Geometry · G.C.A.3

Inscribed polygons and cyclic quadrilaterals

Any three points not on one line sit on exactly one circle. Which four-sided shapes can have all four corners on one circle, and how can you tell from the shape alone?

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CircumcenterGeometry Angle sums of polygonsGeometry

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Proving the inscribed angle theoremGeometry

Where you are 15 in Circles

Geometry11

Circles15

Radius and diameter Chords, secants and tangents

Inscribed polygons and cyclic quadrilaterals

Tangent properties Circumference ▾ Show all 15 (10 more)

Why this exists

Pick any three points that are not on one straight line. There is always a circle through all three, and only one: its center is the Circumcenter, the point equally far from all three. So a triangle can always be fitted into a circle with every corner touching it.

Add a fourth point and that luck runs out. The circle is already fixed by the first three points, and the fourth either lands on it or does not. Most of the time it does not. So a question appears that did not exist for triangles: which quadrilaterals, and which polygons in general, have all their corners on one circle?

Finding the center and measuring every distance answers it for one shape at a time. This lesson finds something better. A quadrilateral sits on a circle exactly when its opposite angles add to 180^\circ, so the angles alone decide, with no center and no measuring. The same idea explains why every rectangle and every regular polygon fits in a circle, and why a parallelogram that is not a rectangle never does.

Definition

A polygon is inscribed in a circle when every one of its vertices lies on the circle. The circle is then circumscribed about the polygon, and is called its circumscribed circle or circumcircle. A polygon that can be inscribed in some circle is called cyclic. A cyclic quadrilateral is a quadrilateral whose four vertices lie on one circle.

Figure 1Inscribed polygons. Every vertex lies on the circle, and the circle is circumscribed about the polygon.

Being on the circle is a statement about distances. If the circle is \odot O with radius r, then a polygon A_1 A_2 \ldots A_n is inscribed in it exactly when

OA_1 = OA_2 = \cdots = OA_n = r

So a polygon is cyclic exactly when some point is the same distance from all of its vertices. That point is the center of the circumscribed circle.

Only the vertices have to touch the circle. The sides are chords of the circle and lie inside it, and the polygon is always named with its vertices in order around the circle.

Intuition

Draw the four radii from the center to the corners of a cyclic quadrilateral. They cut it into four thin triangles, and each one has two sides that are radii, so each one is isosceles: its two angles at the circle are equal. Every corner of the quadrilateral is then built from two of these base angles, one from each triangle that meets there.

Now look at a pair of opposite corners, A and C. Between them they collect one base angle from each of the four triangles, every triangle exactly once. The other two corners, B and D, collect the same four base angles again. So the two pairs of opposite corners share the quadrilateral’s 360^\circ equally, and each pair gets 180^\circ.

The converse has a picture too. Fix A, B and D, which settles the circle, and slide C. Push C in toward the diagonal \overline{BD} and the angle at C opens; pull it out, away from \overline{BD}, and the angle closes. There is exactly one distance along the way at which the angle at C is exactly 180^\circ - m\angle A, and that is where C sits on the circle.

Results

The circumscribed circle of a triangle. Every triangle is cyclic, and it has exactly one circumscribed circle. Its center is the Circumcenter, where the perpendicular bisectors of the sides meet, and its radius is the distance from there to any vertex.

When a polygon is cyclic. A polygon is cyclic exactly when some point is equally far from all its vertices, which is the same as saying that the perpendicular bisectors of all its sides pass through one point. That point is the center, and the circle is the only one through the vertices.

Cyclic quadrilateral theorem. If ABCD is a cyclic quadrilateral, its opposite angles are supplementary:

\begin{aligned} m\angle A + m\angle C &= 180^\circ \\ m\angle B + m\angle D &= 180^\circ \end{aligned}

Converse. If a quadrilateral whose sides do not cross has one pair of opposite angles adding to 180^\circ, it is cyclic. One pair is enough: the four angles add to 360^\circ, so the other pair then adds to 180^\circ as well.

Regular polygons. Every regular polygon is cyclic.

