Diagonals Bisect Each Other

Here, we need to prove that the diagonals of a parallelogram bisect each other. Lets start by drawing in our diagonals

As you'll remember from our opposite sides proof, these diagonals are also transversals. Lets look at diagonal CB first.

This shows us that angles ∠2 and ∠5 are alternate interior angles, so they are congruent.

Now looking at diagonal AD

We see that angles ∠1 and ∠4 are alternate interior angles, so they are congruent

Putting it all together

We can use this picture for the rest of the proof. Notice that if you are good at seeing alternate interior angles, you could actually just start with this picture.

Now lets think about what we know. We have two congruent pairs of angles. We also know that opposite sides of a parallelogram are congruent, so AB and CD are congruent. That means we have an ASA situation, and thus ΔABE ≅ ΔDCE

From here, its simple. If the triangles are congruent, then

• BE ≅ CE (CPCTC)

  • AE ≅ DE (CPCTC)

But wait, if BE ≅ CE, that means BC was bisected. Also if AE ≅ DE, that means AD was bisected. And therefore the diagonals are bisecting each other.

Lets look at how we would write that in a 2-column proof

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Prove that diagonals of a parallelogram bisect each other.

To watch a Khan Academy video of this proof, click here