The inscribed angle theorem
Stand anywhere on a circle and look at two fixed points on it. Why is the angle you see the same wherever you stand?
Where you are 15 in Circles
Why this exists
A central angle has its vertex at the center of the circle, and its measure is the measure of its arc by definition. But most angles that turn up in a circle do not sit at the center. They sit on the circle itself: the corners of a triangle whose three vertices all lie on a circle, or the angle where two chords meet at the rim. Such an angle cuts off an arc just as a central angle does, and until now nothing has said how big it is.
The answer is as simple as it could be: the angle is exactly half its arc. Two facts fall straight out of that and are used constantly. Every angle whose vertex is on one arc and whose sides pass through the same two points has the same size, wherever on the arc the vertex sits. And every angle with its vertex on the circle and its sides through the two ends of a diameter is a right angle.
This lesson takes two days. Part 1 states the theorem, proves it, and uses it to find angles and arcs. Part 2 is the special case of a diameter, the angle in a semicircle, together with its converse, which finds circles from right angles.
Definition
An inscribed angle is an angle whose vertex lies on a circle and whose two sides each cross the circle again, so that each side contains a chord from the vertex. If A, B and C are three points on a circle, then \angle ABC is an inscribed angle with vertex B, and its sides contain the chords \overline{BA} and \overline{BC}.
The arc the angle opens onto is its intercepted arc: the arc from A to C that does not contain the vertex B. Apart from its two endpoints, that arc lies inside the angle. It can be a minor arc, a semicircle or a major arc, depending on where B sits.
In Figure 1, \angle ABC is inscribed and its intercepted arc is the heavy minor arc \overset{\frown}{AC}. The central angle \angle AOC opens onto the same arc. If B were on the short arc instead, the intercepted arc would be the long way round, a major arc, named with three letters through a point on it, such as \overgroup{ADC}.
Two inscribed angles intercept the same arc when their sides pass through the same two points A and C and their vertices lie on the same one of the two arcs from A to C, that is, on the same side of the chord \overline{AC}.
An angle in a semicircle is an inscribed angle whose sides pass through the two ends of a diameter. Its vertex lies on one semicircle, and its intercepted arc is the other.
Intuition
In Figure 2 the points A and C stay fixed, and the vertex of the angle moves along the upper arc from P to Q to R. The triangles look nothing alike: the one with its top at P leans one way, the one with its top at R leans the other, and their sides have different lengths. Yet the angle at the top is the same size every time.
As the vertex slides along the arc, both of its sides turn. They turn in the same direction and by the same amount, so the opening between them never changes. That is why the size of the angle can depend only on the arc it looks at, and not on where on the other arc it is looking from.
Why half? Put the vertex where one side passes through the center. The triangle formed by the vertex, the center and the far end of the other side has two radii for sides, so it is isosceles, and the central angle is the outside angle of that triangle: two equal angles added together. The proof below shows that every other position comes back to this one.
Results
Part 1: an inscribed angle is half its arc
Inscribed Angle Theorem. The measure of an inscribed angle is half the measure of its intercepted arc. For an inscribed angle \angle ABC, with \overset{\frown}{AC} the arc that does not contain B,
When the intercepted arc is a minor arc, its measure is that of the central angle \angle AOC, so the inscribed angle is half the central angle on the same arc:
Inscribed angles on the same arc. Inscribed angles that intercept the same arc are congruent. More generally, inscribed angles whose intercepted arcs have equal measure are congruent, since each one is half of the same number.
Part 2: the angle in a semicircle
Angle in a semicircle. An angle in a semicircle is a right angle. If \overline{AB} is a diameter and C is any other point of the circle, then
This is often called Thales’s theorem.
The converse. If m\angle ACB = 90^\circ, then C lies on the circle with diameter \overline{AB}. Said another way: in every right triangle, the hypotenuse is a diameter of the circle through all three vertices, and the midpoint of the hypotenuse is the same distance from all three vertices.
Together the two say exactly where a segment \overline{AB} is seen at a right angle: from the points of the circle with diameter \overline{AB}, other than A and B themselves, and from nowhere else.
