Parallel Transversal Theorem

Theorem

 * If a straight line be drawn parallel to one of the sides of a triangle, it will cut the sides of the triangle proportionally; and, if the sides of the triangle be cut proportionally (so that the segments adjacent to the third side are corresponding terms in the proportion), the line joining the points of section will be parallel to the remaining side of the triangle.

Proof
Let $\triangle ABC$ be a triangle, and $DE$ be drawn parallel to the side $BC$.

We need to show that $BD : DA = CE : EA$.


 * Euclid-VI-2.png

Let $BE$ and $CD$ be joined.

From Triangles with Equal Base and Same Height have Equal Area the area of $\triangle BDE$ is the same as the area of $\triangle CDE$.

From Ratios of Equal Magnitudes, $\triangle BDE : \triangle ADE = \triangle CDE : \triangle ADE$.

From Areas of Triangles and Parallelograms Proportional to Base, $\triangle BDE : \triangle ADE = BD : DA$.

By the same reasoning, $\triangle CDE : \triangle ADE = CE : EA$.

From Equality of Ratios is Transitive, $BD : DA = CE : EA$.

Now let the sides $AB, AC$ of $\triangle ABC$ be cut proportionally, so that $BD : DA = CE : EA$.

Join $DE$.

We need to show that $DE \parallel BC$.

We use the same construction as above.

From Areas of Triangles and Parallelograms Proportional to Base we have that:
 * $BD : DA = \triangle BDE : \triangle ADE$
 * $CE : EA = \triangle CDE : \triangle ADE$

From Equality of Ratios is Transitive, $\triangle BDE : \triangle ADE = \triangle CDE : \triangle ADE$.

So from Magnitudes with Same Ratios are Equal, the area of $\triangle BDE$ is the same as the area of $\triangle CDE$.

But these triangles are on the same base $DE$.

So from Equal Sized Triangles on Same Base are Same Height, it follows that $DE$ and $BC$ are parallel.