如图直线AF交DC于F,AM平分∠BAE.试探究∠MAN,∠AFG的数量关系

Question

Answer 1

∠AFG=2∠MAN
证明：
设∠BAF=x,∠HAE=y
有AN,AM为角平分线得,
∠BAM=x+1/2(180°-x-y)=90°+1/2x-1/2y
∠BAN=1/2(180°-y)=90°-1/2y
∴∠MAN=∠BAM-∠BAN=1/2x
∵∠AFG=∠BAF=x
∴∠AFG=2∠MAN