Cyclic Quadrilaterals

First take intercepted arcs of $182^\circ$ and $178^\circ$. Halving gives $91^\circ$ and $89^\circ$, and their sum is $180^\circ$. We now follow the same steps without fixing the arc measures.

Let $A,B,C,D$ be distinct points in convex order on a circle. Write $p$ for the measure of arc $BCD$ and $q$ for the measure of arc $DAB$. These are exactly the two arcs from $B$ to $D$, so

The inscribed-angle theorem gives

Adding the two angle measures therefore gives

Using the two arcs from $A$ to $C$ instead gives

Neither step depended on the numerical arcs chosen first.

The comparison can first be rehearsed with numbers. Suppose $ABCE$ is cyclic, its angles at $A,C$ are $85^\circ,95^\circ$, and $X$ is the intersection of its diagonals. A point $D$ strictly between $X$ and $E$ gives smaller angles at both $A$ and $C$, so their sum is less than $180^\circ$. A point beyond $E$ on ray $BX$ gives larger angles at both vertices, so their sum exceeds $180^\circ$. Only $D=E$ can give the required sum. The following argument explains each comparison and constructs $E$.

Let convex $ABCD$ have a supplementary opposite pair. Since its four angles total $360^\circ$, we have

Let diagonals $AC,BD$ meet at $X$. Convexity puts $X$ strictly inside both segments: $A,X,C$ are in order, and $B,X,D$ are in order. Draw the circle through noncollinear $A,B,C$.

First, $X$ is strictly inside this circle. To see this, let $M$ be the perpendicular foot from its center $O$ to chord line $AC$. If $AC$ is not a diameter, $M$ is the chord midpoint, and $MX<MA$. Pythagoras compares the right triangles with common leg $OM$, giving $OX<OA$. If $X=M$, this follows directly from $OM<OA$. If $AC$ is a diameter, $O$ is its midpoint and an interior point $X$ has $OX<OA$ directly.

Line $BX$ therefore has a second circle intersection $E$, with $B,X,E$ in order. Here is the existence and order check: let $H$ be the perpendicular foot from $O$ to this line. Because $X$ is strictly inside, $OH$ is less than the radius. The line-circle cases give two intersections, equidistant from $H$. Pythagoras makes points strictly between those intersections exactly the points of the line inside the circle, so $X$ is between them. One is $B$; the other is $E$. If $H=O$, these are simply the ends of a diameter.

The point $E$ is on the opposite side of $AC$ from $B$. Thus $ABCE$ is convex in that order, and the theorem already proved gives

Both $D$ and $E$ lie on ray $BX$ beyond $X$. If they differ, there are exactly two orders to examine.

Case 1: $X,D,E$ are in order. Then $D$ is strictly inside triangle $ACE$. At $A$, ray $AD$ lies strictly between $AC$ and $AE$. Diagonal $AC$ is inside both convex whole angles $BAD$ and $BAE$, so the positive extra angle $DAE$ gives

At $C$, ray $CD$ lies strictly between $CA$ and $CE$, giving

Their sum is less than $180^\circ$, contradicting the given sum.

Case 2: $X,E,D$ are in order. Now $E$ is strictly inside triangle $ACD$. The same angle additions, with $D,E$ exchanged, give

Their sum exceeds $180^\circ$, again contradicting the hypothesis.

Both unequal positions are excluded, so $D=E$ and $D$ lies on the circle through $A,B,C$. The numerical rehearsal used only these strict comparisons, not special values of the two angles.

For the numerical instance above, the linear pair and the opposite-angle theorem each give $180^\circ-54^\circ$, or $126^\circ$. In general, the linear pair gives

For convex cyclic $ABCD$, the opposite-angle theorem gives

Thus

Conversely, for a convex quadrilateral with the same side extension, this equality and the linear pair make the opposite interior angles supplementary. The converse test then proves cyclicity.