文[1][2][3][4]中都有如下一道几何题.如图1,△ABC中,E、F分别在边AB、AC上,BF与CE相交于点P,且∠1=∠2=1/2∠A,求证:BE=CF.证明作∠A的平分线交BC于点D,连结DE、DF,则∠DAF=∠DAE=1/2∠A,∵∠1=∠2=1/2∠A,∴∠DAF=∠DAE=∠1=∠2,∴A、B、D、F四点与A、E、D、C四点分別共圆,于是BD=DF,DE=DC, In Figure 1, △ ABC, E, F, respectively, in the side AB, AC, BF and CE intersect at the point P, and ∠ 1 = ∠2 = 1/2 ∠A, Proof: BE = CF. Prove that the bisector of ∠A is BC at point D, and link DE and DF, then ∠DAF = ∠DAE = 1 / 2∠A, ∵∠ 1 = ∠2 = 1 / 2∠A, ∴∠DAF = ∠DAE = ∠1 = ∠2, ∴A, B, D, F four points and A, E, D, = DF, DE = DC,