Question

在△ABC中,∠BAC=5.25°,D是BC上一点,AD平分∠BAC,过A作DA的垂线交直线BC的延长线于点M,若BM=

Answer 1

延长B至AE 使AE=AC,连接ME,延长MA交CE于F

AC=AE AEC=ACE=1/2BAC=CAD

AD//CE MA垂直AD MA垂直EC

AE=AC EF=CF

MF垂直CE EM=MC MEC=MCE

AEC=ACE MEB=ACB

BM=AB+AC=BE BME=BEM

MEC+MCE+BEM=180

3EMB+2AEC=180 2AEC=BAC=5.25

EMB=58.25

ACB=MEB=EMB=58.25

ABC=2EMB=116.5