Why it is true
The proof of the theorem splits into three cases, by where the center O lies compared with the angle: on a side, inside, or outside. The first case does the real work; the other two are brought back to it by drawing one diameter.
Case 1: the center lies on a side. In panel (a), O lies on side \overline{BA}, so \overline{BA} is a diameter. Draw the radius \overline{OC}. Triangle OBC has OB = OC, because both are radii, so it is isosceles, and by the base angles theorem its angles at B and at C are equal. Call each of them x.
The angle \angle AOC lies outside triangle OBC, next to its angle at O along the straight line BA. By the exterior angle theorem it equals the sum of the two angles of the triangle that are not next to it:
Since x is a base angle of a triangle, x < 90^\circ, so 2x < 180^\circ and \overset{\frown}{AC} is a minor arc with measure m\angle AOC = 2x. The inscribed angle is x, which is half of it.
Case 2: the center lies inside the angle. In panel (b), draw the diameter from B through O, ending at D. The ray BD runs inside \angle ABC, so by angle addition
Each of the two pieces has the center on one of its sides, so Case 1 applies to each: m\angle ABD = \tfrac12\, m\overset{\frown}{AD} and m\angle DBC = \tfrac12\, m\overset{\frown}{DC}. The point D lies on the intercepted arc, so by arc addition the arcs \overset{\frown}{AD} and \overset{\frown}{DC} together make it up. Adding the two equations,
This case also covers every angle whose intercepted arc is a major arc. When B is on the minor arc, the point D opposite B is on the major arc, so the center is inside the angle.
Case 3: the center lies outside the angle. In panel (c), draw the diameter \overline{BD} again. Now both sides of the angle lie on the same side of it. Name the points so that ray BA lies inside \angle DBC, as drawn. Then the inscribed angle is a difference,
Case 1 applies to both angles on the right, since each has \overline{BD} as a side. On the circle, A lies on the arc \overset{\frown}{DC}, so by arc addition m\overset{\frown}{AC} = m\overset{\frown}{DC} - m\overset{\frown}{DA}. Therefore
That covers every position of the center, so the theorem holds for every inscribed angle.
Same arc. Two inscribed angles that intercept the same arc are each half of that arc’s measure, so they are equal. Where the vertex sits on its own arc never enters the formula, which is why it can slide freely.
Semicircle. If \overline{AB} is a diameter, the arc intercepted by \angle ACB is a semicircle, which measures 180^\circ, so m\angle ACB = \tfrac12 \cdot 180^\circ = 90^\circ.
The converse. Suppose m\angle ACB = 90^\circ, and let M be the midpoint of \overline{AB}. Extend \overline{CM} past M to a point C' with MC' = MC. The quadrilateral ACBC' has diagonals \overline{AB} and \overline{CC'} that cut each other in half, so it is a parallelogram. In a parallelogram neighboring angles add to 180^\circ, so one right angle forces all four, and ACBC' is a rectangle. The diagonals of a rectangle are equal, so CC' = AB, and halving both gives MC = MA = MB. So C is as far from M as A and B are: it lies on the circle with center M through A and B, which is the circle with diameter \overline{AB}.
Euclid proves the halving rule as Proposition 20 of Book III of the Elements, the equal angles on one arc as Proposition 21, and the right angle in a semicircle as Proposition 31. Set out as a formal proof with a reason on every line, the theorem is its own skill: proving the inscribed angle theorem.
Worked examples
Examples 1 to 4 go with Part 1, the theorem itself. Examples 5 to 8 go with Part 2: the angle in a semicircle and its converse, and then the theorem put to work on a regular polygon.
Half the arc, and back again
In \odot O, the inscribed angle \angle PQR intercepts \overset{\frown}{PR}, and m\overset{\frown}{PR} = 84^\circ.
(a) Find m\angle PQR and the central angle m\angle POR.
(b) A different inscribed angle, \angle XYZ, measures 35^\circ. Find the measure of the arc it intercepts.
Solution(a) An inscribed angle is half its intercepted arc:
The arc measures less than 180^\circ, so it is a minor arc, and a minor arc measures the same as its central angle: m\angle POR = 84^\circ. The angle at the rim is half the angle at the center.
