# The inscribed angle theorem and Thales' semicircle

> From every point on one arc of a circle, a chord is seen under the same angle, half the angle at the center. Two isosceles triangles explain why.

Series: Geometry (https://www.monocurl.com/monocurl-essays/geometry/)
Canonical: https://www.monocurl.com/monocurl-essays/geometry/inscribed-angle/
Keywords: inscribed angle theorem, Thales theorem, angle in a semicircle, central angle, angles in the same segment, cyclic quadrilateral, inscribed angle proof

[Figure 1, animation: A point P travels around a circle while two points A and B on the circle stay fixed. The lengths PA and PB, measured in radii, change all the time. The angle at P stays at 50°.]

Two posts, $A$ and $B$, stand on the rim of a circular pond. Walk along the rim and keep looking at them. As you go, one post comes nearer and the other recedes, and the triangle you form with them changes from tall to squat to lopsided. In the animation the two distances swing between about one radius and nearly the full width of the pond. Through all of it the angle between your two lines of sight, the angle at $P$, does not move from 50°.

Nothing in the triangle $APB$ looks constant except that one angle. Why should it be? And why 50°, when the posts could have been placed anywhere?

## Is it the circle that matters?

One reaction is to doubt that the circle has much to do with it. Perhaps the angle under which you see two posts simply does not depend much on where you stand. That is easy to test: step off the rim.

[Figure 2, animation: The point P moves along a straight path from inside the circle to outside it. The angle at P is larger than 55° inside the circle, smaller outside, and equal to 55° only at the moment P crosses the circle.]

The angle is far from constant. Close to the posts it is wide, far away it is narrow, and it passes through every value in between. The circle through $A$ and $B$ is the exact boundary on which it takes one particular value, in this case 55°. So the circle is doing something precise, and we need to find out what.

## The smallest case: a diameter

The plan is to start with the most symmetric arrangement and understand it completely. Put the two posts at opposite ends of a diameter.

To have numbers to work with, use coordinates. Take the circle of radius 5 centered at the origin, with $A = (-5, 0)$ and $B = (5, 0)$. The point $(3, 4)$ is on this circle, because $3^2 + 4^2 = 25$.

**Pause and try.** Let $P = (3, 4)$. Compute $PA^2$, $PB^2$ and $AB^2$ with the distance formula. What does the converse of the [Pythagorean theorem](@geometry/pythagorean-theorem) (if $a^2 + b^2 = c^2$ in a triangle, the angle opposite $c$ is a right angle) say about triangle $APB$? Then repeat with $P = (-4, 3)$.

**Answer.** For $P = (3, 4)$: $PA^2 = (3 + 5)^2 + 4^2 = 80$ and $PB^2 = (3 - 5)^2 + 4^2 = 20$, while $AB^2 = 10^2 = 100$. Since $80 + 20 = 100$, the converse of the Pythagorean theorem says the angle opposite $AB$, which is the angle at $P$, is a right angle. For $P = (-4, 3)$: $PA^2 = 1 + 9 = 10$ and $PB^2 = 81 + 9 = 90$, and again $10 + 90 = 100$.

[Figure 3, image: Three points on a circle of radius 5 whose diameter AB lies along the x-axis. For each one, PA^2 + PB^2 comes to exactly 100 = AB^2, so each triangle has a right angle at the point on the circle.]

All three points give right angles, and the algebra shows this is no accident of the chosen points. If $P = (x, y)$ is any point of the circle, so that $x^2 + y^2 = 25$, then

$$
PA^2 + PB^2 = (x + 5)^2 + y^2 + (x - 5)^2 + y^2 = 2(x^2 + y^2) + 50 = 100.
$$

So the angle at $P$ is always 90°. That settles the fact, but a calculation in which the terms $10x$ and $-10x$ cancel does not show why. The reason is in the one thing a circle gives us for free: every point on it is the same distance from the center.

Draw the segment from the center $O$ to $P$. It has the same length as $OA$ and $OB$, since all three are radii, and it cuts the triangle $APB$ into two triangles that each have two equal sides.

[Figure 4, steps: The proof that an angle standing on a diameter is a right angle. The radius OP cuts triangle APB into two isosceles triangles, and the angle sum of triangle APB then forces the angle at P to be 90°.]

