State and prove Converse of cyclic quadrilateral theorem.
Converse is stated as:
If a pair of opposite angles of a quadrilateral is supplementary, the quadrilateral is cyclic.
Proof: A proof by contradiction is a good approach. Suppose you have a quadrilateral ABCD whose opposite angles are supplementary, but it is not cyclic. The vertices A,B,C determine a circle, and the point D does not lie on this circle, since we assume the quadrilateral is not cyclic.
Suppose for instance that D lies outside the circle, and so the circle intersects ABCD at some point E on CD (try drawing a picture to see this if needed.) Now D is supplementary to B, and since E is the opposite angle of B in the cyclic quadrilateral ABCE, E is supplementary to B by the theorem you already know, and so D and E are congruent. But this contradicts the fact that an exterior angle cannot be congruent to an interior angle, which proves the converse. A similar method works if D lies inside the circle as well. (I abuse notation a bit and refer to a vertex and the angle at that vertex by the same letter.)
If a pair of opposite angles of a quadrilateral is supplementary, the quadrilateral is cyclic.
Proof:
A proof by contradiction is a good approach. Suppose you have a quadrilateral ABCD whose opposite angles are supplementary, but it is not cyclic. The vertices A,B,C determine a circle, and the point D does not lie on this circle, since we assume the quadrilateral is not cyclic.
Suppose for instance that D lies outside the circle, and so the circle intersects ABCD at some point E on CD (try drawing a picture to see this if needed.) Now D is supplementary to B, and since E is the opposite angle of B in the cyclic quadrilateral ABCE, E is supplementary to B by the theorem you already know, and so D and E are congruent. But this contradicts the fact that an exterior angle cannot be congruent to an interior angle, which proves the converse. A similar method works if D lies inside the circle as well. (I abuse notation a bit and refer to a vertex and the angle at that vertex by the same letter.)
