Converse of the Triangle Proportionality Theorem

The Converse of the Triangle Proportionality Theorem says

• • If a line divides two sides of a triangle proportionally, then that line is parallel to the third side

Let's see if this is true.

Demonstration

In the following figure, if

= , prove that line DE is parallel to line AB

This proof is basically the reverse of the proof for the Triangle Proportionality Theorem. It's pretty boring.

Given

=

Take the reciprocal

=

Add 1 to both sides

1 +

= 1 +

Any number divided by itself is 1, so we can replace 1 with CD/CD or CE/CE

+ = +

Combine terms using our common denominator

=

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

On the other hand CB = CE + EB. Combining these parts, we get

=

Now, since both triangles have ∠C included between their corresponding proportional sides, we can use SAS for triangle similarity. Which now means that ΔABD ~ ΔCDE

If the triangles are similar, that means ∠1 ≅ ∠2. since we have a transversal line CA going through DE and AB, then we know ∠1 ≅ ∠2 if DE || AB, because then 1 and 2 would be corresponding angles. But wait, we already know they are congruent. So, by the converse of the corresponding angles theorem, DE || AB. And that's what we wanted to prove all along

~~~~~~~

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

Given

=

Prove that ED || AB,