# Proof: By Euclid

• For let $BE$ and $CD$ have been joined.
• Thus, triangle $BDE$ is equal to triangle $CDE$.
• For they are on the same base $DE$ and between the same parallels $DE$ and $BC$ [Prop. 1.38].
• And $ADE$ is some other triangle.
• And equal (magnitudes) have the same ratio to the same (magnitude) [Prop. 5.7].
• Thus, as triangle $BDE$ is to [triangle] $ADE$, so triangle $CDE$ (is) to triangle $ADE$.
• But, as triangle $BDE$ (is) to triangle $ADE$, so (is) $BD$ to $DA$.
• For, having the same height - (namely), the (straight line) drawn from $E$ perpendicular to $AB$ - they are to one another as their bases [Prop. 6.1].
• So, for the same (reasons), as triangle $CDE$ (is) to $ADE$, so $CE$ (is) to $EA$.
• And, thus, as $BD$ (is) to $DA$, so $CE$ (is) to $EA$ [Prop. 5.11].
• And so, let the sides $AB$ and $AC$ of triangle $ABC$ have been cut proportionally (such that) as $BD$ (is) to $DA$, so $CE$ (is) to $EA$.
• And let $DE$ have been joined.
• I say that $DE$ is parallel to $BC$.
• For, by the same construction, since as $BD$ is to $DA$, so $CE$ (is) to $EA$, but as $BD$ (is) to $DA$, so triangle $BDE$ (is) to triangle $ADE$, and as $CE$ (is) to $EA$, so triangle $CDE$ (is) to triangle $ADE$ [Prop. 6.1], thus, also, as triangle $BDE$ (is) to triangle $ADE$, so triangle $CDE$ (is) to triangle $ADE$ [Prop. 5.11].
• Thus, triangles $BDE$ and $CDE$ each have the same ratio to $ADE$.
• Thus, triangle $BDE$ is equal to triangle $CDE$ [Prop. 5.9].
• And they are on the same base $DE$.
• And equal triangles, which are also on the same base, are also between the same parallels [Prop. 1.39].
• Thus, $DE$ is parallel to $BC$.
• Thus, if some straight line is drawn parallel to one of the sides of a triangle, then it will cut the (other) sides of the triangle proportionally.
• And if (two of) the sides of a triangle are cut proportionally, then the straight line joining the cutting (points) will be parallel to the remaining side of the triangle.
• (Which is) the very thing it was required to show.

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