如图,三角形ABC中,BD和CD分别平分∠ABC和外角∠ACF,求证：∠A=2∠D!

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证明如下;∠ACF=∠A+∠B故1/2∠ACF=1/2∠A+1/2∠B.
又1/2∠B=∠DBC,1/2∠ACF=∠DCF=∠D+∠DBC.
故1/2∠A+1/2∠B=1/2∠ACF=∠D+∠DBC
即1/2∠A+1/2∠B=1/2∠B+∠D,移项得1/2∠A=∠D
即∠A=2∠D得证