Inscribed Angles

⭕ Inscribed Angles

Part 1 of 5 — Arcs, Central Angles & Intercepted Arcs

Topics in This Part

🔑 Key Concept: Every angle theorem in this lesson compares an angle to the arc it cuts off. Before we can master inscribed angles in Part 2, we have to be fluent in arcs — how they're measured and which arc a given angle "catches." That's the whole job of Part 1.

The Language of Circles

A few words appear in every problem. Lock them in now:

An arc is named by its endpoints, like $\overset{\frown}{AB}$. Two points actually split a circle into two arcs:

• a minor arc (the shorter one, less than $180°$), and
• a major arc (the longer one, more than $180°$).

💡 To avoid ambiguity, a major arc is usually named with three letters — $\overset{\frown}{ABC}$ — so you know which way around the circle you mean.

Circle Vocabulary 🔽

Pick the term that fits each description.

Central Angles Measure Their Arcs

A central angle has its vertex at the center of the circle. Its two sides are radii, and they cut off an arc.

$\textbf{The measure of a minor arc} = \textbf{the measure of its central angle.}$

So if central angle $\angle AOB = 70°$ (where $O$ is the center), then the minor arc it opens onto, $\overset{\frown}{AB}$, is also $70°$.

Because a full circle is $360°$:

$\text{minor arc} + \text{major arc} = 360°.$

Example

If $\overset{\frown}{AB} = 70°$, then the major arc $\overset{\frown}{ACB} = 360° - 70° = 290°$.

⚠️ Don't confuse arc measure with arc length. Arc measure is in degrees and depends only on the central angle. Arc length is an actual distance and also depends on the radius. This lesson is entirely about arc measure.

Concept Check 🎯

What an Angle "Intercepts"

An angle intercepts an arc when the arc lies in the interior of the angle, with its endpoints sitting on the two sides of the angle.

Picture an angle with its vertex somewhere and its two sides slicing across the circle. The arc "trapped" between the two sides — the one you'd see looking out from the vertex — is the intercepted arc.

🔑 The single most important habit in this whole lesson: for any angle, find the arc it intercepts first. Once you can name that arc, every theorem becomes a one-step formula. We will say intercepted arc over and over — make sure you can point to it.

In Part 1 we measure arcs with central angles (vertex at the center). Starting in Part 2, the vertex moves onto the circle — and a beautiful pattern appears.

Arc Arithmetic 🧮

A circle is divided by points $A$, $B$, and $C$. Use "central angle $=$ arc" and "arcs sum to $360°$."

1) Central angle $\angle AOB = 130°$. Then $\overset{\frown}{AB} = \,?$ degrees. 2) Minor arc $\overset{\frown}{AB} = 130°$ and minor arc $\overset{\frown}{BC} = 95°$. The remaining arc $\overset{\frown}{CA} = \,?$ degrees. 3) A diameter splits a circle into two arcs. Each semicircle measures $\,?$ degrees.

⭕ Inscribed Angles

Part 2 of 5 — The Inscribed Angle Theorem

🔑 The Big Idea: Move the vertex from the center (Part 1) onto the circle itself, and the angle shrinks to exactly half of the arc it intercepts. That one fact — the Inscribed Angle Theorem — powers the rest of this lesson.

What Is an Inscribed Angle?

An inscribed angle is an angle whose:

• vertex lies ON the circle, and
• two sides are chords of the circle.

So an inscribed angle "sits" on the circle and reaches out with two chords. Those chords cut off the intercepted arc — the arc on the far side, not containing the vertex.

$\boxed{\text{inscribed angle} = \tfrac{1}{2}\,(\text{intercepted arc})}$

Equivalently, the arc is twice the inscribed angle:

$\text{intercepted arc} = 2 \times (\text{inscribed angle}).$

💡 Compare the two vertex positions:

Vertex at...	Angle vs. its arc
the center (central angle)	angle $=$ arc
a point on the circle (inscribed angle)	angle $= \tfrac{1}{2}$ arc

Worked Examples

Example 1 — Find the angle from the arc

Inscribed angle $\angle ABC$ intercepts arc $\overset{\frown}{AC} = 80°$. Then:

$m\angle ABC = \tfrac{1}{2}(80°) = 40°.$

Example 2 — Find the arc from the angle

An inscribed angle measures $54°$. Its intercepted arc is:

$\text{arc} = 2(54°) = 108°.$

Example 3 — Inscribed and central on the same arc

A central angle and an inscribed angle both intercept arc $\overset{\frown}{PQ} = 100°$.

