如图,已知△ABC中,三条内角平分线AD,BE,CF相交于点O,OG⊥BC.求证：∠BOD=∠GOC.

∠BOD
＝∠ABO＋∠OAB
＝1/2∠ABC＋1/2∠BAC
＝1/2(∠ABC＋∠BAC)
＝1/2(180°－∠ACB)
＝90°－1/2∠ACB
＝90°－∠OCG
＝∠COG