Alternate angles of a cyclic polygon with an even number of sides. If A_1 A_2 \ldots A_{2k} is cyclic, with 2k sides, then every other angle adds to the same total:

\begin{aligned} &m\angle A_1 + m\angle A_3 + \cdots + m\angle A_{2k-1} \\ &\qquad = (k-1)\cdot 180^\circ \end{aligned}

and the even-numbered angles m\angle A_2 + m\angle A_4 + \cdots + m\angle A_{2k} add to the same (k-1)\cdot 180^\circ.

For a quadrilateral, k = 2 and this is the cyclic quadrilateral theorem. For a hexagon, k = 3 and alternate angles add to 360^\circ.

Which familiar quadrilaterals are cyclic

The theorem and its converse settle every familiar quadrilateral from its angles alone.

Quadrilateral	Cyclic?
rectangle, square	always
isosceles trapezoid	always (Example 2)
parallelogram	only a rectangle (Example 4)
rhombus	only a square (Exercise 21)
kite	only with two right angles (Example 3)
trapezoid, unequal legs	never (Exercise 22)

A rectangle’s opposite angles are 90^\circ + 90^\circ. A parallelogram’s opposite angles are equal, so being supplementary forces each to be 90^\circ. A kite’s two equal angles are opposite each other, so they too must each be 90^\circ.

Figure 2Three familiar quadrilaterals that fit on a circle.

Why it is true

The triangle. The Circumcenter lesson proves that the perpendicular bisectors of a triangle’s sides meet at one point, and that this point is equally far from all three vertices. The circle with that center through one vertex therefore passes through all three. It is the only one, because the center of any circle through the three vertices is equally far from them, and the perpendicular bisector theorem puts such a point on all three bisectors, which meet in only one point.

The same reasoning covers any polygon. A point equally far from all the vertices is equally far from the two ends of every side, so it lies on every side’s perpendicular bisector; and a point on every bisector is equally far from each pair of neighboring vertices, so it is equally far from all of them.

Opposite angles, center inside. Let ABCD be inscribed in \odot O with O inside ABCD, as in Figure 3. Join O to the four vertices. Since OA = OB = OC = OD, the four triangles OAB, OBC, OCD and ODA are isosceles, and by the Isosceles and equilateral triangles the two base angles of each are equal. Call them a, b, c and d.

Figure 3Four radii cut the quadrilateral into four isosceles triangles. Each corner gets one base angle from each side.

The center is inside the quadrilateral, so the ray from each vertex through O runs inside that vertex’s angle and splits it into two base angles. By angle addition,

\begin{aligned} m\angle A &= d + a, & m\angle B &= a + b, \\ m\angle C &= b + c, & m\angle D &= c + d. \end{aligned}

The angles of a quadrilateral add to 360^\circ (Angle sums of polygons), so

\begin{aligned} 2(a + b + c + d) &= 360^\circ \\ a + b + c + d &= 180^\circ \end{aligned}

Then the opposite angles add up as claimed:

\begin{aligned} m\angle A + m\angle C &= (d + a) + (b + c) = 180^\circ \\ m\angle B + m\angle D &= (a + b) + (c + d) = 180^\circ \end{aligned}

If O lies on a side, say on \overline{AB}, then that side is a diameter and triangle OAB is flat. Take a = 0 and the same lines go through unchanged.

Opposite angles, center outside. If O is outside ABCD, then exactly one side separates it from the quadrilateral; name the vertices so that this side is \overline{AB}, as in Figure 4. At C and at D the ray toward O still runs inside the angle, so m\angle C = b + c and m\angle D = c + d as before. At A, though, the ray \overrightarrow{AB} lies between \overrightarrow{AD} and \overrightarrow{AO}, so d is the sum of \angle DAB and a, and m\angle A = d - a. In the same way m\angle B = b - a.

Figure 4The center outside. Triangle OAB now lies outside ABCD, so its base angles are taken away instead of added.

