-kcapm@(ZH33!DNlJ-K^{UUUffffffwwwwww 0PQ%Given: Triangle ABC with medians AD, BE, and CF Show: AD, BE, and CF concurrent By the definition of median, AF=FB, BD=DC, and CE=EA. Therefore, each of the ratios AF/FB, BD/DC, and CE/EA equals 1. So their product equals 1 as well. Therefore AD,BE, and CF are concurrent by Ceva's Theorem. ,VhM[[ UT[`[@A [P[ [ NlJ-K^{C[pHH[0[C?CC .3[ NlJ-K^{B[`HH[0[B@C2 %33 NlJ-K^{AUUffffffwwwwww CB [P  NlJ-K^{l[ebd',ysgm  C?CCCB?-[  NlJ-K^{kZrDD[[ UTwwwww[ B@C2C?CC?-e DV  NlJ-K^{jle System Profiler Calculaor ChooserControl PanelsCBB@C2?h`me NlJ-K^{F[`0[P[[` BB qv NlJ-K^{E[P0[@[[P C)B v{  NlJ-K^{D D;G@7C[`UTC@DB?CB>B[EBC: g_e DV  NlJ-K^{ole System Profiler Calculaor ChooserControl PanelsC?CCBB?p-e DV  NlJ-K^{nle System Profiler Calculaor ChooserControl PanelsB@C2C)B? u33  NlJ-K^{mUUffffffwwwwww BC:CB? e DV NlJ-K^{Gle System Profiler Calculaor ChooserControl PanelsCC