Question

已知△ABC,AD为角平分线,∠BAC=60°,将△ADC沿直线AD翻折,点C"与C相对应,若AC=2,BC"=1,求△

Answer 1

没有图,有两种情况（C'在AB延长线上或AB边上,因AD是角平分线)

如在AB边上:

AB=AC'+BC'=AC+BC'=2+1=3

S=AB*(AC*sin∠BAC)/2=3*2*根3/2*1/2=(3根3)/2

如在AB延长线上:

AB=AC'-BC'=AC-BC'=2-1=1

S=AB*(AC*sin∠BAC)/2=1*2*根3/2*1/2=根3/2 (此时ABC实际是个直角三角形)