olymmath

IR-TST 2007 Problem 1

  • Day 1

In triangle ABCABC , MM is midpoint of ACAC , and DD is a point on BCBC such that DB=DMDB=DM . We know that 2BC2−AC2=AB.AC2BC^{2}-AC^{2}=AB.AC . Prove that BD.DC=AC2.AB2(AB+AC)BD.DC=\frac{AC^{2}.AB}{2(AB+AC)}

Dataset text AI-MO AoPS. Quoted with source.

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