Take any point O in the triangle ABC, connect Ao, Bo, CO respectively, and extend the opposite edge to a ', B', C '. Prove: OA' / AA '+ ob' / BB '+ OC' / CC '= 1

Take any point O in the triangle ABC, connect Ao, Bo, CO respectively, and extend the opposite edge to a ', B', C '. Prove: OA' / AA '+ ob' / BB '+ OC' / CC '= 1

Through O, Mn is parallel to BC, AB is parallel to m, AC is parallel to n,
Then ob '/ BB' = on / BC
OC'/CC'=MO/BC
Add the two formulas,
OB '/ BB' + pg / CG = an / AC
Because OA '/ AA' = CN / AC
So OA '/ AA' + ob '/ BB' + OC '/ CC' = 1