Advanced Geometric Problem & Solution (Twice a week update) Now 4502 problems posted.
<Tri-2024-2151>, hint
Since DE is bisector of <BEC BE/EC=BD/DC
and BF=BE AF=CE. BF/FA=BD/DC ---> FD//AC ---> <FDE=<CED=<BED
DF=BF(DF//AC, )=BE, DE is common. Triangles BDE and FED are congruent
Hence BD=EF .
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