1.过P作直线CD平行AM,则CD‖BN
∵MA‖NB
∴∠MAB+∠NBA=180°
又∵∠PAB+∠PBA+∠APB=180°
且∠PAB+∠MAP+∠PBA+∠NBP=180°
∴∠APB=∠MAP+∠NBP