Question

如图所示，AB、CD相交于E,CF、BF分别∠ACD和∠ABD的平分线且相交于F。求证∠F=1/2(∠A+∠D)问题中。

你回答的∠A+∠D+2(∠BAD+∠ADC)+2(∠FCD+∠ABF)=360

是什么意思啊。求助，快啦。

Answer 1

<CEB=<A+<ACD=<D+<ABD

AB、CF相交于M, CD、BF相交于N。
<F+<FME+<FNE+<MEN=360
<FME=<AMC=180-<A-<ACF=180-<A-<ACD/2
<FNE=<BND=180-<D-<DBN=180-<D-<ABD/2
<MEN=<CEB=<A+<ACD
<F=360-<FME-<FNE-<MEN
=360-(180-<A-<ACD/2)-(180-<D-<ABD/2)-(<A+<ACD)
=360-180+<A+<ACD/2-180+<D+<ABD/2-<A-<ACD
=<D+<ABD/2-<ACD/2

<A+<ACD=<D+<ABD
<ABD=<A+<ACD-<D
<ABD/2=<A/2+<ACD/2-<D/2

<F=<D+<ABD/2-<ACD/2
=<D+<A/2+<ACD/2-<D/2-<ACD/2
=<A/2+<D/2
=(<A+<D)/2

Answer 2

都没有图。。。