The angle sum now adds (d - a), (b - a), (b + c) and (c + d). Each of b, c and d appears twice and a is taken away twice, so

\begin{aligned} 2(b + c + d - a) &= 360^\circ \\ b + c + d - a &= 180^\circ \end{aligned}

and once more

\begin{aligned} m\angle A + m\angle C &= (d - a) + (b + c) = 180^\circ \\ m\angle B + m\angle D &= (b - a) + (c + d) = 180^\circ \end{aligned}

This is Proposition 22 of Book III of Euclid’s Elements. Euclid reaches it through angles that stand on the same arc; the argument here needs only isosceles triangles and the angle sum.

The converse. Its proof compares the angle at C with the angle at a point of the circle, which makes precise the sliding picture in the Intuition section. It is written out in proving the inscribed angle theorem, once the angles on a circle are measured by their arcs.

Regular polygons. Let A_1 A_2 A_3 \ldots A_n be regular, with every side s and every angle \theta. Triangle A_1 A_2 A_3 is isosceles, so its circumcenter O lies on its line of symmetry, the bisector of \angle A_2; let R = OA_2. Then \angle OA_2A_3 is \tfrac{\theta}{2}. Triangle OA_2A_3 is isosceles (OA_2 = OA_3), so \angle OA_3A_2 is \tfrac{\theta}{2} too, which leaves \tfrac{\theta}{2} for \angle OA_3A_4. Triangles OA_3A_2 and OA_3A_4 share \overline{OA_3}, have A_3A_2 = A_3A_4 = s, and have equal angles \tfrac{\theta}{2} between those sides, so they are congruent by SAS (SSS, SAS, ASA, AAS) and OA_4 = OA_2 = R. The same step carries on from A_4 to A_5 and all the way round, so every vertex is R from O.

Alternate angles. Join the center O to every vertex of a cyclic 2k-gon and call the base angles of triangle OA_iA_{i+1} each t_i (with A_{2k+1} meaning A_1). With O inside, the angle at A_i is t_{i-1} + t_i, one base angle from the triangle on each side. Each triangle OA_iA_{i+1} has one end at an odd-numbered vertex and one at an even-numbered vertex, so the odd-numbered angles together contain every t_i exactly once, and so do the even-numbered ones. All 2k angles together contain every t_i twice and add to (2k - 2) \cdot 180^\circ, so each half is (k - 1) \cdot 180^\circ. With O outside, one triangle’s base angle is taken away at both of its ends instead of added, exactly as in the quadrilateral, and the same count holds.

Worked examples

Example 1

The circle around a rectangle

A rectangle measures 8 by 6. Show that it is cyclic and find the radius of its circumscribed circle.

Solution

The diagonals of a rectangle are equal and bisect each other (Rectangles, rhombuses and squares). So the point M where they cross, in the left panel of Figure 2, is half a diagonal from each of the four vertices. A point equally far from all four vertices is the center of a circle through them, so the rectangle is cyclic.

Each diagonal is the hypotenuse of a right triangle with legs 8 and 6, so by the The Pythagorean theorem

\sqrt{8^2 + 6^2} = \sqrt{100} = 10

and the radius is half of that, r = 5.

Check. The theorem agrees: opposite angles of a rectangle are 90^\circ + 90^\circ = 180^\circ. The center of the circle around any rectangle is where its diagonals cross, and each diagonal is a diameter.

Example 2

An isosceles trapezoid

In trapezoid ABCD, \overline{AB} \parallel \overline{DC}, the legs \overline{AD} and \overline{BC} are equal and not parallel, and m\angle A = m\angle B = 68^\circ, as in the middle panel of Figure 2. Find m\angle C and m\angle D, and show that ABCD is cyclic.

Solution

\overline{AD} crosses the two parallel sides, so \angle A and \angle D are same-side interior angles and add to 180^\circ:

m\angle D = 180^\circ - 68^\circ = 112^\circ

In the same way \overline{BC} crosses the parallels, so m\angle C = 180^\circ - 68^\circ = 112^\circ.

Now add a pair of opposite angles: m\angle A + m\angle C = 68^\circ + 112^\circ = 180^\circ. By the converse, ABCD is cyclic.

