Question

DC和BE相交于点A,EF平分∠DEA,CF平分∠ACB,求证：∠F＝0.5（∠B+∠D）

Answer 1

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

:)