已知直角三角形abc中,角c=90度,角a,角b的角平分线交于点p,求角apb的度数

135°∵直角三角形ABC∴∠C = 90°∴∠BAC+∠CBA=90° ∵AD平分∠BAC∴∠CAD=∠BAD=1\2∠CAB∵BE平分∠CBA∴∠CBE=∠ABE=1\2∠CBA∴∠ DAB+∠EBA=1\2∠CAB+1\2∠CBA=1\2(∠CAB+∠CBA)=1\2×90 ° =45°∴∠ APB=180°-∠DAB-∠EBA=180°-(∠DAB+∠EBA)=180°-45°=135°