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A theorem in Euclid's Elements states that the line segment between the midpoints of two sides of a triangle is parallel to the third side and is half its length. I have looked in various textbooks in geometry and have not found a generalization of this. If $0 < p < 1$, such as $2/3$ or $\sqrt{2}/2$, and if $E$ and $F$ are points on sides $\overline{AB}$ and $\overline{AC}$, respectively, of $\triangle{ABC}$ such that \begin{equation*} \bigl\vert \overline{AE} \bigr\vert = p\bigl\vert \overline{AB} \bigr\vert \qquad \text{and} \qquad \bigl\vert \overline{AF} \bigr\vert = p\bigl\vert \overline{AC} \bigr\vert , \end{equation*} $\overline{EF}$ is parallel to $\overline{BC}$, and \begin{equation*} \bigl\vert \overline{EF} \bigr\vert = p\bigl\vert \overline{BC} \bigr\vert . \end{equation*} I would appreciate an explanation in the case that $p$ is a proper fraction.

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    $\begingroup$ It seems like it would suffice to consider similar triangles. Do you know what those are? Can you see why you get a triangle similar to your original one, just with sidelengths scaled by p? $\endgroup$ – Improve Dec 13 '15 at 2:21
  • $\begingroup$ @Improve I am looking for an explanation that would fit just after the cited theorem regarding the line segment between the midpoints of two sides of a triangle. The argument for that theorem involved the congruence of two triangles. $\endgroup$ – user74973 Dec 13 '15 at 2:26
  • $\begingroup$ Do you agree why the line EF should be parallell to BC? $\endgroup$ – Improve Dec 13 '15 at 2:27
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    $\begingroup$ Perhaps this is what you are looking for: proofwiki.org/wiki/… $\endgroup$ – Improve Dec 13 '15 at 2:32
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    $\begingroup$ For rational $p$ ---that is, $p = q/r$ where $q$ and $r$ are integers--- you can craft an argument that sub-divides the each side of the triangle into $r$ equal pieces. Connecting the various division points together appropriately forms what appears to be a triangular grid of $r^2$ congruent triangles, although it falls to you to prove that all the "grid lines" meet in the appropriate places and that the sub-triangles are in fact congruent. (The Midpoint Theorem itself helps show that, when applied systematically to various $2$-by-$2$ triangles in the "grid".) The result follows. $\endgroup$ – Blue Dec 13 '15 at 2:43

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