∠1,∠CBD=∠DBE=∠2.∠ABC=180-(∠C+2∠1),而,∠ABC=180-2∠2,则有∠C+2∠1=2∠2,∠2-∠1=∠C/2
∵∠BAC=∠DAE∴∠BAC+∠BAE=∠DAE+∠BAE∴∠CAE=∠BAD∵AB=AC,AD=AE∴△CAE≌△BAD∴∠AEC=∠ADE∵AD=AE∴∠AED=∠ADE=∠AEC∵∠CAB=∠
∠ACD=∠EAB=∠CBF∠CAB=∠CAD+∠EAB=∠CAD+ACD=∠EDF=50°△DEF的各内角与△ABC的各内角有着∠CAB=∠EDF,∠CBA=∠FED,∠ACB=∠DFE证法与题1相
sinB/30=sin120°/70sinB=3sin120°/7=3sin60°/7=3*√3/14≈0.3712, ∠B=21.8°,∠C=180°-120°-21.8°=38.2°.A
40CC'//AB故∠ACC"=70AC"=AC故∠cac"=40就是∠BAB'=40
∵▱ABCD∴∠ABC=∠ADC=125°,∠CAD=∠ACB=34°∴∠CAB=180°-∠ACB-∠ABC=180°-34°-125°=21°故答案为125,21.
∵AD平分∠CAB,且∠C=90°,DE⊥AB,∴CD=DE=5∵∠CAB=45°,∠C=∠DEB=90°,∴∠BDE=∠B=45°,∴DE=BE=5,∴DB=52,∴BC=5+52.