It is known that in the triangle ABC, O is any point in the triangle, and the extension lines of Ao, Bo, CO intersect the opposite sides of D, e, F Verification: AE / EC + AF / FB = AO / OD

It is known that in the triangle ABC, O is any point in the triangle, and the extension lines of Ao, Bo, CO intersect the opposite sides of D, e, F Verification: AE / EC + AF / FB = AO / OD

According to Menelaus theorem, because FOC is the section of △ abd, so (AF / BF) * (BC / DC) * (do / OA) = 1, that is AF / BF = (OA / OD) * (DC / BC);
Similarly, BOE is the intercept of △ ADC, so (AE / EC) * (CB / DB) * (do / OA) = 1, that is AE / EC = (OA / do) * (BD / BC)
Add the two formulas to get: AE / EC + AF / FB = (AO / do) * (DC / BC + BD / BC), DC + BD = BC, so AE / EC + AF / FB = AO / OD
Well, I can tell you another way:
Let a be the parallel line L of BC, extend Bo and l to m, because am ∥ BC, ∥ AE / EC = am / BC; lengthen CO and l to N, then ∥ an ∥ BC, ∥ AF / FB = an / BC; add the above two formulas to get AF / FB + AE / EC = Mn / BC; also ∥ Mn ∥ BC, ∥ Mn / BC = Mo / ob, similarly ∥ am ∥ BD, ∥ OM / Bo = AO / OD
(but if you want to learn plane geometry better, there is no harm in learning some famous theorems:)