# Properties of the Inscribed Angle

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## Contents |

# Basic Description

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]}

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."

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

A **secant 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.

Proof:

Connect *AO* and extend it to meet the circle at *X*

Since the radii of a circle are equal,

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

Because the base angles of an isosceles triangle are equal,

Thus,

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:

Proof:

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

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

By the angle addition property,

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:

Proof:

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

Repeating steps from the previous two proofs, we get:

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

Q.E.D.

^{[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

**[...]**

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]}

Proof:

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

and

Since equals of equals are equal,

Q.E.D.

^{[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]}

Proof:

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

By Proposition 21, since and correspond to the same arc,

Similarly,

By the angle addition property,

Thus,

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

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

Hence, by **Equa (3)**,

Q.E.D.

^{[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]}

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*,

Thus, as point *B* shifts around the perimeter of the circle, is always a right triangle.

Q.E.D.

# References

- ↑
^{1.0}^{1.1}^{1.2}^{1.3}^{1.4}^{1.5}^{1.6}Health, L, Thomas. (2002).*Euclid's Elements*. Place of publication: Green Lion Press. ISBN-13: 978-1888009187 - ↑ Wikipedia. All concepts (words/phrases in bold) are extracted from their Wikipedia pages.
- ↑ Wikipedia, http://en.wikipedia.org/wiki/Thales'_theorem.