(b) Run the theorem backwards. The angle is half the arc, so the arc is twice the angle:
Two angles on one arc
In Figure 5, B and C lie on the same arc from A to D, with m\angle ABD = (3x + 5)^\circ and m\angle ACD = (5x - 21)^\circ. Find x, the two angles, and m\overset{\frown}{AD}.
Both angles have their sides through A and D, and both vertices are on the upper arc, so both intercept the lower arc \overset{\frown}{AD}. Inscribed angles on the same arc are congruent, so the two expressions are equal:
Substitute into both expressions, since they must agree: 3(13) + 5 = 44 and 5(13) - 21 = 44. Each angle measures 44^\circ.
The arc is twice either angle: m\overset{\frown}{AD} = 2 \cdot 44^\circ = 88^\circ.
The angles of an inscribed triangle
Points A, B and C on a circle cut it into three arcs. Taking each arc to be the one that does not contain the third point, m\overset{\frown}{AB} = 110^\circ, m\overset{\frown}{BC} = 150^\circ and m\overset{\frown}{CA} = 100^\circ. Find the three angles of triangle ABC.
SolutionEach angle of the triangle has its vertex on the circle and its sides along two sides of the triangle, so each is an inscribed angle. Its intercepted arc is the one between the other two vertices, the arc opposite it:
Check. 55 + 75 + 50 = 180, as the angles of a triangle must. That is no accident: the three arcs go once round the circle, so they add to 360^\circ, and halving gives 180^\circ. It is a check rather than a new proof of the triangle angle sum, because the theorem itself was proved using the exterior angle theorem.
An angle that opens onto the major arc
In Figure 6, the minor arc \overset{\frown}{AC} measures 100^\circ. The point B lies on that minor arc, and D lies on the major arc. Find m\angle ADC and m\angle ABC.
The vertex D is on the major arc, so the arc from A to C that does not contain D is the minor arc:
The vertex B is on the minor arc, so the arc that does not contain B is the major arc \overgroup{ADC}, the long way round. A major arc and its minor arc fill the circle, so
and
The angle at B is obtuse, which matches the picture: B sits close to the chord \overline{AC} and sees it nearly flat.
Notice that 50^\circ + 130^\circ = 180^\circ. The two angles look at the two arcs that together make the whole circle, so together they make half of 360^\circ. Since A, B, C and D all lie on the circle, this is the opening fact about quadrilaterals inscribed in a circle: their opposite angles add to 180^\circ.
A triangle on a diameter
\overline{AB} is a diameter of a circle, and C is a point on the circle with AC = 6 cm and BC = 8 cm. Find m\angle ACB, the length of the diameter, and the radius.
Solution\angle ACB has its sides through the two ends of a diameter, so it is an angle in a semicircle: m\angle ACB = 90^\circ. Triangle ACB is therefore a right triangle with hypotenuse \overline{AB}, and the Pythagorean theorem gives the diameter:
The radius is half the diameter, 5 cm.
Two acute angles on a diameter
\overline{AB} is a diameter of a circle and C is another point on it. m\angle CAB = (2x + 6)^\circ and m\angle CBA = (3x - 1)^\circ. Find x, the two angles, and the arcs \overset{\frown}{BC} and \overset{\frown}{AC} that do not contain the third vertex.
SolutionThe angle at C is in a semicircle, so it is 90^\circ. The three angles of triangle ABC add to 180^\circ, so the other two add to 90^\circ:
So m\angle CAB = 2(17) + 6 = 40^\circ and m\angle CBA = 3(17) - 1 = 50^\circ.
Each of these angles is inscribed too. \angle CAB intercepts \overset{\frown}{BC} and \angle CBA intercepts \overset{\frown}{AC}, so
Check. The two arcs together run from A through C to B, which is a semicircle, and 80 + 100 = 180.
Testing a right angle with coordinates
Let A = (-5, 0) and B = (5, 0).
(a) Is \angle ACB a right angle for C = (3, 4)? Is C on the circle with diameter \overline{AB}?
(b) Answer the same questions for D = (4, 2).
SolutionThe circle with diameter \overline{AB} has its center at the midpoint (0, 0) and radius 5, so a point is on it when the sum of the squares of its coordinates is 25.
