Question

在三角形ABC中,D,E是BC上的点,角BDA＝角BAD,角CAE＝角CAE,角DAE＝三分之一角BAC,求角BAC的度

Answer 1

应该是∠CAE=∠CEA

∵∠BAD=∠BAD,∠CAE=∠CEA

∴∠BAD=(180°-∠B)/2=90°-1/2∠B

∠CAE=(180°-∠C)/2=90°-1/2∠C

∴∠BAD+∠CAE=180°-1/2(∠B+∠C)=180°-1/2(180°-∠BAC)=90°+1/2∠BAC

∵∠BAD+∠CAE=∠BAC+∠DAE=∠BAC+1/3∠BAC

∴∠BAC+1/3∠BAC=90°+1/2∠BAC

5/6∠BAC=90°

∠BAC=108°