点D和点E在三角形ABC的边BC上,若BD=AD=AE=EC,且∠B=2∠DAE,则∠BAC=

设∠DAE=x,则∠B=2x,因为：BD=AD,所以∠DAB=∠B=2x∠ADE=∠DAB+∠B=4x因为：AD=AE,所以∠AED=∠ADE=4x因为：AE=EC,所以∠EAC=∠C=1/2∠AED=2x所以：∠BAC=∠BAD+∠DAE+∠EAC=z2+x+2x=5x因为∠BAC+∠B+∠C=5x+2x+2x=9...