Nothing here used the number 68. In every isosceles trapezoid the base angles at A and B are equal (Proving properties of an isosceles trapezoid), and \angle B and \angle C always add to 180^\circ, so \angle A and \angle C do too. Every isosceles trapezoid is cyclic.

Check. 68 + 68 + 112 + 112 = 360.

Example 3

A kite on a circle

Kite ABCD has AB = AD and CB = CD, and it is inscribed in a circle with m\angle BAD = 70^\circ, as in the right panel of Figure 2. Find the other three angles.

Solution

In a kite the diagonal \overline{AC} splits the shape into triangles ABC and ADC, which are congruent by SSS (AB = AD, CB = CD, and \overline{AC} is shared). So \angle B \cong \angle D.

The kite is cyclic, so \angle B and \angle D, which are opposite, add to 180^\circ. Two equal angles adding to 180^\circ are each half of it:

\begin{aligned} 2\, m\angle B &= 180^\circ \\ m\angle B = m\angle D &= 90^\circ \end{aligned}

\angle C is opposite \angle A, so m\angle C = 180^\circ - 70^\circ = 110^\circ.

Check. 70 + 90 + 110 + 90 = 360. A kite fits on a circle only when its two equal angles are right angles, and then \overline{AC} is a diameter.

Example 4

Why a parallelogram on a circle is a rectangle

Parallelogram PQRS is inscribed in a circle. Prove that it is a rectangle.

Solution

Opposite angles of a parallelogram are equal (Properties of a parallelogram), so m\angle P = m\angle R; call the common value x. The parallelogram is inscribed, so by the cyclic quadrilateral theorem \angle P and \angle R are also supplementary:

\begin{aligned} x + x &= 180^\circ \\ x &= 90^\circ \end{aligned}

The same argument with \angle Q and \angle S makes them right angles too, so all four angles are 90^\circ and PQRS is a rectangle.

Each theorem alone allows many shapes. Together, equal and supplementary, they leave only 90^\circ. A rhombus is a parallelogram, so a rhombus on a circle is a rectangle with four equal sides, which is a square.

Example 5

Angles from the center

Quadrilateral ABCD is inscribed in \odot O with O inside it, as in Figure 3. You know m\angle OAB = 20^\circ, m\angle OBC = 35^\circ and m\angle OCD = 50^\circ. Find m\angle ODA and the four angles of ABCD.

Solution

The triangles OAB, OBC, OCD and ODA are isosceles, because their sides from O are radii. So their base angles are a = 20^\circ, b = 35^\circ, c = 50^\circ, and an unknown d, which is m\angle ODA. The angle sum of the quadrilateral is 360^\circ, and it contains every base angle twice:

\begin{aligned} 2(20 + 35 + 50 + d) &= 360 \\ 105 + d &= 180 \\ d &= 75 \end{aligned}

So m\angle ODA = 75^\circ. Each angle of ABCD is the sum of the two base angles that meet there:

\begin{aligned} m\angle A &= d + a = 75^\circ + 20^\circ = 95^\circ \\ m\angle B &= a + b = 20^\circ + 35^\circ = 55^\circ \\ m\angle C &= b + c = 35^\circ + 50^\circ = 85^\circ \\ m\angle D &= c + d = 50^\circ + 75^\circ = 125^\circ \end{aligned}

Check. 95 + 85 = 180 and 55 + 125 = 180, as the theorem says. The angles at O check too: each is 180^\circ minus twice a base angle, giving 140^\circ, 110^\circ, 80^\circ and 30^\circ, which go once round O with 140 + 110 + 80 + 30 = 360.

Example 6

Testing four points in coordinates

The points are A(0, 0), B(6, 0), C(6, 4) and D(1, 5). Is ABCD cyclic? What about ABCE with E(2, 5) in place of D?

