如图，AF是△ABC的高线，AD、BE分别是△ABC的角平分线，AD、BE交于点O，且∠ABC=36°，∠C=76°．求∠DAF

∵△ABC中，∠ABC=36°，∠C=76°，∴∠BAC=180°-36°-76°=68°．∵AD平分∠BAC，∴∠BAD=∠DAC=34°．∵AF⊥BC，∴∠CAF=90°-∠C=90°-76°=14°，∴∠DAF=∠DAC-∠CAF=34°-14°=20°；∵BE平分∠ABC，∠ABC=36°，∴∠ABO=12∠ABC=18°，∴∠AOB=180°-∠ABO-∠BAD=180°-18°-34°=128°，∴∠DOE=128°．