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

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

Given