Question

如图,在△ABC中,∠ABC与∠ACB的平分线相交于点O,过点O作DE平行于BC,分别交AB,AC

Answer 1

因为BO,CO分别是∠ABC、∠ACB的角分线所以∠OBC=∠OBD,∠BCO=∠OCE因为DE‖BC所以∠OBC=∠DOB=∠DBO∠BCO=∠COE=∠OCE即BD=OD,OE=CE所以DE=DO+OE=BD+EC