# Parallelogram geometry proof [duplicate]

$\rm ABCD$ is a parallelogram (in the plane) and $\rm E$ is the midpoint of $\rm AB.$ Prove that $\rm AC$ and $\rm DE$ trisect one another. In other words, that their point of intersection $\rm X$ divides each of them in the ratio $2 : 1.$

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## marked as duplicate by Phira, Ｊ. Ｍ., t.b., Henning Makholm, DidOct 31 '11 at 13:35

Hint: Do the triangles $\bigtriangleup{AXE}$ and $\bigtriangleup{CXD}$ have anything in common?

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Medians of the triangle $ABD$ (see picture bellow) intersects in point $X$ and divide each other in ratio $2$:$1$ so we may write following:

$EX:DX=1:2$

$AX=\frac{2}{3}\frac{AC}{2}=\frac{AC}{3} , XF=\frac{1}{3}\frac{AC}{2}=\frac{AC}{6}$

Since diagonals of the parallelogram divide each other in ratio $1:1$ we may write:

$XC=XF+\frac{AC}{2}=\frac{AC}{6}+\frac{AC}{2}=\frac{2AC}{3}$ , so we may conclude:

$AX:XC=\frac{AC}{3}:\frac{2AC}{3}=1:2$

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Ah thank you very much! –  Paul Oct 19 '11 at 15:52