# A line which bisects two sides of a triangle is parallel to the third.

The answers to this question imply that a line which bisects two sides of a triangle must be parallel to the third side. Why is this true? There must be a simple proof.

More generally: Let D and E be points on $\overline{AB}$ and $\overline{CB}$, respectively, such that $AD:DB=CE:EB$. Then, $\overleftrightarrow{DE} \parallel \overline{AC}$. (Thanks @Isaac)

We could also look at the contrapositive. Suppose that $DF$ is not parallel to $AC$. Then the triangles $DBF$ and $ABC$ won't be similar.

-
More generally: Let $D$ and $E$ be points on $\overline{AB}$ and $\overline{CB}$, respectively, such that $AD:DB=CE:EB$. Then, $\overleftrightarrow{DE}\parallel\overline{AC}$. – Isaac Feb 2 '12 at 20:39
Also, the generalization I mentioned is the converse of the "Side-Splitting Theorem" in *UCSMP Geometry*—the author and supposed originator of that specific theorem name has said it's a very funny theorem. – Isaac Feb 2 '12 at 22:41

This follows from the Intercept Theorem .

Let $DE$ be the line segment joining the midpoint $D$ of $AB$ and the midpoint $E$ of $BC$. Draw a line parallel to to $DE$ that passes through $A$. Extend the side $BC$ so that it intersects this line in the point $F$. By the intercept theorem, $${DB\over DA}={BE\over EF}$$

But $DB=DA$, so, $EF=BE=EC$. It follows that $F=C$, and, thus, $AC$ is parallel to $DE$.

(Of course, you could argue using similar triangles too. The intercept Theorem is equivalent to "the similar triangle business".)

-

Use "like" triangles, i.e. side angle side, in proportion.

-