Solution

Three of the points fix the circle, so find the circle through A, B and C and ask whether the fourth point is on it. \overline{AB} is horizontal and \overline{BC} is vertical, so \angle ABC is a right angle, and the circumcenter of a right triangle is the midpoint of its hypotenuse (Circumcenter). The midpoint of \overline{AC} is (3, 2), and with the The distance formula the squared radius is

(0 - 3)^2 + (0 - 2)^2 = 9 + 4 = 13

Now measure the fourth point the same way. For D(1, 5):

(1 - 3)^2 + (5 - 2)^2 = 4 + 9 = 13

For E(2, 5):

(2 - 3)^2 + (5 - 2)^2 = 1 + 9 = 10

D is exactly \displaystyle \sqrt{13} from (3, 2), so it is on the circle and ABCD is cyclic. E is only \displaystyle \sqrt{10} away, inside the circle, so ABCE is not cyclic.

Check. If ABCD is cyclic, \angle D must be 180^\circ - 90^\circ = 90^\circ. From D the sides go along (-1, -5) to A and (5, -1) to C, and (-1)(5) + (-5)(-1) = 0, so the two sides are perpendicular and \angle D is a right angle, as predicted.

Example 7

A regular hexagon in a circle

A regular hexagon ABCDEF is inscribed in a circle of radius 7, as in Figure 5. Find its perimeter.

Figure 5Six equal central angles of 60^\circ cut a regular hexagon into six triangles.

Solution

A regular polygon is cyclic, and its center O is equally far from all six vertices. The six radii cut the full turn at O into six equal angles (the six triangles are congruent by SSS), so each is

\frac{360^\circ}{6} = 60^\circ

which is the central angle of the hexagon. Triangle OAB is isosceles with OA = OB = 7, so its two base angles are equal, and they share 180^\circ - 60^\circ = 120^\circ, which makes each 60^\circ. All three angles are 60^\circ, so the triangle is equilateral and AB = 7.

Every side is 7, so the perimeter is 6 \cdot 7 = 42.

Check. Each angle of the hexagon is two base angles, 60^\circ + 60^\circ = 120^\circ, which is the interior angle of a regular hexagon, \tfrac{(6 - 2) \cdot 180^\circ}{6}.

Example 8

Four angles of a cyclic hexagon

Hexagon ABCDEF is inscribed in a circle, as in Figure 6, with m\angle A = 110^\circ, m\angle B = 120^\circ, m\angle C = 130^\circ and m\angle D = 125^\circ. Find m\angle E and m\angle F.

Figure 6Four angles of a cyclic hexagon. The other two are forced.

Solution

The angle sum of a hexagon, 720^\circ, gives only m\angle E + m\angle F. The alternate-angle result gives each one: in a cyclic hexagon (k = 3) every other angle adds to (3 - 1) \cdot 180^\circ = 360^\circ.

\angle A, \angle C and \angle E are alternate angles, so

\begin{aligned} 110 + 130 + m\angle E &= 360 \\ m\angle E &= 120^\circ \end{aligned}

\angle B, \angle D and \angle F are the other three, so

\begin{aligned} 120 + 125 + m\angle F &= 360 \\ m\angle F &= 115^\circ \end{aligned}

Check. All six add to 110 + 120 + 130 + 125 + 120 + 115 = 720, the angle sum of any hexagon. That check would pass for any split of 235^\circ between E and F; it is the circle that decides the split.

Boundaries

Four points need not lie on a circle. Three points not on one line always do, and they fix the circle; a fourth point is on it only by the condition in this lesson. A parallelogram that is not a rectangle, a rhombus that is not a square, and a trapezoid with unequal legs never fit on a circle, whatever their size.

Three points on one line lie on no circle. Their two perpendicular bisectors are parallel and never meet, so no point is equally far from all three. A circle and a line meet in at most two points.

The vertices must be taken in order round the circle. The theorem is about the quadrilateral ABCD whose sides are \overline{AB}, \overline{BC}, \overline{CD} and \overline{DA}, with A, B, C, D in that order round the circle. Four points on a circle joined in a different order make a figure whose sides cross, and its “opposite angles” are not supplementary in general. A quadrilateral inscribed in a circle with its vertices in order is always convex.

The converse needs sides that do not cross. For a quadrilateral whose sides do not cross, one supplementary pair of opposite angles is enough, and then none of its angles can be more than 180^\circ, so it is automatically convex.

