Parallel Transversal Theorem

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 have Same Height, it follows that $DE$ and $BC$ are parallel.