  - step: A triangle APB whose side AB is a diameter of the circle. We want the angle at P.
  - step: Draw the radius OP. Now OA, OP and OB are three radii, all of the same length r.
  - step: Triangle AOP has two equal sides, so its angles at A and at P are equal. Call each of them α.
  - step: Triangle BOP also has two equal sides. Call its equal angles β.
  - step: The angles of triangle APB are α at A, β at B and α + β at P. They add to 180°, so α + β is 90°.
The figure uses one fact about triangles with two equal sides. A triangle with two equal sides is called **isosceles**, and its angles opposite the equal sides are equal. (This is Euclid's Proposition 5 of Book I. One way to see it: flip the triangle over so that the two equal sides trade places. It lands exactly on itself, so the two angles at the base trade places as well, and must be equal.)

**Theorem.** **(Thales' theorem.)** Let $AB$ be a diameter of a circle and let $P$ be a point of the circle other than $A$ and $B$. Then $\angle APB = 90°$.

**Proof.** Let $O$ be the center, which is the midpoint of $AB$, and let $r$ be the radius. Then $OA = OP = OB = r$.

In triangle $AOP$ the sides $OA$ and $OP$ are equal, so the angles opposite them are equal: $\angle OAP = \angle OPA$. Call this angle $\alpha$. In triangle $BOP$ the sides $OB$ and $OP$ are equal, so $\angle OBP = \angle OPB$. Call this angle $\beta$.

The point $O$ lies on the side $AB$ between $A$ and $B$, so the segment $PO$ lies inside the angle $\angle APB$ and splits it: $\angle APB = \alpha + \beta$. The angles of triangle $APB$ at $A$ and $B$ are $\angle OAP = \alpha$ and $\angle OBP = \beta$. Since [the angles of a triangle add to 180°](@geometry/triangle-angle-sum),

$$
\alpha + (\alpha + \beta) + \beta = 180°,
$$

so $2(\alpha + \beta) = 180°$ and $\angle APB = \alpha + \beta = 90°$.

Tradition credits the theorem to Thales of Miletus, who lived around 600 BC.

### The converse, and a rectangle

The theorem has a converse, and the neatest way to see both at once is to complete the picture. Continue the radius $PO$ through the center to the opposite point $P'$ of the circle.

[Figure 5, image: The diameter AB and a second diameter PP'. The four-sided figure APBP' has diagonals of equal length that cut each other in half at O, and it is a rectangle.]

Thales' theorem applies at each of the four corners, since each corner is a point of the circle looking at a diameter. So $APBP'$ has four right angles and is a rectangle. Read backward, the picture says that every right triangle is half of a rectangle, and a rectangle's diagonals are equal and cut each other in half. That gives the converse.

**Theorem.** **(Converse of Thales' theorem.)** If $\angle APB = 90°$, then $P$ lies on the circle that has $AB$ as a diameter.

**Proof.** Draw through $A$ the line parallel to $PB$ and through $B$ the line parallel to $PA$, and let them meet at $P'$. Then $APBP'$ is a parallelogram, so its opposite sides are equal, $AP' = PB$, and its diagonals $AB$ and $PP'$ cut each other in half at a point $O$ (two standard facts about parallelograms, each proved with congruent triangles).

Because $AP'$ is parallel to $PB$, and $PB$ is perpendicular to $PA$, the angle $\angle PAP'$ is a right angle too. So $APB$ and $PAP'$ are right triangles with the same two legs: $PA$ in both, and $PB = AP'$. They are congruent (side-angle-side), and their hypotenuses are equal: $AB = PP'$. Therefore $OP = \frac{1}{2}PP' = \frac{1}{2}AB = OA$, and $P$ lies on the circle centered at the midpoint $O$ of $AB$ that passes through $A$, which is the circle with diameter $AB$.

Together the two theorems say that the circle with diameter $AB$ is exactly the set of points from which $AB$ is seen at a right angle, apart from $A$ and $B$ themselves.

**Pause and try.** The front of a building is a straight wall 30 m long. A photographer wants to stand where the wall spans an angle of exactly 90° in her view. Describe all the places she can stand. How far is each of them from the middle of the wall?

**Answer.** Call the ends of the wall $A$ and $B$. By Thales' theorem and its converse, the points with $\angle APB = 90°$ are the points of the circle with diameter $AB$. She can stand anywhere on the half of that circle in front of the building. Every such point is a radius away from the center of the circle, which is the middle of the wall: exactly 15 m, wherever she stands.