The angle rule is for quadrilaterals and for even numbers of sides. A cyclic pentagon has no rule beyond the one every pentagon has, that all five angles add to 540^\circ: no smaller set of its angles adds to a fixed total. For a cyclic polygon with 2k sides, it is the alternate angles that add to (k - 1) \cdot 180^\circ, not just any k of them.

Where the center sits changes the bookkeeping, not the theorem. With the center inside, each angle is the sum of two base angles; with it outside, one triangle’s base angles are subtracted at its two ends. The result is the same in both cases, but the angle sums d + a and so on only hold as written when the center is inside.

Equal sides alone do not make a polygon cyclic. A rhombus has four equal sides and is cyclic only when it is a square. A regular polygon needs equal sides and equal angles both.

Where this goes next

The proof here measured every angle against the center. Measuring angles on a circle against the arcs they cut off gives a second, shorter road: an angle with its vertex on the circle is half the arc it opens onto. That is the inscribed angle theorem, and its proof gives this lesson’s theorem in one line, two opposite angles opening onto arcs that together make the whole circle, and also proves the converse.

Regular polygons are cyclic, so each one has a center, a circumscribed circle and a central angle of \tfrac{360^\circ}{n}. The hexagon in Example 7 is the start of finding the area of a regular polygon, by cutting it into those isosceles triangles.

Constructing the circumscribed circle draws the circle through three given points with compass and straightedge, using the two perpendicular bisectors from this lesson. And the converse becomes a tool for proving that four points lie on one circle: a quadrilateral with two opposite right angles, such as the one formed by a point outside a circle, the two points where its tangents touch the circle, and the center, is always cyclic.

Exercises

Which shapes are cyclic

For each shape, say whether it is always cyclic, sometimes cyclic, or never cyclic, and give the reason: (a) a rhombus; (b) an isosceles trapezoid; (c) a parallelogram; (d) a trapezoid whose legs have different lengths.

Circles around rectangles and regular polygons

A rectangle measures 5 by 12. Find the radius of its circumscribed circle.
A square has side 4. Find the radius of its circumscribed circle.
A rectangle is inscribed in a circle of radius 10, and one of its sides is 12. Find the other side.
A rectangle is inscribed in a circle of radius 13, and one of its sides is 10. Find its area.
A square is inscribed in a circle of radius 6. Find its side and its area.
A regular hexagon is inscribed in a circle of radius 9. Find its perimeter.
A regular octagon is inscribed in a circle with center O. Find the angle at O between the radii to two neighboring vertices, the base angles of the isosceles triangle those radii make with a side, and each angle of the octagon.

Angles of cyclic quadrilaterals

An isosceles trapezoid has angles of 75^\circ at each end of its longer base. Find its other two angles, and show from the angles that it is cyclic.
Kite ABCD, with AB = AD and CB = CD, is inscribed in a circle, and m\angle C = 48^\circ. Find m\angle A, m\angle B and m\angle D.
In cyclic quadrilateral ABCD, \angle A is three times as large as \angle C. Find both.
In cyclic quadrilateral ABCD, \angle B is 40^\circ larger than \angle D. Find both.
Trapezoid ABCD, with \overline{AB} \parallel \overline{DC}, is inscribed in a circle, and m\angle A = 63^\circ. Find the other three angles.

Working from the center

Quadrilateral ABCD is inscribed in \odot O with O inside it. m\angle OAB = 32^\circ, m\angle OBC = 41^\circ and m\angle OCD = 28^\circ. Find m\angle ODA and the four angles of ABCD.
Quadrilateral ABCD is inscribed in \odot O with O inside it. m\angle AOB = 100^\circ, m\angle BOC = 80^\circ and m\angle COD = 110^\circ. Find m\angle DOA, and then the four angles of ABCD.

Coordinates

Show that the quadrilateral with vertices P(-5, 0), Q(3, -4), R(4, 3) and S(0, 5) is cyclic, by finding the center and radius of its circle.
The points are A(0, 0), B(4, 0) and C(4, 2). Which of D(1, 3) and E(2, 3) makes a cyclic quadrilateral with them, as ABCD or ABCE?

