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°