## From a diameter to any chord

Now return to the pond, where the posts were not opposite each other. A segment joining two points of a circle is called a **chord**. In the first animation the chord $AB$ was seen from the rim at 50°, and the angle at the center, $\angle AOB$, was 100°. In the second, the rim angle was 55° and the angle at the center was 110°. For a diameter the angle at the center is a straight angle, 180°, and Thales gives 90°.

| angle at the center | angle at the rim |
|---|---|
| 100° | 50° |
| 110° | 55° |
| 180° | 90° |

The pattern is hard to miss, and the next animation tests it by moving one of the posts.

[Figure 6, animation: The point B moves around the circle while A and P stay fixed. The angle at the center O and the angle at P both change, and the angle at P is always half of the angle at the center.]

If the angle at the rim is always half the angle at the center, the opening puzzle is solved at once. The angle at the center depends only on $A$, $B$ and $O$. It knows nothing about where $P$ is. Half of it is then the same for every $P$.

So the claim to prove is "half". The proof of Thales' theorem worked by drawing the radius to $P$, which produced isosceles triangles. The same radius is available here.

**Pause and try.** In the first slide of the next figure, the chord $AB$ does not pass through $O$, so the radius $OP$ no longer splits triangle $APB$ into two triangles. Extend it instead: draw the full diameter from $P$ through $O$ to the far side of the circle at $Q$. Let $\alpha = \angle APQ$. Using the isosceles triangle $AOP$, find $\angle AOQ$ in terms of $\alpha$.

**Answer.** In triangle $AOP$, $OA = OP$, so the angle at $A$ equals the angle at $P$, which is $\alpha$. The third angle is $\angle AOP = 180° - 2\alpha$. The angles $\angle AOP$ and $\angle AOQ$ lie side by side along the straight line $PQ$, so $\angle AOQ = 180° - (180° - 2\alpha) = 2\alpha$. The angle at the center is twice the angle at $P$.

[Figure 7, steps: The proof that an inscribed angle is half the central angle, in the case where the center lies inside the angle. The diameter through P splits both angles in two, and on each side an isosceles triangle shows the central part is twice the inscribed part.]

  - step: An inscribed angle APB and the central angle AOB, both standing on the arc from A to B.
  - step: Draw the diameter from P through O to Q. It splits the angle at P and the angle at O into a left part and a right part.
  - step: OA and OP are radii, so triangle AOP is isosceles. Its angles at A and at P are equal: call them α.
  - step: The third angle of that triangle is 180° − 2α, so the angle AOQ next to it on the straight line PQ is 2α.
  - step: The same argument on the right, with the isosceles triangle BOP, gives angle QOB = 2β.
  - step: Add the two parts. The central angle is 2α + 2β, twice the inscribed angle α + β.
This is Thales' proof with one change. There, the two central pieces $2\alpha$ and $2\beta$ lay along a straight line and added to 180°. Here they add to whatever the angle at the center happens to be.

### The statement

To state the result properly we need names for the two kinds of angle, and some care about which arc is meant, because two points cut a circle into two arcs.

**Definition.** Let $A$ and $B$ be two points of a circle with center $O$.

An **inscribed angle** on the chord $AB$ is an angle $\angle APB$ whose vertex $P$ is a third point of the circle. It **intercepts** the arc from $A$ to $B$ that does not contain $P$.

The **central angle** of an arc from $A$ to $B$ is the angle through which the radius turns as its end moves along that arc from $A$ to $B$. It is between 0° and 360°. For an arc shorter than a semicircle it is the ordinary angle $\angle AOB$.

**Theorem.** **(Inscribed angle theorem.)** An inscribed angle is half the central angle of the arc it intercepts.

**Proof.** Let $\angle APB$ be the inscribed angle and let $Q$ be the other end of the diameter through $P$.

*A lemma.* Let $X$ be a point of the circle other than $P$ and $Q$. Then $\angle XOQ = 2\,\angle XPQ$. Indeed, $OX = OP$, so triangle $XOP$ is isosceles and $\angle OXP = \angle OPX = \angle XPQ$. Write $\alpha$ for this angle. The angles of triangle $XOP$ add to 180°, so $\angle XOP = 180° - 2\alpha$. Since $P$, $O$, $Q$ lie on a line with $O$ between, $\angle XOQ = 180° - \angle XOP = 2\alpha$.

Now there are three cases, according to where $A$ and $B$ lie relative to the line $PQ$.

*Case 1: $A$ and $B$ are on opposite sides of $PQ$.* Then the ray $PQ$ lies inside the angle $\angle APB$, so $\angle APB = \angle APQ + \angle QPB$. The intercepted arc runs from $A$ through $Q$ to $B$, so its central angle is $\angle AOQ + \angle QOB$. By the lemma this is $2\,\angle APQ + 2\,\angle QPB = 2\,\angle APB$.

*Case 2: one of the points, say $B$, is $Q$.* Then the intercepted arc runs from $A$ to $Q$, its central angle is $\angle AOQ$, and the lemma with $X = A$ is the claim.

*Case 3: $A$ and $B$ are on the same side of $PQ$.* Name them so that $B$ is the farther from $Q$ along the circle. Then $A$ lies on the arc from $Q$ to $B$ that avoids $P$, and the ray $PA$ lies inside the angle $\angle QPB$, so $\angle APB = \angle QPB - \angle QPA$. The intercepted arc runs from $A$ to $B$ and is what remains of the arc from $Q$ to $B$ when the arc from $Q$ to $A$ is removed, so its central angle is $\angle QOB - \angle QOA$. By the lemma this is $2\,\angle QPB - 2\,\angle QPA = 2\,\angle APB$.

Case 1 is the one in the stepped figure. In the other two the center is on a side of the angle or outside it, and the same lemma does the work with a subtraction in place of the addition.

[Figure 8, image: The other two cases of the proof. Left: one side of the inscribed angle passes through the center, and a single isosceles triangle is enough. Right: the center lies outside the angle, and the angle is a difference of two angles measured from the diameter PQ.]

In the right-hand picture the arcs from $Q$ to $B$ and from $Q$ to $A$ have central angles 140° and 60°, so the inscribed angles $\angle QPB$ and $\angle QPA$ are 70° and 30°, and their difference is 40°, half of $\angle AOB = 80°$.

**Key idea.** An angle with its vertex on a circle is half the central angle of the arc it intercepts. So all the inscribed angles that intercept the same arc are equal.

Thales' theorem is the special case in which the intercepted arc is a semicircle: half of 180° is 90°.

## The other side of the chord

The definition was careful about which arc is intercepted, and the reason shows up when $P$ crosses to the other side of the chord.

**Pause and try.** A chord $AB$ cuts off a short arc with central angle 130°. A point $P$ on the long arc sees the chord at $\frac{1}{2} \cdot 130° = 65°$. What angle does a point $P'$ on the short arc see? Which arc does its angle intercept?

**Answer.** The angle at $P'$ intercepts the arc from $A$ to $B$ that does not contain $P'$, which is the long arc. Its central angle is $360° - 130° = 230°$, so $\angle AP'B = 115°$. Notice that $65° + 115° = 180°$.

[Figure 9, image: A chord AB seen from a point P on the long arc and from a point P' on the short arc. The two angles intercept the two arcs that make up the whole circle, so they add to 180°.]

This always happens. The two arcs make up the whole circle, so their central angles add to 360°, and the inscribed angles on them, being halves, add to 180°. The four points $A$, $P$, $B$, $P'$ form a four-sided figure with all its corners on a circle, called a **cyclic quadrilateral**, and we have shown that its opposite angles at $P$ and $P'$ add to 180°. The same argument with the chord $PP'$ covers the other pair, at $A$ and $B$.

So the opening claim needs a qualification. Walking around the pond, the angle stays the same as long as you stay on one side of the chord. If you pass a post and continue along the short arc, the angle jumps to 180° minus its former value.

**Common mistake.** **Halving the wrong arc.** With $P'$ on the short arc in the figure above, it is tempting to say $\angle AP'B = \frac{1}{2} \cdot 130° = 65°$, because 130° is "the" angle $\angle AOB$. An inscribed angle intercepts the arc on the far side of the chord from its vertex. A quick check catches the error: an angle whose vertex sits close to the chord is obviously obtuse.

**Common mistake.** **Applying the theorem to a vertex that is not on the circle.** The theorem is about angles whose vertex lies on the circle. As the second animation showed, a point inside the circle sees the chord at a larger angle than a point on the arc beyond it, and a point outside sees it at a smaller one. Before halving a central angle, check that the vertex is on the circle and that both sides end on it.

## Using the theorem

**Worked example.** **Reading off angles.** The points $A$, $B$, $C$ lie on a circle with center $O$, and $\angle AOB = 96°$ and $\angle BOC = 140°$, where the arcs $AB$ and $BC$ do not overlap. Find the three angles of triangle $ABC$.

Each angle of the triangle is an inscribed angle, so we need the central angle of the arc opposite each corner. Two are given. The third arc, from $C$ to $A$, is what is left of the circle: $360° - 96° - 140° = 124°$.

The angle at $C$ intercepts the arc $AB$, so it is $\frac{1}{2} \cdot 96° = 48°$. The angle at $A$ intercepts the arc $BC$: $\frac{1}{2} \cdot 140° = 70°$. The angle at $B$ intercepts the arc $CA$: $\frac{1}{2} \cdot 124° = 62°$.

*Check.* $48° + 70° + 62° = 180°$, as it must be, since the three arcs add to 360° and each angle is half of one of them.

**Worked example.** **A chord from an angle.** A triangle is inscribed in a circle of radius 6, and one of its angles is 30°. How long is the side opposite that angle?

Call the side $AB$ and the center $O$. We know nothing else about the triangle, so the only handle on $AB$ is the triangle $AOB$, in which two sides are radii. The 30° angle is an inscribed angle on the chord $AB$, so the central angle is $\angle AOB = 60°$. Triangle $AOB$ is isosceles with $OA = OB = 6$, so its other two angles are equal and add to $180° - 60° = 120°$: each is 60°. A triangle with three angles of 60° is equilateral, so $AB = 6$.

The answer did not depend on the rest of the triangle. In a circle of radius 6, every chord seen from the rim at 30° has length 6.

The next example is a practical trick, and it runs the theorem in the reverse direction: from a known inscribed angle to the arc.

[Figure 10, image: Finding the center of a circle with a sheet of paper. With the corner of the sheet on the circle, its edges cross the circle at the two ends of a diameter. A second placement gives a second diameter, and the two diameters cross at the center.]

**Worked example.** **Finding a lost center.** You have a circular plate and want to mark its center. All you have is a pencil and a rectangular sheet of paper. How can you do it?

A sheet of paper has a reliable right angle at each corner, so the question is what a right angle can tell us about a circle. Place a corner of the sheet at a point $P$ on the rim of the plate, and mark the points $A$ and $B$ where the two edges of the sheet cross the rim.

The angle $\angle APB$ is an inscribed angle of 90°. By the inscribed angle theorem, the arc it intercepts has central angle $2 \times 90° = 180°$. An arc with central angle 180° is a semicircle, so $A$, $O$, $B$ lie on a straight line and $AB$ is a diameter. Draw it with the edge of the sheet.

One diameter passes through the center but does not say where on it the center is. Move the corner to a different point of the rim and repeat. The second diameter also passes through the center, and two different diameters meet only there.

*Check.* Measure from the crossing point to the rim along each of the four half-diameters. The four distances should agree.

## Where the theorem stops

The theorem needs the vertex on the circle, and the second animation showed what happens otherwise. The problems below ask you to prove that an interior point sees a chord at a larger angle and an exterior point at a smaller one.

There is also a limiting case. As $P$ slides along the arc toward $A$, the side $PB$ turns into the chord $AB$, and the side $PA$ shrinks to nothing while its direction approaches that of the tangent line at $A$. The angle keeps its value all the way, which suggests that the angle between a chord and the tangent at one of its ends equals the inscribed angle on the far side. That is true, and the last but one problem proves it directly.

Finally, the whole lesson takes place in a flat plane. The proof used the angle sum of a triangle at every step. On a sphere, where [the angle sum is more than 180°](@geometry/triangle-angle-sum), the proof breaks down, and neither Thales' theorem nor the inscribed angle theorem holds there.

## Looking back

The hard part was finding anything in the moving triangle $APB$ to hold on to. The answer was a point that is not part of the triangle at all, the center $O$. Joining it to the three points on the circle produces radii, equal radii produce isosceles triangles, and isosceles triangles produce equal angles. Almost every proof about angles in a circle starts with that move: draw the radii.

The second idea was to solve the symmetric case first. Thales' theorem looked like a special fact about diameters, and the general theorem turned out to be the same argument with the diameter drawn through $P$ in place of the one through $A$ and $B$.

**Check yourself.** Points $A$ and $B$ on a circle with center $O$ satisfy $\angle AOB = 80°$. A point $P$ lies on the shorter of the two arcs from $A$ to $B$. What is $\angle APB$?
Answer: 140° $P$ is on the short arc, so its angle intercepts the long arc, whose central angle is $280°$. Half of that is $140°$, which is also $180° - 40°$.

**Recap.** - Thales: if $AB$ is a diameter and $P$ is another point of the circle, then $\angle APB = 90°$. Conversely, every point that sees $AB$ at a right angle is on the circle with diameter $AB$.
- Inscribed angle theorem: an angle with its vertex on the circle is half the central angle of the arc it intercepts, which is the arc on the far side from the vertex.
- Inscribed angles on the same arc are equal. Inscribed angles on the two opposite arcs of a chord add to 180°, so opposite angles of a cyclic quadrilateral add to 180°.
- The proof: draw the diameter through the vertex. Each radius to an end of the chord makes an isosceles triangle, and the central angle beside it is twice the base angle.
- Inside the circle the chord is seen at a larger angle, outside at a smaller one.

## Problems

**Problem 1.** Points $A$ and $B$ on a circle with center $O$ satisfy $\angle AOB = 84°$. Find the angle $\angle APB$ for a point $P$ on the longer arc, and for a point $P'$ on the shorter arc.

**Solution.** For $P$ on the longer arc, the intercepted arc is the shorter one, with central angle 84°, so $\angle APB = 42°$. For $P'$ on the shorter arc, the intercepted arc is the longer one, with central angle $360° - 84° = 276°$, so $\angle AP'B = 138°$. Check: $42° + 138° = 180°$.

**Problem 2.** $AB$ is a diameter of a circle of radius 6.5, and $P$ is a point of the circle with $PA = 5$. Find $\angle APB$ and the length $PB$.

**Solution.** By Thales' theorem $\angle APB = 90°$. So $APB$ is a right triangle with hypotenuse $AB = 13$ and one leg 5. By the Pythagorean theorem $PB^2 = 13^2 - 5^2 = 169 - 25 = 144$, so $PB = 12$.

**Problem 3.** On a clock face, a triangle has its corners at the 12, the 4 and the 9. Find its three angles.

**Solution.** The twelve hour marks divide the circle into arcs of $360° / 12 = 30°$. The angle at the 9 intercepts the arc from 12 to 4 that avoids the 9, which spans 4 hours, or 120°. So the angle at the 9 is 60°. The angle at the 12 intercepts the arc from 4 to 9 that avoids the 12: 5 hours, 150°, giving 75°. The angle at the 4 intercepts the arc from 9 to 12 that avoids the 4: 3 hours, 90°, giving 45°. Check: $60° + 75° + 45° = 180°$.

**Problem 4.** (a) A cyclic quadrilateral $ABCD$ has $\angle A = 70°$ and $\angle B = 95°$. Find $\angle C$ and $\angle D$. (b) Show that a parallelogram whose four corners lie on a circle must be a rectangle.

**Solution.** (a) Opposite angles of a cyclic quadrilateral add to 180°. The angle opposite $A$ is $C$, so $\angle C = 110°$, and the angle opposite $B$ is $D$, so $\angle D = 85°$. Check: the four angles add to $70° + 95° + 110° + 85° = 360°$, as the angles of any quadrilateral do.

(b) In a parallelogram, opposite angles are equal. If its corners lie on a circle, opposite angles also add to 180°. Two equal angles that add to 180° are each 90°. So all four angles are right angles and the parallelogram is a rectangle.

**Problem 5.** Let $A$ and $B$ be points of a circle, and let $\theta$ be the inscribed angle on the chord $AB$ from points of one of the two arcs. Let $P$ be a point on the same side of the line $AB$ as that arc. (a) Suppose $P$ is inside the circle, and let the line $AP$, continued beyond $P$, meet the arc at $P'$. Show that $\angle APB > \theta$. (b) Suppose $P$ is outside the circle. The segment from $P$ to the midpoint $M$ of $AB$ crosses the arc at a point $P'$. Continue the line $AP'$ beyond $P'$ until it meets the segment $PB$ at $N$, and show that $\angle APB < \theta$. You may use the fact that an exterior angle of a triangle equals the sum of the two interior angles at the other corners.

**Solution.** (a) The point $P'$ is on the arc, so $\angle AP'B = \theta$. Since $P$ lies on the segment $AP'$, this is also the angle of triangle $PP'B$ at $P'$. The angle $\angle APB$ is an exterior angle of triangle $PP'B$ at $P$, because $A$, $P$, $P'$ lie on a line with $P$ between. So

$$
\angle APB = \angle PP'B + \angle PBP' = \theta + \angle PBP' > \theta.
$$

(b) The point $P'$ lies on the segment $PM$, which joins a corner of triangle $APB$ to a point of the opposite side. So $P'$ is inside that triangle, and the line from $A$ through $P'$ leaves it through the side $PB$, at $N$. Now $\theta = \angle AP'B$ is an exterior angle of triangle $P'NB$ at $P'$, and $\angle ANB$ is an exterior angle of triangle $APN$ at $N$. So

$$
\theta = \angle P'NB + \angle P'BN > \angle ANB = \angle APN + \angle PAN > \angle APB.
$$

Together with the theorem, this shows that among points on one side of the line $AB$, the arc is exactly the set of points that see the chord at the angle $\theta$. (The exterior angle fact follows from the angle sum: the exterior angle and the interior angle beside it add to 180°, and so do all three interior angles.)

**Problem 6.** A line that touches a circle at a single point $A$ is called a tangent, and it is perpendicular to the radius $OA$. Let $AB$ be a chord whose shorter arc has central angle $2\theta$, with $2\theta < 180°$, and let $T$ be a point of the tangent at $A$ on the same side of the line $OA$ as $B$. Show that $\angle TAB = \theta$, the inscribed angle on the chord $AB$ from the longer arc.

**Solution.** Triangle $AOB$ is isosceles with $OA = OB$ and angle $2\theta$ at $O$. Its other two angles are equal and add to $180° - 2\theta$, so $\angle OAB = 90° - \theta$. The tangent is perpendicular to $OA$, so $\angle OAT = 90°$. The point $B$ lies on the same side of the tangent as $O$ (the whole circle does) and on the same side of $OA$ as $T$, so the ray $AB$ lies inside the right angle $\angle OAT$, and

$$
\angle TAB = \angle OAT - \angle OAB = 90° - (90° - \theta) = \theta.
$$

A point of the longer arc sees the chord at half the central angle $2\theta$, which is the same $\theta$.

[Figure 11, image: Two chords AB and CD of a circle cross at the point X inside it. The triangles XAC and XDB are shaded, and the angles at A and D are marked.]

**Problem 7.** Two chords $AB$ and $CD$ of a circle cross at a point $X$ inside the circle, as in the figure above. (a) Show that triangles $XAC$ and $XDB$ are similar. (b) Deduce that $XA \cdot XB = XC \cdot XD$. (c) If $XA = 3$, $XB = 8$ and $XC = 4$, find $XD$.

**Solution.** (a) The angle of triangle $XAC$ at $A$ is $\angle BAC$, an inscribed angle that intercepts the arc from $B$ to $C$ not containing $A$. The angle of triangle $XDB$ at $D$ is $\angle CDB$, an inscribed angle that intercepts the arc from $C$ to $B$ not containing $D$. Because the chords cross, the points come in the order $A$, $C$, $B$, $D$ around the circle, so the arc between $C$ and $B$ that avoids $A$ also avoids $D$: both angles intercept the same arc, and they are equal. The angles of the two triangles at $X$ are equal as well, because they are opposite angles where two lines cross. Two pairs of equal angles make the triangles [similar](@geometry/similar-triangles): $\triangle XAC \sim \triangle XDB$, with $X$ matching $X$, $A$ matching $D$ and $C$ matching $B$.

(b) Matching sides are in the same ratio: $\dfrac{XA}{XD} = \dfrac{XC}{XB}$. Multiplying both sides by $XD \cdot XB$ gives $XA \cdot XB = XC \cdot XD$.

(c) $3 \cdot 8 = 4 \cdot XD$, so $XD = 6$. Check: $3 \cdot 8 = 24 = 4 \cdot 6$.