Polygons with more sides

Hexagon ABCDEF is inscribed in a circle, with m\angle A = 100^\circ and m\angle C = 140^\circ. Find m\angle E.
Octagon A_1 A_2 \ldots A_8 is inscribed in a circle. What do m\angle A_1 + m\angle A_3 + m\angle A_5 + m\angle A_7 add to?

Reasoning

Explain why every triangle has exactly one circumscribed circle, and why three points on one straight line have none.
Prove that a rhombus inscribed in a circle is a square.
Trapezoid ABCD, with \overline{AB} \parallel \overline{DC}, is inscribed in a circle. Prove that m\angle A = m\angle B, and that the legs \overline{AD} and \overline{BC} are equal.
Kite ABCD has AB = AD and CB = CD. Show that it is cyclic exactly when \angle B and \angle D are right angles.
In the proof with the center inside, m\angle A + m\angle C uses each of a, b, c, d exactly once. What happens if you add two neighboring angles, m\angle A + m\angle B, instead, and why does that sum not come out to a fixed number?
Explain why a polygon is cyclic exactly when the perpendicular bisectors of all its sides pass through one point.

Answers

(a) Sometimes: only when it is a square, since a rhombus is a parallelogram and a parallelogram on a circle is a rectangle (Example 4). (b) Always: its base angles are equal, so opposite angles are supplementary (Example 2). (c) Sometimes: only when it is a rectangle (Example 4). (d) Never: a trapezoid on a circle has equal legs (Exercise 22).
\tfrac{13}{2} = 6.5. The diagonal is \displaystyle \sqrt{5^2 + 12^2} = 13, and it is a diameter.
\displaystyle 2\sqrt{2}. The diagonal is \displaystyle \sqrt{4^2 + 4^2} = 4\sqrt{2}, and the radius is half of it.
16. The diagonal is a diameter, 20, so the other side is \displaystyle \sqrt{20^2 - 12^2} = \sqrt{256} = 16.
240. The diagonal is 26, so the other side is \displaystyle \sqrt{26^2 - 10^2} = \sqrt{576} = 24, and the area is 10 \cdot 24 = 240.
Side \displaystyle 6\sqrt{2}, area 72. The diagonal is the diameter 12, and a square’s diagonal is \displaystyle \sqrt{2} times its side, so the side is \displaystyle \tfrac{12}{\sqrt{2}} = 6\sqrt{2} and the area is \displaystyle (6\sqrt{2})^2 = 72.
54. The six central angles are 60^\circ, so each triangle from the center is equilateral and each side equals the radius, 9; then 6 \cdot 9 = 54.
45^\circ, 67.5^\circ and 135^\circ. The central angle is \tfrac{360^\circ}{8} = 45^\circ; the base angles share 180^\circ - 45^\circ = 135^\circ, so each is 67.5^\circ; each angle of the octagon is two base angles, 135^\circ, which agrees with \tfrac{(8 - 2) \cdot 180^\circ}{8} = 135^\circ.
105^\circ and 105^\circ. Each leg crosses the parallel bases, so the angles along it add to 180^\circ: 180^\circ - 75^\circ = 105^\circ. Opposite angles are 75^\circ + 105^\circ = 180^\circ, so by the converse it is cyclic.
m\angle A = 132^\circ, m\angle B = m\angle D = 90^\circ. \angle B \cong \angle D in a kite and they are supplementary, so each is 90^\circ; \angle A is opposite \angle C, so it is 180^\circ - 48^\circ = 132^\circ.
m\angle A = 135^\circ, m\angle C = 45^\circ. They are opposite, so 3x + x = 180 and x = 45.
m\angle B = 110^\circ, m\angle D = 70^\circ. They are opposite, so (x + 40) + x = 180 and x = 70.
m\angle B = 63^\circ, m\angle C = 117^\circ, m\angle D = 117^\circ. Along leg \overline{AD}, m\angle D = 180^\circ - 63^\circ; \angle C is opposite \angle A, so it is also 117^\circ; \angle B is opposite \angle D, so it is 180^\circ - 117^\circ = 63^\circ.
m\angle ODA = 79^\circ; the angles are m\angle A = 111^\circ, m\angle B = 73^\circ, m\angle C = 69^\circ, m\angle D = 107^\circ. The base angles add to 180^\circ, so d = 180 - 32 - 41 - 28 = 79. Then A = 32 + 79, B = 32 + 41, C = 41 + 28, D = 28 + 79, and 111 + 69 = 73 + 107 = 180.
