In any triangle the sum of any two angles is less than two right angles.

Let *ABC* be a triangle.

I say that the sum of any two angles of the triangle *ABC* is less than two right angles.

Produce *BC* to *D.*

Since the angle *ACD* is an exterior angle of the triangle *ABC,* therefore it is greater than the interior and opposite angle *ABC.* Add the angle *ACB* to each. Then the sum of the angles *ACD* and *ACB* is greater than the sum of the angles *ABC* and *BCA.*

But the sum of the angles *ACD* and *ACB* is equal to two right angles. Therefore the sum of the angles *ABC* and *BCA* is less than two right angles.

Similarly we can prove that the sum of the angles *BAC* and *ACB* is also less than two right angles, and so the sum of the angles *CAB* and *ABC* as well.

Therefore *in any triangle the sum of any two angles is less than two right angles.*

Q.E.D.

- ... the angle

- If

This proposition is strengthened in Proposition I.32 to say the sum of all three angles in a triangle equals two right angles.