(2) As shown in the figure: ∫AM, CM divides ∠BAD and ∠BCD equally,
∴∠BAM=∠MAD,∠MCB=∠MCD,
∠∠ANC =∠b+∠BAM =∠M+∠MCB,∠AEC=∠MCD+∠D=∠MAD+∠M,
∴∠m=∠b+∠bam-∠mcb①,∠m=∠mcd+∠d-∠mad②,
∴ ①+②: 2∠M=∠B+∠D,
∴∠M= 1/2(∠B+∠D)。