what is the proof that a centroid of a triangle intersects each median in a 2:1 ratio?

Answer 1

Say, you have a triangle ABC, and AD, BE, CF are the medians intersected at point O. Please draw a line through C, such that it parallels to BE and intersects the extended AD at G. Try now to see if you can solve the problem without read the rest.
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Since GC // OE and AE = EC, OE is the midline in triangle AGC and thus AO = OG.
Also we can prove that triangles DBO and DCG congruent to one another (with angle-side-angle), since BD = DC, angle BDO = angle CDG, and angle OBD = angle GCD. Therefore OD = DG. Finally:
AO = OG = OD + DG = 2*OD

Answer 2

Construction: Let the three medians of a triangle ABC meet in G. Let Q be the midpoint of GB, P the midpoint of AG, K the midpoint of CG PROOF: PQ is parallel to AB and also PQ=1/2 AB (mid segment theorem) ON is parallel to AB and also ON=1/2 AB (mid segment theorem) This means that PQ=ON and PQ is parallel to ON. We can conclude here that the quad OPQN is a parallelogram (Theorem). Also we know that diagonals of a parallelogram bisect each other. In this case PN intersects with OQ in G. We can also argue that the median AN intersects with the median BO at G. Using the same argument that we used to show that OPQM is a parallelogram, we can show that KPMN is also a parallelogram. Similarly the diagonals of KPMN intersect and bisect each other at the point G, also the median CM intersects with the median BO in G. From above we know that the median BO intersects with the median AN in G, therefore G must be the common point where all the three medians are meeting. So the three medians are concurrent. For us to show that the medians intersect each other in the ratio 2:1, we can just use the results from above. We have already proved that OPQM and KPMN are parallelograms. We know also that diagonals of a parallelogram bisect each other. This means then that in parallelogram OPQN, OG =GQ. But we also know through construction that GQ =QB. Therefore OG=GQ=QB=1/3 OB. ( or BG=2/3OB) Using the same argument we can say that AP=PG=GN = 1/3 AN(or AG=2/3AN). In the same manner we can show that CG=2/3 CM. The conclusion is therefore that the medians of a triangle intersect each other in the ratio 2:1