Me. i really need with this proof.

Usually you prove the medians are concurrent way before you get the Ceva's Theorem.  But we do the problem as given.  Next time try to include the whole problem in the photo.

FA=FB, BE=EC, CG=GA

C. Definition of median

The medians go to the midpoints of the sides, so we have congruent segments from the foot.

FA/FB=1, BE/EC=1, CG/GA=1

A. Division of equal segment lengths yields 1

This one is cut off so we can't read the full justification, but we have equal segments so their ratio is 1.

(FA/FB)(BE/EC)(CG/GA)=1

E. Substitution property

Again the justification is cut off, but we have three things equal to 1, when we multiply them together we still get 1.

Medians are Concurrent

B. Ceva's Theorem

This is really the converse of Ceva's Theorem; we have the required ratio product equal to 1, so we know the medians are concurrent.   Technically they might be parallel (which is concurrent in the projective sense, meeting at infinity).  So we really need to show they're not parallel as well, which they can't be, but they left that part out.

I don't like this proof very much.

rx and sx

step-by-step explanation: