# Basic Description

Figure 1:
Example of an Inscribed Angle

An inscribed angle is formed by two chords sharing an endpoint on a circle. (See Figure 1) The basic properties of inscribed angles were originally described in Euclid's Elements (300 BC), Book III, Propositions 20 through 22. Euler's three propositions are: [1]:

• The inscribed angle is half the central angle.
• Inscribed angles on the same arc of a circle are equal.
• The sum of opposite angles of inscribed quadrilaterals in a circle is equal to 180 degrees.

This helper page will first define some basic related concepts, then prove the three propositions in Elements, and finally prove Thales' Theorem, which is a special case of Proposition 20.

# A Few Concepts [2]

Figure 2-1: Terminology inside a circle:click to enlarge

A circle is defined many ways. Euclid's definition is: A circle is a plane figure contained by one line such that all the straight lines falling upon it from one point among those lying within the figure are equal to one another. (Euclid uses "line" where today we would use "curve." He uses "straight line" where we would often just use "line.") He then defines that point as the center of the circle.

Modern definitions come in two varieties. The static version is: "A circle is a simple shape of Euclidean geometry consisting of those points in a plane that are equidistant from a given point, the center." The dynamic version is: "A circle can be defined as the curve traced out by a point that moves so that its distance from a given point is constant."

Figure 2-2: Intercepting and subtending:click to enlarge

The radius of a circle is any line segment from the circle's center to its perimeter.

A line of a circle is a line that intersects two points on the circle.

A chord is the portion of a secant that lies within the curve.

Intercept/Subtend: In Figure 2-2, we say that the inscribed angle BAC intercepts or subtends the arc BC. A central angle BOC also intercepts or subtends that same arc.

A central angle is an angle whose vertex is the center of a circle, and whose sides pass through a pair of points on the circle, thereby subtending (previously defined) an an arc between those two points whose angle is equal to the central angle itself.

# Propositions 20, 21, 22

## Proposition 20

An inscribed angle is equal to one-half the central angle intercepting the same arc.[1]

Proof Outline:

Notice under Proposition 20, different relative positions of the inscribed angle and the center of the circle lead to similar but different proofs. The following three cases will be considered for a comprehensive proof of Proposition 20:

• Case 1: One side of the angle is the diameter of the circle.
• Case 2: The center of the circle is inside the inscribed angle.
• Case 3: The center of the circle is outside the inscribed angle.

We shall start with Case 1, where one side of the angle passes through the center of the circle, as shown in Figure 3-1.

Figure 3-1: Inscribed angle passing through center of circle

Connect AO and extend it to meet the circle at X

Since the radii of a circle are equal,

$AO=OB$

Because an exterior angle of a triangle is equal to the sum of the opposite interior angles,

$\angle BOX= \angle BAO + \angle OBA$

Because the base angles of an isosceles triangle are equal,

$\angle BAO= \angle OBA$

Thus,

$\angle BOX = 2 \angle BAO$

Q.E.D.

Case 2: The center of the circle is inside the inscribed angle (as demonstrated in Figure2-2). We offer the following proof:

Figure 3-2: Center of circle inside the inscribed angle

Proof:

By connecting AO and extending it to intercept the circle at point X and repeating the steps of the previous proof, we get:

Equa (1)        $\angle BOX = 2 \angle BAO$

Equa (2)        $\angle COX = 2 \angle CAO$

Using the angle addition property, we add Equa (1) and Equa (2) to get:

$\angle BOX + \angle COX= 2 (\angle BAO + \angle CAO)$

$\angle BOC = 2\angle BAC$

Q.E.D.

Case 3: The center of the circle is outside the inscribed angle (as demonstrated in Figure2-2). We offer the following proof:

Figure 3-3: Center of circle outside the inscribed angle

Proof:

We first connect AO and extend it to intercept the circle at point X.

Repeating steps from the previous two proofs, we get:

Equa (1)        $\angle BOX = 2\angle BAX = 2(\angle BAC + \angle CAX)$

Equa (2)        $\angle COX = 2\angle CAX$

Using the angle addition property, we subtract Equa (2) from Equa (1) to get:

$\angle BOC = 2\angle BAC$

Q.E.D.

If you're interested, here's a literal translation of Euclid's proof of the proposition by Sir Thomas H. Health.[1]

Notice the elegance in his flow of logic and how his style of presentation differs from that of the present day. Also notice how he consistently proved [...]

Figure 3-2, points relabeled

Let ABC be a circle, let the angle BDC be an angle at its center, and the angle BAC an angle at the circumference, and let them have the same circumference BC as base.

I say that the angle BDC is double the angle BAC.

Join AD, and draw it through to E.

Then, since DA equals DB, the angle DAB also equals the angle DBA. Therefore the sum of the angles DAB and DBA is double the angle DAB.

For the same reason the angle CDE is also double the angle CAD.

Therefore the whole angle CDB is double the whole angle CAB.

Therefore in a circle the angle at the center is double the angle at the circumference when the angles have the same circumference as base.

Q.E.D.

## Proposition 21

In a circle, different inscribed angles subtending the same arc are equal in measure.[1]

Figure 4:click to enlarge

Proof:

Let ABC be a circle and point D its center. According to Proposition 20,

$\angle BAC = \frac{1}{2} \angle BDC$

and

$\angle BEC = \frac{1}{2} \angle BDC$

Since equals of equals are equal,

$\angle BEC = \angle BAC$

Q.E.D.

Again, a literal translation of Euclid's proof of the proposition.[1]

Let ABCE be a circle, and let the angles BAD and BED be angles in the same segment BAEC.

I say that the angles BAC and BEC equal one another.

Take the center D of the circle ABCE, and join DB and DC.

Now, since the angle BDC is at the center, and the angle BAC at the circumference, and they have the same circumference BAC as base, therefore the angle BDC is double the angle BAC. III.20

For the same reason the angle BDC is also double the angle BEC.

Therefore the angle BAC equals the angle BEC.

Therefore in a circle the angles in the same segment equal one another.

Q.E.D.

## Proposition 22

The sum of opposite angles of quadrilaterals in circles is equal to two right angles.[1]

Figure 5:click to enlarge

Proof:

Let ABCD be a circle, and let ABCD be a quadrilateral in it.

By Proposition 21, since $\angle BAC$ and $\angle BDC$ correspond to the same arc,

$\angle BAC= \angle BDC$

Similarly,

$\angle CAD= \angle CBD$

$\angle BAD=\angle BAC + \angle CAD$

Thus,

Equa (3)        $\angle BAD + \angle DCB= (\angle BAC + \angle CAD)+ \angle DCB$

Substituting angle BCD for BAC and angle CBD for CAD since they correspond to the same arc,

$\angle DCB + (\angle BAC + \angle CAD) = \angle DCB + (\angle BDC + \angle CBD)$

Now look at triangle CBD, because the measures of the interior angle of a triangle always add up to 180 degrees,

$\angle DCB + (\angle BDC + \angle CBD)= 180^\circ$

Hence, by Equa (3),

$\angle BAD + \angle DCB=180^\circ$

Q.E.D.

Again, our proof is followed by a literal translation of Euclid's proof.[1]

Let ABCD be a circle, and let ABCD be a quadrilateral in it.

I say that the sum of the opposite angles equals two right angles.

Join AC and BD.

Then, since in any triangle the sum of the three angles equals two right angles, the sum of the three angles CAB, ABC, and BCA of the triangle ABC equals two right angles. I.32

But the angle CAB equals the angle BDC, for they are in the same segment BADC, and the angle ACB equals the angle ADB, for they are in the same segment ADCB, therefore the whole angle ADC equals the sum of the angles BAC and ACB. III.21

(Note: Here Euclid's way of expression may be confusing to us modern readers. What he means by "segment BADC" is actually just "arc AC".)

Add the angle ABC to each. Therefore the sum of the angles ABC, BAC, and ACB equals the sum of the angles ABC and ADC. But the sum of the angles ABC, BAC, and ACB equals two right angles, therefore the sum of the angles ABC and ADC also equal two right angles.

Similarly we can prove that the sum of the angles BAD and DCB also equals two right angles.

Therefore the sum of the opposite angles of quadrilaterals in circles equals two right angles.

Q.E.D.

# Thales' Theorem

In geometry, Thales' theorem states that if A, B, and C are points on a circle where the line AC is a diameter of the circle, then the angle ABC is a right angle.[3]

Figure 6:click to enlarge

Proof:

By observation, angle AOC (Note: AOC is a straight line) and angle ABC subtend to the same arc AC of the circle.

By proposition 20, Book III, Elements, $\angle ABC = \frac {1}{2} \angle AOC = 90^\circ$

Thus, as point B shifts around the perimeter of the circle, $\vartriangle ABC$ is always a right triangle.

Q.E.D.

# References

1. Health, L, Thomas. (2002). Euclid's Elements. Place of publication: Green Lion Press. ISBN-13: 978-1888009187
2. Wikipedia. All concepts (words/phrases in bold) are extracted from their Wikipedia pages.
3. Wikipedia, http://en.wikipedia.org/wiki/Thales'_theorem.