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Prove using similar triangle properties that "any two medians of a triangle divide each other in the ratio $2:1$. I do not know which criteria of similar triangle must be used

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Here's a diagram:

enter image description here

$|AB|:|ED|=2:1$ and $AB\parallel ED$ (midpoint theorem) $\Longrightarrow \triangle ABC \sim \triangle EDC$.

Now use alternate interior angles to prove $\triangle ABS \sim \triangle DES$. Use the fact that $|AB|:|ED|=2:1$ to conclude.

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  • $\begingroup$ Your answer is essentially correct. I just add in some details. The first two results are based on the “midpoint theorem”. From those facts, we conclude that △ABC∼△EDC. $\endgroup$ – Mick Aug 24 '14 at 18:10
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There are several properties of similar triangles that you can use, including:

  1. If two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar.

  2. The three sides of one triangle are proportional to the three corresponding sides of another triangle if and only if the triangles are similar.

  3. If two sides of one triangle are proportional to two sides of another triangle and their included angles are congruent, then the triangles are similar.

  4. If a line divides two sides of a triangle proportionally, then it is parallel to the third side.

In this diagram, point $D$ is the midpoint of $\overline {AC} $, point $E$ is the midpoint of $\overline {BC} $, point $H$ is the midpoint of $\overline {AF} $, and point $G$ is the midpoint of $\overline {BF} $.

enter image description here

Examine this carefully, find the similar triangles, and use their properties. Don't forget to look for congruent triangles as well.

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