(a) Compare the slopes of the two sides:
Their product is -1, so \overline{AC} \perp \overline{BC} and m\angle ACB = 90^\circ. And 3^2 + 4^2 = 25, so C is on the circle, just as the converse says it must be.
(b) Here
and the product is -\tfrac{4}{9}, not -1, so \angle ADB is not a right angle. And 4^2 + 2^2 = 20, less than 25, so D is inside the circle, not on it. The two facts agree: only points on the circle see \overline{AB} at exactly 90^\circ. Exercise 27 shows that a point inside the circle always sees it at more than 90^\circ.
The angles of a regular pentagon
A regular pentagon ABCDE is inscribed in a circle, as in Figure 7. Find m\angle CAD and the interior angle m\angle BAE, and show that the diagonals \overline{AC} and \overline{AD} cut \angle BAE into three equal parts.
The five sides are equal chords. Equal chords of one circle cut off equal arcs, since the triangles they make with the center are congruent by side-side-side, which makes their central angles equal. Five equal arcs fill the circle, so each measures 360^\circ \div 5 = 72^\circ.
\angle CAD is inscribed and intercepts the arc from C to D, one of the five:
In the same way \angle BAC intercepts the arc from B to C and \angle DAE the arc from D to E, so each of them is 36^\circ as well. The three parts are equal.
The whole angle \angle BAE intercepts the arc from B to E through C and D, which is three of the five arcs, 3 \cdot 72^\circ = 216^\circ:
Check. 3 \cdot 36 = 108, and the angle sum of a polygon gives each angle of a regular pentagon as (5 - 2) \cdot 180^\circ \div 5 = 108^\circ.
Boundaries
The vertex must be on the circle. Take an angle with its sides through A and C and its vertex on the same side of \overline{AC} as an inscribed angle \angle ABC. If the vertex is inside the circle, the angle is larger than \angle ABC; if it is outside, the angle is smaller. Those angles have their own rules, worked out with angles made by chords, secants and tangents, and this theorem is the tool that proves them.
The intercepted arc is the one away from the vertex. The arc from A to C that counts is the one that does not contain B. When B is on the minor arc, that is the major arc, and the angle is obtuse, as in Example 4.
Same arc means same side of the chord. Two inscribed angles with sides through A and C but with vertices on opposite arcs intercept different arcs. They are not equal in general: they add to 180^\circ. The only time they are equal is when \overline{AC} is a diameter and both are right angles.
The vertex cannot be an endpoint of the arc. If B slides all the way to A, the side \overline{BA} shrinks to nothing and there is no inscribed angle left. What remains in the limit is an angle between a chord and a tangent line, which is also half its arc, but that needs a proof of its own and belongs with angles made by chords, secants and tangents.
Equal measure, not equal length. The angle depends only on the arc’s measure in degrees. A 40^\circ inscribed angle in a small circle and one in a large circle both intercept 80^\circ arcs, but those arcs have very different lengths.
The converse needs exactly 90^\circ. If m\angle ACB is anything else, C is not on the circle with diameter \overline{AB}. A point inside that circle sees \overline{AB} at more than 90^\circ (Exercise 27), and a point on line AB makes no angle at all.
Where this goes next
Almost every other angle fact about circles is proved by drawing one extra chord and using this theorem. Opposite angles of a quadrilateral inscribed in a circle add to 180^\circ, because their two arcs fill the circle, which is what Example 4 found. The angle between two chords that cross inside a circle, or between two secants that meet outside it, comes from an inscribed angle and the exterior angle theorem, and those are the angles made by chords, secants and tangents.
Equal inscribed angles also make similar triangles. When two chords cross, as in Figure 8, the angles on equal arcs pair up, the two small triangles at the crossing are similar, and that gives the segment length rules in circles.
The converse finds circles. The circle through the three vertices of a right triangle has its center at the midpoint of the hypotenuse, which is a quick case of constructing the circumscribed circle, and a carpenter’s square laid on a round board finds a diameter (Exercise 25). The arguments here, radii giving isosceles triangles and one extra line bringing a hard case back to an easy one, are the pattern for proofs about circles in general.
