In any triangle, if one of the sides is produced, then the exterior angle equals the sum of the two interior and opposite angles, and the sum of the three interior angles of the triangle equals two right angles.

Let *ABC* be a triangle, and let one side of it *BC* be produced to *D.*

I say that the exterior angle *ACD* equals the sum of the two interior and opposite angles *CAB* and *ABC,* and the sum of the three interior angles of the triangle *ABC, BCA,* and *CAB* equals two right angles.

Draw *CE* through the point *C* parallel to the straight line *AB.*

Since *AB* is parallel to *CE,* and *AC* falls upon them, therefore the alternate angles *BAC* and *ACE* equal one another.

Again, since *AB* is parallel to *CE,* and the straight line *BD* falls upon them, therefore the exterior angle *ECD* equals the interior and opposite angle *ABC.*

But the angle *ACE* was also proved equal to the angle *BAC.* Therefore the whole angle *ACD* equals the sum of the two interior and opposite angles *BAC* and *ABC.*

Add the angle *ACB* to each. Then the sum of the angles *ACD* and *ACB* equals the sum of the three angles *ABC, BCA,* and *CAB.*

But the sum of the angles *ACD* and *ACB* equals two right angles. Therefore the sum of the angles *ABC, BCA,* and *CAB* also equals two right angles.

Therefore *in any triangle, if one of the sides is produced, then the exterior angle equals the sum of the two interior and opposite angles, and the sum of the three interior angles of the triangle equals two right angles.*

Q.E.D.

**Corollary 1.**
The sum of the interior angles of a convex rectilinear figure equals twice as many angles as the figure has sides, less four.

**Corollary 2.**
The sum of the exterior angles of any convex rectilinear figure together equal four right angles.