10.1 Central, Inscribed, Tangent & Chord Angles and Arcs

10.1 Central, Inscribed, Tangent & Chord Angles and Arcs

Metric Foundations: Central Angles and Intercepted Arcs

In Euclidean plane geometry, a circle $\mathcal{C}(O, r)$ is defined as the locus of all points in a plane equidistant from a fixed center point $O$ by radius $r$. An arc is a continuous portion of the circle's circumference. The angular measure of an arc is defined directly by the central angle that subtends it. A central angle is an angle whose vertex is located at the center $O$ of the circle and whose sides are radii intersecting the circle at distinct points $A$ and $B$. By fundamental geometric postulate, the measure of a central angle is identically equal to the degree measure of its intercepted minor arc: $m(\angle AOB) = m(\widehat{AB})$ A full circle spans $360^\circ$ (or $2\pi$ radians), while a semicircle spans exactly $180^\circ$ (or $\pi$ radians). Arcs measuring strictly less than $180^\circ$ are classified as minor arcs and are typically designated by their two endpoints (e.g., $\widehat{AB}$), whereas arcs measuring strictly greater than $180^\circ$ are major arcs, designated by three points (e.g., $\widehat{ACB}$) to clarify the traversal direction along the circumference. The Arc Addition Postulate asserts that if point $C$ lies on $\widehat{AB}$, then $m(\widehat{AC}) + m(\widehat{CB}) = m(\widehat{AB})$. It is critical to distinguish between arc measure (the angular rotation in degrees or radians, invariant under scale dilation) and arc length (the linear distance along the curved boundary, which scales proportionally with radius $r$).

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Inscribed Angles and Thales's Semicircle Theorem

An inscribed angle is an angle whose vertex lies on the circle and whose sides are chords of the circle intersecting the circumference at distinct points. The core foundation of circular angle measurement is the Inscribed Angle Theorem: $m(\angle ABC) = \frac{1}{2}m(\widehat{AC})$

Deductive Proof of the Inscribed Angle Theorem

The proof is partitioned into three exhaustive geometric cases depending on the position of the circle's center $O$ relative to the angle:

1. Case 1 (Center on one side of the angle): Assume side $\overline{BC}$ passes through center $O$, forming a diameter. Draw radius $\overline{OA}$. Because $\overline{OA}$ and $\overline{OB}$ are both radii, triangle $\triangle AOB$ is isosceles with base angles $\angle OBA \cong \angle OAB$. Let $m(\angle ABC) = x$. Then $m(\angle OAB) = x$. Central angle $\angle AOC$ is an exterior angle to $\triangle AOB$. By the Triangle Exterior Angle Theorem, an exterior angle equals the sum of the two remote interior angles: $m(\angle AOC) = m(\angle OBA) + m(\angle OAB) = x + x = 2x$ Because $\angle AOC$ is a central angle intercepting arc $\widehat{AC}$, $m(\widehat{AC}) = m(\angle AOC) = 2x$. Substituting $x = m(\angle ABC)$ yields $m(\angle ABC) = \frac{1}{2}m(\widehat{AC})$.
2. Case 2 (Center in the interior of the angle): Construct an auxiliary diameter from vertex $B$ through center $O$ intersecting the circle at point $D$. Diameter $\overline{BD}$ divides $\angle ABC$ into two adjacent angles $\angle ABD$ and $\angle DBC$, each having a side along a diameter. Applying Case 1 to both sub-angles: $m(\angle ABC) = m(\angle ABD) + m(\angle DBC) = \frac{1}{2}m(\widehat{AD}) + \frac{1}{2}m(\widehat{DC}) = \frac{1}{2}(m(\widehat{AD}) + m(\widehat{DC})) = \frac{1}{2}m(\widehat{AC})$
3. Case 3 (Center in the exterior of the angle): Construct diameter $\overline{BD}$ through $O$. Here, $\angle ABC = \angle ABD - \angle CBD$. Applying Case 1 to each angle yields $m(\angle ABC) = \frac{1}{2}m(\widehat{AD}) - \frac{1}{2}m(\widehat{CD}) = \frac{1}{2}m(\widehat{AC})$.