Exercises
Part 1: an angle and its arc
- An inscribed angle intercepts an arc of 96^\circ. Find the angle.
- An inscribed angle measures 37^\circ. Find the measure of its intercepted arc.
- In \odot O, m\angle AOC = 130^\circ, and B lies on the major arc from A to C. Find m\angle ABC.
- In \odot O, B lies on the major arc from A to C and m\angle ABC = 62^\circ. Find m\angle AOC.
- The minor arc \overset{\frown}{AC} of a circle measures 140^\circ, and B lies on that minor arc. Find m\angle ABC.
- An inscribed angle measures 115^\circ. Find the measure of its intercepted arc, and say whether that arc is minor or major.
Part 1: angles on the same arc
Use Figure 8 for Exercises 7 to 9. The points A, B, C and D lie on the circle in that order, and the chords \overline{AC} and \overline{BD} cross at E.
- m\angle CAD = 28^\circ. Find m\angle CBD and m\overset{\frown}{CD}.
- The arc \overset{\frown}{AB} that contains neither C nor D measures 100^\circ. Find m\angle ACB and m\angle ADB.
- m\angle ACB = (4x - 7)^\circ and m\angle ADB = (2x + 19)^\circ. Find x, the two angles, and the arc \overset{\frown}{AB} that contains neither C nor D.
- Three points on a circle cut it into three arcs whose measures are in the ratio 3 : 4 : 5. Find the angles of the triangle the three points make.
- Triangle ABC is inscribed in a circle, with m\angle A = 70^\circ and m\angle B = 45^\circ. Find the measures of the three arcs \overset{\frown}{BC}, \overset{\frown}{CA} and \overset{\frown}{AB}, each taken as the arc that does not contain the third vertex.
Part 1: reasoning
- Explain why every inscribed angle that intercepts a minor arc is acute, and every inscribed angle that intercepts a major arc is obtuse.
- Two inscribed angles in one circle intercept two different arcs that have the same measure. Explain why the two angles are congruent, even though they do not open onto the same arc.
- Chords \overline{AB} and \overline{CD} of a circle are parallel, and going round the circle the four points come in the order A, B, D, C. Draw \overline{AD} and show that the arcs \overset{\frown}{AC} and \overset{\frown}{BD} between the chords have equal measure. Then explain why AC = BD.
Part 2: angles in a semicircle
- \overline{AB} is a diameter of a circle, C is another point on the circle, and m\angle CAB = 27^\circ. Find m\angle ACB and m\angle CBA.
- \overline{AB} is a diameter, C is on the circle, AC = 5 and BC = 12. Find the diameter and the radius.
- A circle has a diameter \overline{AB} of length 10. A point C on the circle has AC = 6. Find BC.
- \overline{AB} is a diameter and C is on the circle. m\angle CAB = (x + 14)^\circ and m\angle CBA = (2x + 7)^\circ. Find x and both angles.
- Let A = (-13, 0), B = (13, 0) and C = (5, 12). Use slopes to show that \angle ACB is a right angle. Then show that C is on the circle with diameter \overline{AB}, and explain why the theorem said the angle had to be right.
- Let A = (1, 2) and B = (7, 10). Every point C with m\angle ACB = 90^\circ lies on one circle. Find its center and radius. Then check that C = (1, 10) is on that circle and that \angle ACB is a right angle there.
Part 2: polygons and crossing chords
- A regular hexagon ABCDEF is inscribed in a circle. Find m\angle ADB, m\angle ACE and m\angle ABD.
- A regular polygon with nine sides is inscribed in a circle. From one vertex, the two sides and the diagonals to every other vertex are drawn. Find the angle between any two neighboring lines of these, and the interior angle of the polygon.
- In Figure 8, m\angle BAC = 40^\circ and m\angle ABD = 65^\circ. Find m\angle BDC, m\angle ACD, m\angle AEB and m\angle CED.
Part 2: reasoning
- A right triangle is inscribed in a circle. Explain why its hypotenuse must be a diameter, and say where the center of the circle is.
- A carpenter’s square has two straight edges that meet at a right angle. Explain how to find the center of a round board using only the square and a pencil.