$\text{central angle} = 100°, \qquad \text{inscribed angle} = \tfrac{1}{2}(100°) = 50°.$

So the inscribed angle is exactly half of the central angle that sees the same arc.

⚠️ Mind the direction of the formula. Going arc → angle you halve; going angle → arc you double. Mixing these up is the #1 error on this topic.

Concept Check 🎯

Halve or Double? 🔽

Decide the relationship and the result in each case.

Apply the Theorem 🧮

Use inscribed angle $= \tfrac12(\text{intercepted arc})$. Give answers in degrees (no ° symbol).

1) Intercepted arc $= 150°$. Inscribed angle $= \,?$ 2) Inscribed angle $= 45°$. Intercepted arc $= \,?$ 3) Inscribed angle $= 25°$. The matching central angle on the same arc $= \,?$

⭕ Inscribed Angles

Part 3 of 5 — Three Powerful Corollaries

🔑 Why this part matters: Three facts fall straight out of "angle $= \tfrac12$ arc." Recognizing them lets you read an answer off a figure in seconds, with no arithmetic at all.

Corollary 1 — Same Arc ⇒ Equal Angles

If two (or more) inscribed angles intercept the same arc, they are congruent.

Why? Each one equals $\tfrac12$ of that same arc, so they must be equal.

$\angle ADB = \angle ACB = \tfrac12\,\overset{\frown}{AB}.$

This is true no matter where the vertices $C$ and $D$ sit on the major arc — every "viewpoint" on the same side sees the chord $AB$ under the same angle.

💡 You'll often see two triangles sharing a chord; the angles "looking at" that chord from the same side are automatically equal. This is a classic way to prove triangles similar.

Corollary 2 — Angle in a Semicircle is $90°$ (Thales' Theorem)

If the intercepted arc is a semicircle ($180°$) — which happens exactly when the chord is a diameter — then the inscribed angle is:

$\tfrac12(180°) = 90°.$

So any angle inscribed in a semicircle is a right angle. Equivalently: if $\overline{AB}$ is a diameter and $C$ is any other point on the circle, then $\angle ACB = 90°$.

This is Thales' Theorem, and it's one of the most useful results in all of geometry — it manufactures right angles for free.

⚠️ The right angle is at the vertex on the circle ($C$), not at the endpoints of the diameter.

Concept Check 🎯

Putting Thales to Work

Because a triangle inscribed in a semicircle has a right angle, it's a right triangle — so the Pythagorean Theorem applies.

Worked Example

$\overline{AB}$ is a diameter, so $\angle C = 90°$. If $AC = 6$ and $BC = 8$, find the diameter $AB$.

$AB^2 = AC^2 + BC^2 = 6^2 + 8^2 = 36 + 64 = 100$ $AB = \sqrt{100} = 10.$

So the diameter is $10$, and the radius is $5$.

💡 Whenever a problem mentions a triangle with one side as a diameter, immediately mark the opposite angle $90°$ — half the battle is won.

Use the Corollaries 🧮

Answers in degrees or units (no symbols).

1) $\overline{AB}$ is a diameter; $C$ is on the circle. $m\angle ACB = \,?$ degrees. 2) Two inscribed angles see arc $\overset{\frown}{XY} = 88°$. Each angle $= \,?$ degrees. 3) A triangle is inscribed in a semicircle with legs $9$ and $12$. The diameter (hypotenuse) $= \,?$

Name the Corollary 🔽

Match each situation to the result it guarantees.

⭕ Inscribed Angles

Part 4 of 5 — Inscribed (Cyclic) Quadrilaterals

🔑 Big Payoff: When all four vertices of a quadrilateral lie on a circle, its opposite angles are supplementary — they add to $180°$. This single rule cracks open a whole class of figures.

