$∵AB=AC,AD平分∠BAC$ $∴AD⊥BC$ $∴∠ADC=90°$ $又∵E是AC中点,∴DE=\frac {1}{2}AC,AE=\frac {1}{2}AC$ $∴DE=AE,∴△ADE是等腰三角形$ $解:∵OC平分∠BCA,OA平分∠BAC$ $∴∠OCE=∠OCA,∠OAD=∠OAC$ $∵DE//AC,∴∠COE=∠ACO,∠AOD=∠CAO$ $∴∠ECO=∠EOC,∠DAO=∠DOA$ $∴EC=EO,DO=DA$ $∴C_{△BDE}=BD+DO+OE+BE=BD+DA+EC+BE=BA+BC=18$
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