Prove that centroid of a triangle divides its median in the ratio of 2:1.
You need to sketch an acute triangle ABC. Mark on this sketch the midpoint f each side such that: D denotes the midpoint of BC, E denotes the midpoint of AC, F denotes the midpoint of AB.
Joining the midpoints E and F yields the midline EF that is parallel to BC and it is half as long.
You need to focus on two triangles: GBC and GEF (G denotes the centroid of triangle ABC). The line `EF || BC =gt ltGEF-=ltGBC ` (alternate interior angles); `ltGFE-=ltGCB` (alternate interior angles); `ltEGF-=ltBGC` (opposite angles) => `Delta GBC-= DeltaGEF` .
Considering midline theorem yields that `EF = BC/2` . Since `Delta GBC-= DeltaGEF =gt (BG)/(GE)=(GC)/(GF)=(GA)/(GD) = 2/1` .
Considering congruent triangles DFG and GAC => `AG/GD = 2/1` .
Hence, considering all the above yields that centroid of a triangle divides its median in the ratio of 2:1.