The Cyclic Quadrilateral Theorem

A cyclic (or inscribed) quadrilateral has all four vertices on one circle. Label it $ABCD$ in order around the circle. Then:

$m\angle A + m\angle C = 180° \qquad\text{and}\qquad m\angle B + m\angle D = 180°.$

Opposite angles are supplementary.

Why it's true

$\angle A$ and $\angle C$ are inscribed angles that intercept the two arcs that together make the whole circle. So:

$m\angle A + m\angle C = \tfrac12\big(\text{arc}_1\big) + \tfrac12\big(\text{arc}_2\big) = \tfrac12\big(\text{arc}_1 + \text{arc}_2\big) = \tfrac12(360°) = 180°.$

💡 Notice you only need this once: the moment you know one angle, its opposite is $180°$ minus it.

Worked Example

In cyclic quadrilateral $ABCD$, $m\angle A = 95°$ and $m\angle B = 110°$. Find $\angle C$ and $\angle D$.

Opposite to $A$ is $C$: $m\angle C = 180° - 95° = 85°.$

Opposite to $B$ is $D$: $m\angle D = 180° - 110° = 70°.$

✅ Check: All four interior angles of a quadrilateral sum to $360°$: $95° + 110° + 85° + 70° = 360°.\;✓$

Algebra version

If $m\angle A = (2x + 10)°$ and its opposite $m\angle C = (3x - 30)°$, then:

$(2x + 10) + (3x - 30) = 180 \;\Rightarrow\; 5x - 20 = 180 \;\Rightarrow\; 5x = 200 \;\Rightarrow\; x = 40.$

Concept Check 🎯

Cyclic Quadrilateral Practice 🧮

$ABCD$ is inscribed in a circle (vertices in order). Answer in degrees (no ° symbol).

1) $m\angle A = 64°$. Find $m\angle C$ (its opposite). 2) $m\angle B = 121°$. Find $m\angle D$ (its opposite). 3) $m\angle A = (2x)°$ and $m\angle C = (x + 30)°$. Solve for $x$.

Fill the Quadrilateral 🔽

$ABCD$ is cyclic with $m\angle A = 80°$ and $m\angle B = 100°$. Read off the rest.

⭕ Inscribed Angles

Part 5 of 5 — Tangent–Chord Angles, Mixed Practice & Mastery Check

You can now handle central angles, inscribed angles, the three corollaries, and cyclic quadrilaterals. One last vertex position completes the picture — the vertex right on the circle where a tangent meets a chord.

The Tangent–Chord Angle

A tangent is a line that touches the circle at exactly one point. When a tangent and a chord meet at that point of tangency, the angle they form follows the same half-the-arc rule as an inscribed angle:

$\text{tangent–chord angle} = \tfrac12\,(\text{intercepted arc}).$

The intercepted arc is the one "inside" the angle, cut off by the chord.

Worked Example

A chord cuts off an arc of $130°$, and a tangent is drawn at one end of the chord. The angle between the tangent and the chord is:

$\tfrac12(130°) = 65°.$

💡 Same rule, new vertex. Central angle $=$ arc; inscribed angle $= \tfrac12$ arc; tangent–chord angle $= \tfrac12$ arc. The vertex position changes, but the "half the arc" pattern for angles on the circle stays the same.

Tangent–Chord Practice 🧮

A tangent meets a chord at a point on the circle. Use angle $= \tfrac12(\text{intercepted arc})$. Answers in degrees (no ° symbol).

1) Intercepted arc $= 200°$. Tangent–chord angle $= \,?$ 2) Tangent–chord angle $= 48°$. Intercepted arc $= \,?$ 3) The chord is a diameter and the tangent is at its end. The tangent–chord angle $= \,?$

Quick Reference

⚠️ Three habits that prevent every common error: (1) always identify the intercepted arc first; (2) for a vertex on the circle, halve; for a vertex at the center, don't; (3) for a diameter, instantly write $90°$.

Which Rule Applies? 🔽

Match each vertex position to the right relationship between the angle and its intercepted arc.

Mixed Practice 🎯

Exit Quiz ✅

Answer all three to finish the lesson.