(by Polymath) This page is part of the Advanced High School Math Project.

A good number of theorems from advanced geometry rely on finding cyclic quadrilaterals (that is, quadrilaterals that can be inscribed in a circle) in a complex diagram. Although less-studied in basic geometry courses than their more famous cousins (rhombuses, trapezoids, etc.), they seem to occur proportionately more often in advanced math because they often relate two seemingly unrelated objects—quadrilaterals and circles, obviously.

It is a well-known theorem that the measure of an angle inscribed in a circle is half of the measure of its subtended (intercepted) arc. This further means that two angles that intercept the same arc of a circle must have the same measure. But, of course, if you ever do find two angles that intercept the same arc, then the four points in question (the two vertices of the angles and the two endpoints of the arc) form a cyclic quadrilateral. That's where a lot of the power of cyclic quadrilaterals lies. We'll see how that plays out below.

The most basic theorem about cyclic quadrilaterals is that their opposite angles are supplementary. Note the red and green angles in the picture below.

The two angles subtend arcs that total the entire circle, or 360°. But if their measure is half that of the arc, then the angles must total 180°, so they are supplementary. And of course, since the total measure of the angles in the quadrilateral is 360°, the other two angles are supplementary as well.

We can use that theorem to prove its own converse: that if two opposite angles of a quadrilateral are supplementary, then the quadrilateral is cyclic. The proof is by contradiction. Suppose there is a quadrilateral that *does* have supplementary opposite angles, but *is not* cyclic. That quadrilateral is ABCD' below.

Since it is not cyclic, the circle that goes through A, B, and C must not also go through D'. (The argument is almost identical if D' falls inside the circle rather than outside.) Now locate point D on CD'. By assumption, D' is supplementary to B. But by the theorem we proved above, D is also supplementary to B and therefore congruent to D'. But this is clearly a contradiction: the exterior angle of triangle ADD' can't be the same size as an interior angle. This contradiction proves the theorem. Thus a quadrilateral having opposite supplementary angles is equivalent to its being cyclic. (In particular, it is often useful to remember that a quadrilateral with a pair of opposite right angles is cyclic, and in that case, the diagonal of the quadrilateral that separates the right angles is the diameter of the circle.)

Typically, though, we know more about which angles in a diagram are congruent, not which are supplementary. So a very useful extension of this theorem begins with the following diagram:

Suppose the red angle (8) and the green angle (3) are congruent. You can think of them as two spotlights shining on a stage (segment AB) from points X and Y. Since the vertical angles at T are congruent, we know that angles 7 and 4 are also congruent, and thus that triangle AXT and triangle BYT are similar. Then XT and YT must stand in the same ratio as AT and BT. But since the angles between those pairs of proportional segments are congruent (the other pair of vertical angles at T), triangles XTY and ATB must also be similar. This makes angles 1 and 6 congruent and angles 2 and 5 congruent.

Finally, consider angles 1, 8, 4, and 5. Since they are each congruent (respectively) to angles 6, 3, 7, and 2, they must comprise exactly half the total degree measures of the interior angles of the whole quadrilateral; that is, they must be supplementary. And thus the quadrilateral is cyclic by the previous theorem!

This theorem is hard to state, but easier to visualize: if two congruent angles with their vertices in the same half-plane subtend the same segment on the boundary of that half-plane, then the endpoints of that segment together with the vertices of the angles form a cyclic quadrilateral.

Since the congruent angles appear to be spotlights shining on a stage, I propose that this theorem be called the "spotlight theorem". My students tell me this image helps them see when it can be used.

The reason this theorem is so useful is that if you learn that only two of the angles in that diagram are congruent (3 and 8), suddenly many other pairs are congruent (since they subtend the same arc in the now-implied circle). Often, it won't even be necessary to draw the circle itself, since the congruence of the angles will be what you were looking for.

Cyclic quadrilaterals are a surprisingly rich topic, and they show up in unexpected places. The theorems on this page should provide a good start to studying and using them.