∵∠A+∠B+∠C+∠D=360° 且∠B+∠D=70°
∴∠A+∠C=290°
∵AM,CM平分∠BAD和∠BCD
∴∠BAM+∠DCM=∠MAC+∠MCA=(∠A+∠C)/2=145°
又∵∠AMC中,∠MAC+∠MCA+∠M=180°
所以∠M=180°-(∠MAC+∠MCA)=180°-145°=35°
∵∠A+∠B+∠C+∠D=360° 且∠B+∠D=70°
∴∠A+∠C=290°
∵AM,CM平分∠BAD和∠BCD
∴∠BAM+∠DCM=∠MAC+∠MCA=(∠A+∠C)/2=145°
又∵∠AMC中,∠MAC+∠MCA+∠M=180°
所以∠M=180°-(∠MAC+∠MCA)=180°-145°=35°