- Triangle ABC has three acute angles. A' is the foot of the altitude from A, the point where the perpendicular from A meets \overline{BC}, and B' is the foot of the altitude from B on \overline{AC}. Show that A, B, A' and B' lie on one circle, and find its center.
- \overline{AB} is a diameter of a circle, and D is a point inside the circle that is not on line AB. Show that m\angle ADB > 90^\circ. (Extend \overline{AD} beyond D until it meets the circle at C.)
Answers
- 48^\circ. An inscribed angle is half its arc: \tfrac12 \cdot 96^\circ = 48^\circ.
- 74^\circ. The arc is twice the angle: 2 \cdot 37^\circ = 74^\circ.
- 65^\circ. B is on the major arc, so \angle ABC intercepts the minor arc \overset{\frown}{AC}, which measures the same as its central angle, 130^\circ. Half of that is 65^\circ.
- 124^\circ. The intercepted arc measures 2 \cdot 62^\circ = 124^\circ. It is the minor arc, because B is on the major one, and a minor arc measures the same as its central angle.
- 110^\circ. With B on the minor arc, the intercepted arc is the major arc, which measures 360^\circ - 140^\circ = 220^\circ, and half of that is 110^\circ.
- 230^\circ, a major arc. The arc is 2 \cdot 115^\circ = 230^\circ, which is more than 180^\circ.
- m\angle CBD = 28^\circ and m\overset{\frown}{CD} = 56^\circ. \angle CAD and \angle CBD both intercept \overset{\frown}{CD}, so they are equal, and the arc is twice either one.
- m\angle ACB = m\angle ADB = 50^\circ. Both intercept \overset{\frown}{AB}, and each is half of 100^\circ.
- x = 13, both angles measure 45^\circ, and m\overset{\frown}{AB} = 90^\circ. The two angles intercept the same arc, so 4x - 7 = 2x + 19, giving 2x = 26. Check: 4(13) - 7 = 45 and 2(13) + 19 = 45.
- 45^\circ, 60^\circ and 75^\circ. Write the arcs as 3k, 4k and 5k. They fill the circle, so 12k = 360 and k = 30, giving arcs of 90^\circ, 120^\circ and 150^\circ. Each angle of the triangle is half the arc opposite it.
- m\overset{\frown}{BC} = 140^\circ, m\overset{\frown}{CA} = 90^\circ and m\overset{\frown}{AB} = 130^\circ. The third angle is m\angle C = 180^\circ - 70^\circ - 45^\circ = 65^\circ, and each arc is twice the angle opposite it. Check: 140 + 90 + 130 = 360.
- A minor arc measures less than 180^\circ, so an inscribed angle on it, being half, measures less than 90^\circ and is acute. A major arc measures more than 180^\circ and less than 360^\circ, so half of it is more than 90^\circ and less than 180^\circ, which is obtuse.
- Each angle measures half of its own intercepted arc. The two arcs have the same measure, so the two halves are the same number and the angles are congruent. The theorem uses only the arc’s measure, never which arc it is.
- \overline{AD} is a transversal of the parallel chords, with B and C on opposite sides of it, so the alternate interior angles \angle BAD and \angle ADC are equal. \angle BAD is inscribed and intercepts \overset{\frown}{BD}; \angle ADC is inscribed and intercepts \overset{\frown}{AC}. Equal inscribed angles have intercepted arcs of equal measure (each is twice the angle), so m\overset{\frown}{AC} = m\overset{\frown}{BD}. These two equal arcs together with \overset{\frown}{AB} and \overset{\frown}{CD} fill the circle, so each is less than 180^\circ, a minor arc, and their central angles \angle AOC and \angle BOD are equal. Triangles AOC and BOD then have two sides that are radii and an equal angle between them, so they are congruent by side-angle-side, and AC = BD.
- m\angle ACB = 90^\circ and m\angle CBA = 63^\circ. The angle at C is in a semicircle, and the triangle’s angles add to 180^\circ: 180^\circ - 90^\circ - 27^\circ = 63^\circ.
- The diameter is 13 and the radius is \tfrac{13}{2} = 6.5. The angle at C is right, so \displaystyle AB = \sqrt{5^2 + 12^2} = \sqrt{169} = 13.
