Question

三角行ABC的两外角平分线BF、CF交于F,求证F=90°-1/2∠A

Answer 1

∠F=180°－∠CBF－∠BCF=180°－(1/2)[180°－∠ABC]－(1/2)[180°－∠ACB]=180°－90°＋(1/2)∠ABC－90°＋(1/2)∠ACB=(1/2)[∠ABC＋∠ACB]=(1/2)[180°－∠A]=90°－(1/2)∠A.