解：∵AC=AD∴∠ACD=90-1/2∠A∵BC=BE∴∠BCE=90-1/2∠B∴∠ACD+∠BCE=180-1/2(∠A+∠B)∵∠A+∠B=90°∴∠ACD+∠BCE=180-1/2(∠A+∠B)=135°∴∠DCE=∠ACD+∠BCE-∠C=135°-90°=45°
