如图,DC和BE相交于点A,EF平分∠DEA,CF平分∠ACB.求证：∠F=2\1(∠B+∠D）.

如图,DC和BE相交于点A,EF平分∠DEA,CF平分∠ACB.求证：∠F=2\1(∠B+∠D）.
如图,DC和BE相交于点A,EF平分∠DEA,CF平分∠ACB.求证：∠F=2\1(∠B+∠D）.

如图,DC和BE相交于点A,EF平分∠DEA,CF平分∠ACB.求证：∠F=2\1(∠B+∠D）.
设DC交EF于点M EB交FC于点N.所以,
1.∠F+∠FNE+1/2∠DEB=180
2.∠F+∠FMC+1/2∠DCB=180
3.∠D+∠FMC+1/2∠DEB=180
4.∠B+∠FNE+1/2∠DCB=180
1+2
5.2∠F+∠FNE+ ∠FMC+1/2∠DCB+1/2∠DEB=360
3+4
6.∠D+∠B+∠FNE+1/2∠DCB+∠FMC+1/2∠DEB=360
5=6
7.2∠F+∠FNE+ ∠FMC+1/2∠DCB+1/2∠DEB=∠D+∠B+∠FNE+1/2∠DCB+∠FMC+1/2∠DEB
化简7
2∠F=∠D+∠B