Question

如图12,在三角形ABC中,三个内角的角平分线交于点O,OE⊥BC于点E,求∠ABO+∠BCO+∠CAO的度数

Answer 1

∠ABC+∠BCA+∠CAB=180°

三个内角的角平分线交于O

∠ABO=1/2∠ABC

∠BCO=1/2∠BCA

∠CAO=1/2∠CAB

∠ABO+∠BCO+∠CAO=1/2(∠ABC+∠BCA+∠CAB)=1/2×180°=90°