Triangle Proportionality Theorem

The Triangle Proportionality Theorem says

• A line parallel to one side of a triangle divides the other two proportionally.

Let's see if this is true.

Demonstration

In the following figure, line ED and line segment AB are parallel. Prove that

=

If we see DE and AB as parallel lines, then line DA is a transversal. It's the classic "two parallel lines and a transversal". In that case, angles 1 and 2 are corresponding angles, so ∠1 ≅ ∠2

If we see DE and AB as parallel lines, then line CB is a transversal. It's the classic "two parallel lines and a transversal". In that case, angles 3 and 4 are corresponding angles, so ∠3 ≅ ∠4

Since we now have two pairs of congruent angles, we can say that ΔABD ~ ΔCDE by the AA criteria for triangle similarity.

If the triangles are similar, that means their sides are proportional. In other words:

=

We know that CA is made up of CD and DA. In other words, CA = CD + DA.

On the other hand CB = CE + EB. Breaking these apart, we get

=

Distribute the denominators

+ = +

Any number divided by itself = 1

1 +

= 1 +

Subtract 1 from both sides

=

Take the reciprocal of both sides

=

and we've got it.

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Here's how to do it in a 2-column proof

If ED || AB, prove that:

=