如图:△ABC中,∠CBA=2∠CAB,∠CBA的角平分线BD与∠CAB的角平分线AD相交于点D,且BC=AD,求证:∠ACB=60゜
证明:过D点作DE∥AB交AC于E,连BE,易证四边形DEAB为等腰梯形,BE=AD,又BC=AD,则BE=BC,设∠DAB=x,∠EAB=2x,∠EBA=x,则∠CBE=3x,∠CEB=3x,∠ACB=3x,△CBE为等边三角形,∴∠ACB=60゜
,如图:△ABC中,∠CBA=2∠CAB,∠CBA的角平分线BD与∠CAB的角平分线AD相交于点D,且BC=AD,求证:∠ACB=60゜
证明:过D点作DE∥AB交AC于E,连BE,易证四边形DEAB为等腰梯形,BE=AD,又BC=AD,则BE=BC,设∠DAB=x,∠EAB=2x,∠EBA=x,则∠CBE=3x,∠CEB=3x,∠ACB=3x,△CBE为等边三角形,∴∠ACB=60゜
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