Question

数学题,写清楚点http://wenwen.soso.com/z/q214882258.htm

Answer 1

设∠DAE=x

∠BAC=3x

∠B+∠C=180-∠BAC=180-3x

∵BD=AB,CE=AC

∴∠BDA=(180°-∠B)/2,∠CEA=(180°-∠C)/2

∠DAE=x=180°-∠BDA-∠CEA

=180°-[360°-(∠B+∠C)]/2=180°-(180°+3x)/2

即：x=90°-3x/2

x=36°

∴∠BAC=108°