证明:∵在△AFB和△EFC中,∠A+1/2 ∠ABD=∠E+1/2 ∠ACD,①
又∵在△AOB和△DOC中,∠D+1/2∠ACD=∠E+1/2 ∠ABD,②
∴①+②,得:2∠E=∠A+∠D,
∴∠E=1/2(∠A+∠D).
清晰过程如下: