如图,ABC=90°,AD⊥AB,AD=AB,BE⊥CD于E,AF⊥AC交EB于F,求证:CF平分∠ACB
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![如图,ABC=90°,AD⊥AB,AD=AB,BE⊥CD于E,AF⊥AC交EB于F,求证:CF平分∠ACB](/uploads/image/z/3736104-24-4.jpg?t=%E5%A6%82%E5%9B%BE%2CABC%3D90%C2%B0%2CAD%E2%8A%A5AB%2CAD%3DAB%2CBE%E2%8A%A5CD%E4%BA%8EE%2CAF%E2%8A%A5AC%E4%BA%A4EB%E4%BA%8EF%2C%E6%B1%82%E8%AF%81%3ACF%E5%B9%B3%E5%88%86%E2%88%A0ACB)
如图,ABC=90°,AD⊥AB,AD=AB,BE⊥CD于E,AF⊥AC交EB于F,求证:CF平分∠ACB
如图,ABC=90°,AD⊥AB,AD=AB,BE⊥CD于E,AF⊥AC交EB于F,求证:CF平分∠ACB
如图,ABC=90°,AD⊥AB,AD=AB,BE⊥CD于E,AF⊥AC交EB于F,求证:CF平分∠ACB
∠ACB=90°吧?
AD⊥AB→∠DAB=90°;
BE⊥CD→∠E=90°;
AF⊥AC→∠CAF=90°
∠DAB=∠CAF→∠FAB=∠CAD;
四边形ABED中,∠DAB=90°,∠E=90°,所以∠D+∠ABE=180°,又∠ABF+∠ABE=180°,所以∠D=∠ABF;
∠FAB=∠CAD;
∠ABF=∠D;
AB=AD→△AFB≌△ACD→AF=AC→∠AFC=∠ACF
∠FAC=∠ACB=90°→AF‖CB→∠AFC=∠BCF
∠AFC=∠BCF,∠AFC=∠ACF→∠ACF=∠BCF→CF平分∠ACB
AD⊥AB→∠DAB=90°;
BE⊥CD→∠E=90°;
AF⊥AC→∠CAF=90°
∠DAB=∠CAF→∠FAB=∠CAD;
四边形ABED中,∠DAB=90°,∠E=90°,所以∠D+∠ABE=180°,又∠ABF+∠ABE=180°,所以∠D=∠ABF;
∠FAB=∠CAD;
∠ABF=∠D;
AB=AD→△AFB≌△ACD→AF=...
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AD⊥AB→∠DAB=90°;
BE⊥CD→∠E=90°;
AF⊥AC→∠CAF=90°
∠DAB=∠CAF→∠FAB=∠CAD;
四边形ABED中,∠DAB=90°,∠E=90°,所以∠D+∠ABE=180°,又∠ABF+∠ABE=180°,所以∠D=∠ABF;
∠FAB=∠CAD;
∠ABF=∠D;
AB=AD→△AFB≌△ACD→AF=AC→∠AFC=∠ACF
∠FAC=∠ACB=90°→AF‖CB→∠AFC=∠BCF
∠AFC=∠BCF,∠AFC=∠ACF→∠ACF=∠BCF→CF平分∠ACB
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