Corollaries: Congruent Arcs and Thales's Theorem

Two direct consequences follow immediately from this theorem:

• Inscribed Angles Intercepting the Same Arc: Any two inscribed angles that subtend identical or congruent arcs are congruent: if $\angle ABC$ and $\angle ADC$ intercept $\widehat{AC}$, then $m(\angle ABC) = m(\angle ADC) = \frac{1}{2}m(\widehat{AC})$.
• Thales's Theorem (Inscribed Right Triangle): An angle inscribed in a semicircle intercepts an arc of $180^\circ$. Therefore, its measure is identically $\frac{1}{2}(180^\circ) = 90^\circ$. Conversely, if an inscribed triangle has a right angle, its hypotenuse must be a diameter of the circumscribed circle.

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Inscribed (Cyclic) Quadrilaterals and Chord-Tangent Angle Theorems

A polygon is inscribed in a circle if all of its vertices lie on the circumference; such a polygon is termed cyclic. For quadrilaterals, this property imposes a strict metric constraint:

Inscribed Quadrilateral Theorem

A convex quadrilateral $ABCD$ can be inscribed in a circle if and only if its opposite angles are supplementary: $m(\angle A) + m(\angle C) = 180^\circ \quad \text{and} \quad m(\angle B) + m(\angle D) = 180^\circ$

Proof: Let quadrilateral $ABCD$ be inscribed in circle $O$. Angle $\angle A$ intercepts arc $\widehat{BCD}$, while opposite angle $\angle C$ intercepts arc $\widehat{DAB}$. Together, arcs $\widehat{BCD}$ and $\widehat{DAB}$ cover the entire circumference without overlap, so $m(\widehat{BCD}) + m(\widehat{DAB}) = 360^\circ$. By the Inscribed Angle Theorem: $m(\angle A) = \frac{1}{2}m(\widehat{BCD}) \quad \text{and} \quad m(\angle C) = \frac{1}{2}m(\widehat{DAB})$ Adding these equations yields: $m(\angle A) + m(\angle C) = \frac{1}{2}(m(\widehat{BCD}) + m(\widehat{DAB})) = \frac{1}{2}(360^\circ) = 180^\circ$ Because the sum of all interior angles of a quadrilateral is $360^\circ$, it immediately follows that $m(\angle B) + m(\angle D) = 360^\circ - 180^\circ = 180^\circ$.

Tangent-Chord Theorem

A line tangent to a circle intersects the circle at exactly one point of tangency $T$ and is perpendicular to the radius drawn to that point ($OT \perp \text{tangent}$). When a chord $\overline{TA}$ is drawn from the point of tangency, it forms an angle with the tangent line. The Tangent-Chord Theorem states that the measure of the angle formed by a tangent and a chord equals half the measure of the intercepted arc: $m(\angle PTA) = \frac{1}{2}m(\widehat{TA})$ This theorem can be proven by drawing diameter $\overline{TD}$. Because $OT \perp PT$, $\angle PTD = 90^\circ$. Inscribed $\triangle TAD$ has $\angle TAD = 90^\circ$ by Thales's theorem. Subtracting complementary angles proves that the angle between the tangent and chord matches the inscribed angle intercepting arc $\widehat{TA}$, confirming the half-arc relationship.

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Angles Formed by Chords, Secants, and Tangents: Interior vs. Exterior Configurations

When intersecting lines form angles with a circle, the formula for the angle depends entirely on the location of the vertex relative to the circle:

Interior Vertex: Intersecting Chords

When two chords $\overline{AC}$ and $\overline{BD}$ intersect at point $E$ in the interior of the circle (not at the center), the vertical angles formed intercept two opposite arcs $\widehat{AB}$ and $\widehat{CD}$. The angle measure is the arithmetic mean of the two intercepted arcs: $m(\angle AEB) = \frac{1}{2}\left(m(\widehat{AB}) + m(\widehat{CD})\right)$ Proof: Draw auxiliary chord $\overline{AD}$. In triangle $\triangle ADE$, angle $\angle AEB$ is an exterior angle at vertex $E$. By the Triangle Exterior Angle Theorem, $m(\angle AEB) = m(\angle ADE) + m(\angle DAE)$. Inscribed angle $\angle ADE$ intercepts arc $\widehat{AB}$, so $m(\angle ADE) = \frac{1}{2}m(\widehat{AB})$. Inscribed angle $\angle DAE$ intercepts arc $\widehat{CD}$, so $m(\angle DAE) = \frac{1}{2}m(\widehat{CD})$. Summing these expressions yields $m(\angle AEB) = \frac{1}{2}(m(\widehat{AB}) + m(\widehat{CD}))$.

