The sum of the opposite angles of quadrilaterals in circles equals two right angles.

Let *ABCD* be a circle, and let *ABCD* be a quadrilateral in it.

I say that the sum of the opposite angles equals two right angles.

Join *AC* and *BD.*

Then, since in any triangle the sum of the three angles equals two right angles, the sum of the three angles *CAB, ABC,* and *BCA* of the triangle *ABC* equals two right angles.

But the angle *CAB* equals the angle *BDC,* for they are in the same segment *BADC,* and the angle *ACB* equals the angle *ADB,* for they are in the same segment *ADCB,* therefore the whole angle *ADC* equals the sum of the angles *BAC* and *ACB.*

Add the angle *ABC* to each. Therefore the sum of the angles *ABC, BAC,* and *ACB* equals the sum of the angles *ABC* and *ADC.* But the sum of the angles *ABC, BAC,* and *ACB* equals two right angles, therefore the sum of the angles *ABC* and *ADC* also equal two right angles.

Similarly we can prove that the sum of the angles *BAD* and *DCB* also equals two right angles.

Therefore *the sum of the opposite angles of quadrilaterals in circles equals two right angles. *

Q.E.D.