Question

已知：如图,在三角形ABC中,AB=8,BC=10,角BAC,角BCA的角平分线相交于点O的直线DE平行AC分别交AB,

Answer 1

∵OA平分∠BAC,∴∠OAB=∠OAC,

∵DE∥AC,∴∠OAC=∠AOD,

∴∠AOD=∠OAD,

∴OD=AD,

同理：CE=OE,

∴ΔBDE周长=AB+BC=18.