Exterior Vertex: Secants and Tangents

When two lines intersect at point $P$ strictly outside the circle, they intercept two distinct arcs: a far arc $\widehat{\text{arc}}{\text{far}}$ and a near arc $\widehat{\text{arc}}{\text{near}}$. Whether the lines are two secants, a secant and a tangent, or two tangents, the angle measure is half the positive difference of the intercepted arcs: $m(\angle P) = \frac{1}{2}\left(m(\widehat{\text{arc}}_{\text{far}}) - m(\widehat{\text{arc}}_{\text{near}})\right)$ Proof (Two Secants): Let secants $\overline{PAB}$ and $\overline{PCD}$ intersect the circle at near points $A, C$ and far points $B, D$. Draw chord $\overline{AD}$. In $\triangle PAD$, inscribed angle $\angle BAD$ is an exterior angle: $m(\angle BAD) = m(\angle P) + m(\angle ADC)$. Rearranging gives $m(\angle P) = m(\angle BAD) - m(\angle ADC) = \frac{1}{2}m(\widehat{BD}) - \frac{1}{2}m(\widehat{AC}) = \frac{1}{2}(m(\widehat{BD}) - m(\widehat{AC}))$.

Circumscribed Angles

A circumscribed angle is formed by two tangent lines meeting at an exterior point $P$. Because tangents intercept the entire circle partitioned into a minor arc $x$ and a major arc $360^\circ - x$, the exterior angle formula simplifies: $m(\angle P) = \frac{1}{2}\left((360^\circ - x) - x\right) = \frac{1}{2}(360^\circ - 2x) = 180^\circ - x$ Consequently, a circumscribed angle and its intercepted minor central angle (or minor arc) are always supplementary.

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Circle Angle and Arc Theorems Reference Table

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Worked Computational Exemplar: Exterior Secant and Tangent System

Problem: From an exterior point $P$, a tangent segment $\overline{PT}$ touches circle $O$ at $T$, and a secant line passes through the circle intersecting it at near point $A$ and far point $B$. The circle is partitioned into three arcs: $\widehat{TA}$, $\widehat{AB}$, and $\widehat{BT}$. Suppose the measure of $\widehat{BT}$ is three times the measure of $\widehat{TA}$, so $m(\widehat{BT}) = 3x$, and the far secant arc $m(\widehat{AB}) = 140^\circ$. Determine the measure of parameter $x$, the measure of near arc $\widehat{TA}$, and the measure of exterior angle $\angle P$.

• Step 1: Arc Sum Constraint: A complete circle measures $360^\circ$. Summing all component arcs: $m(\widehat{TA}) + m(\widehat{AB}) + m(\widehat{BT}) = 360^\circ$ $x + 140^\circ + 3x = 360^\circ$ $4x + 140^\circ = 360^\circ \implies 4x = 220^\circ \implies x = 55^\circ$
• Step 2: Calculate Arc Measures:
  • Near intercepted arc: $m(\widehat{TA}) = x = 55^\circ$.
  • Far intercepted arc: $m(\widehat{BT}) = 3x = 3(55^\circ) = 165^\circ$.
  • Check sum: $55^\circ + 140^\circ + 165^\circ = 360^\circ$.
• Step 3: Apply Exterior Angle Formula: The tangent and secant rays intercept far arc $\widehat{BT}$ and near arc $\widehat{TA}$: $m(\angle P) = \frac{1}{2}\left(m(\widehat{BT}) - m(\widehat{TA})\right) = \frac{1}{2}(165^\circ - 55^\circ) = \frac{1}{2}(110^\circ) = 55^\circ$ Notice that in this specific configuration, $m(\angle P)$ equals $m(\widehat{TA}) = 55^\circ$.