m\angle DOA = 70^\circ; the angles are m\angle A = 95^\circ, m\angle B = 90^\circ, m\angle C = 85^\circ, m\angle D = 90^\circ. The angles at O make 360^\circ, so m\angle DOA = 360 - 100 - 80 - 110 = 70. Each base angle is half of 180^\circ minus the angle at O: a = 40, b = 50, c = 35, d = 55. Then A = d + a, B = a + b, C = b + c, D = c + d.
Center (0, 0), radius 5. The perpendicular bisector of \overline{PS} is y = -x and that of \overline{SR} is y = 2x; they meet at the origin. Each vertex satisfies x^2 + y^2 = 25: 25 + 0, 9 + 16, 16 + 9 and 0 + 25.
ABCD is cyclic and ABCE is not. \angle B is a right angle, so the circle through A, B, C has center (2, 1), the midpoint of \overline{AC}, and squared radius 4 + 1 = 5. For D: (1 - 2)^2 + (3 - 1)^2 = 5, on the circle. For E: (2 - 2)^2 + (3 - 1)^2 = 4, inside it.
120^\circ. Alternate angles of a cyclic hexagon add to 360^\circ, so m\angle E = 360^\circ - 100^\circ - 140^\circ.
540^\circ. With 2k = 8 sides, k = 4, and alternate angles add to (4 - 1) \cdot 180^\circ = 540^\circ, half the octagon’s 1080^\circ.
The perpendicular bisectors of two sides of a triangle meet in exactly one point, the circumcenter, which is equally far from all three vertices; a circle through the vertices must have its center on both bisectors, so it is that circle and no other. For three points on one line the two bisectors are both perpendicular to that line, so they are parallel and never meet; no point is equally far from all three, so no circle passes through them.
Opposite angles of a rhombus are equal, since it is a parallelogram; on a circle they are also supplementary, so each is 90^\circ. A rhombus with right angles has four equal sides and four right angles, so it is a square.
\angle A and \angle D are same-side interior angles along \overline{AD}, so they add to 180^\circ. \angle B and \angle D are opposite angles of a cyclic quadrilateral, so they add to 180^\circ as well. Both are 180^\circ - m\angle D, so m\angle A = m\angle B. For the legs, drop perpendiculars from D and C to line AB; they are equal, because the distance between two parallel lines is the same everywhere. If \angle A is a right angle, both legs are those perpendiculars and are equal. Otherwise the two right triangles have equal legs (the heights) and equal angles at A and B (each is m\angle A = m\angle B, or 180^\circ minus it when that angle is obtuse), so they are congruent by AAS and the hypotenuses \overline{AD} and \overline{BC} are equal.
In the kite, triangles ABC and ADC are congruent by SSS, so \angle B \cong \angle D. If the kite is cyclic, these equal opposite angles are supplementary, so each is 90^\circ. If \angle B and \angle D are right angles, they add to 180^\circ, and by the converse the kite is cyclic.
m\angle A + m\angle B = (d + a) + (a + b) = 2a + b + d, which counts a twice and leaves out c. The fixed fact is only a + b + c + d = 180^\circ; changing how the four base angles are shared out (moving the vertices round the circle) changes 2a + b + d, so neighboring angles have no fixed sum. In Example 5, m\angle A + m\angle B = 95^\circ + 55^\circ = 150^\circ, not 180^\circ.
If the polygon is cyclic, its center is equally far from the two ends of every side, so by the perpendicular bisector theorem it lies on every side’s perpendicular bisector. Conversely, if all the bisectors pass through one point, that point is equally far from the two ends of each side; going round the polygon side by side, it is the same distance from every vertex, so the circle with that center through one vertex passes through all of them.