- BC = 8. The angle at C is right and \overline{AB} is the hypotenuse, so BC^2 = 10^2 - 6^2 = 64.
- x = 23, m\angle CAB = 37^\circ and m\angle CBA = 53^\circ. The right angle at C leaves 90^\circ for the other two: (x + 14) + (2x + 7) = 90, so 3x = 69.
- The slope of AC is \tfrac{12}{18} = \tfrac{2}{3} and the slope of BC is \tfrac{12}{-8} = -\tfrac{3}{2}. Their product is -1, so the sides are perpendicular. The circle with diameter \overline{AB} has center (0, 0) and radius 13, and 5^2 + 12^2 = 169 = 13^2, so C is on it. Then \angle ACB has its sides through the ends of a diameter, an angle in a semicircle, which the theorem says is 90^\circ.
- Center (4, 6), radius 5. By the converse the circle has diameter \overline{AB}, so its center is the midpoint of \overline{AB}, which is (4, 6), and its radius is half of \displaystyle AB = \sqrt{6^2 + 8^2} = 10. The point (1, 10) is \displaystyle \sqrt{3^2 + 4^2} = 5 from the center, so it is on the circle. \overline{CA} is vertical (both points have x = 1) and \overline{CB} is horizontal (both have y = 10), so the angle at C is right.
- m\angle ADB = 30^\circ, m\angle ACE = 60^\circ and m\angle ABD = 90^\circ. The six sides cut six equal arcs of 60^\circ. \angle ADB intercepts one of them. \angle ACE intercepts the arc from E to A through F, two arcs or 120^\circ. \angle ABD intercepts the arc from A to D through F and E, three arcs or 180^\circ, so \overline{AD} is a diameter and the angle is right.
- 20^\circ and 140^\circ. The nine equal sides cut nine equal arcs of 360^\circ \div 9 = 40^\circ. Two neighboring lines from the vertex intercept exactly one of those arcs, so the angle between them is 20^\circ. The interior angle intercepts the seven arcs that do not touch the vertex, 280^\circ, so it is 140^\circ. Check: (9 - 2) \cdot 180^\circ \div 9 = 140^\circ.
- m\angle BDC = 40^\circ, m\angle ACD = 65^\circ, and m\angle AEB = m\angle CED = 75^\circ. \angle BAC and \angle BDC both intercept \overset{\frown}{BC}; \angle ABD and \angle ACD both intercept \overset{\frown}{AD}. In triangle ABE the third angle is 180^\circ - 40^\circ - 65^\circ = 75^\circ, and \angle CED is vertical to \angle AEB.
- The right angle is an inscribed angle, so its intercepted arc measures 2 \cdot 90^\circ = 180^\circ, a semicircle. The two ends of a semicircle are the ends of a diameter, and those are the ends of the hypotenuse. The center is the midpoint of the hypotenuse.
- Lay the square on the board with its corner touching the edge, and mark the two points where its edges cross the rim. The angle at the corner is an inscribed right angle, so it intercepts a semicircle and the two marks are the ends of a diameter; draw it. Move the square to another place on the rim and draw a second diameter the same way. Every diameter passes through the center, so the center is where the two lines cross.
- The circle with diameter \overline{AB}, centered at the midpoint of \overline{AB}. Since \overline{AA'} \perp \overline{BC}, m\angle AA'B = 90^\circ, so by the converse A' lies on the circle with diameter \overline{AB}. In the same way m\angle AB'B = 90^\circ, so B' lies on that circle too. A and B are on it as the ends of the diameter. Because the triangle is acute, the feet land inside the sides and differ from A and B, so each right angle is a real angle.
- C is on the circle and differs from A and B, since D is not on line AB, so \angle ACB is an angle in a semicircle and measures 90^\circ. The point D lies between A and C, so \angle ADB is the exterior angle at D of triangle DCB, and it equals the sum of the two angles of that triangle not next to it, so m\angle ADB = m\angle DCB + m\angle DBC. Here m\angle DCB = 90^\circ, so m\angle ADB = 90^\circ + m\angle DBC, which is more than 90^\circ.