7.5 Other Angles
The measures of angles other than those formed by two radii are also related to the measures of the arcs they intercept on a circle. Two chords that intersect at a point on the circle form an inscribed angle. In the figure below, $\angle ABC$ is an inscribed angle and intercepts the minor arc $\overparen{AC}$. Also $\angle ABC$ is inscribed in the major arc $\overparen{ABC}$. It is important to note that the vertex of an inscribed angle is always a point on the circle.
We start with the following theorem that forms the basis for other results about angle and arc measures.
- Theorem 7.11:
The Inscribed Angle Theorem
The measure of an inscribed angle is half the measure of its intercepted arc.
We divide the proof of this theorem into three cases, because so far the only information we have connecting angle measures and arc measures relates to central angles.
Case I: The center of the circle lies on one side of the angle.
Proof: $\overline{PA}\cong \overline{PB}$ since all radii of a circle are congruent. Then $\angle 1\cong \angle 2$ by the Isosceles Triangle Theorem. By the Exterior Angle Theorem, $m\angle 3=m\angle 1+m\angle 2$, so $m\angle 1=\dfrac{1}{2}m\angle 3$. But $\angle 3$ is a central angle of the circle, so $m\angle 3=m\overparen{AC}$. By substitution, we then have $m\angle 1=\dfrac{1}{2}m\overparen{AC}$. |
Case II: The center of the circle lies in the interior of the angle.
Proof: Draw diameter
$\overline{BD}$. Then from Case I, we have $m\angle
1=\dfrac{1}{2}m\overparen{AC}$and $m\angle
2=\dfrac{1}{2}m\overparen{DC}$. By the Angle Addition Postulate,
$m\angle 1+m\angle 2=m\angle ABC$; and by the Arc Addition Postulate,
$m\overparen{AD}+m\overparen{DC}=m\overparen{AC}$. Then we have \begin{align} m\angle ABC&=m\angle 1+m\angle 2\\ &=\frac{1}{2}m\overparen{AD}+\frac{1}{2}m\overparen{DC}\\ &=\frac{1}{2}\left(m\overparen{AD}+m\overparen{DC}\right)\\ &=\frac{1}{2}m\overparen{AC} \end{align} |
Case III: The center of the circle lies in the exterior of the angle.
Proof: This situation is handled in much the same way as Case II. Draw diameter $\overline{BD}$, and use Case I with subtraction instead of addition. Again we obtain $$m\angle ABC=\frac{1}{2}m\overparen{AC}.$$ |
Example: In the figure $\overline{AB}\parallel \overline{CD}$. If $\overparen{AC}=72$, determine $m\angle B$ and $m\overparen{BD}$. |
Solution: $\angle B$ is an inscribed angle, so its measure is half the measure of its intercepted arc. Hence $m\angle B=36$. Since $\overline{AB}\parallel \overline{CD}$, $m\angle C=m\angle B$ because those are alternate interior angles. Finally, $\angle C$ is an inscribed angle intercepting $\overparen{BD}$. The measure of the intercepted arc is double the measure of the inscribed angle, so $\overparen{BD}=72$.
While we usually refer to the arc that is intercepted by an inscribed angle (rather than to the arc within which the angle is inscribed), there is one instance in which we refer to the inscribing arc. We have the following corollaries to the Inscribed Angle Theorem; their proofs are called for in the exercises.
- Corollaries:
An angle inscribed in a semicircle is a right angle.
Two inscribed angles that intercept the same arc are congruent.
The opposite angles of an inscribed quadrilateral are supplementary.
Example: Given:
Chords $\overline{AB}$ and $\overline{CD}$ intersect at point $P$
inside a circle. Prove: $\triangle APC\sim \triangle DPB$ |
Solution: Proof: $\angle A$ and $\angle D$ are inscribed angles that intercept the same arc — $\overparen{BC}$. By the second of the corollaries above, those angles are congruent. In the same way, $\angle C$ and $\angle B$ are inscribed angles that both intercept $\overparen{AD}$, so they are congruent. Then by Angle-Angle, $\triangle APC\sim \triangle DPB$.
Two chords that intersect inside a circle determine four arcs that overlap only at their endpoints. Pairs of these are used in finding the measures of the angles formed.
- Theorem 7.12: The measure of an angle formed by two chords intersecting inside a circle is half the sum of the measures of its intercepted arc and the intercepted arc of its vertical angle.
Given:
$\overline{AB}$ and $\overline{CD}$ intersect at $P$. Prove: $m\angle APD=\dfrac{1}{2}\left(m\overparen{AD}+m\overparen{BC}\right)$ |
\begin{align}
m\angle APD&=m\angle A+m\angle C\\
&=\frac{1}{2}m\overparen{BC}+\frac{1}{2}m\overparen{AD}\\
&=\frac{1}{2}\left(m\overparen{AD}+m\overparen{BC}\right)
\end{align}
Example: An angle formed by two intersecting chords is $56^\circ$, and one of the intercepted arcs is $48^\circ$. Find the degree measure of the other intercepted arc.
Solution: Let
$x=m\overparen{AD}$. Then, \begin{align} 56&=\frac{1}{2}(x+48)\\ 112&=x+48\\ x&=64 \end{align} |
We conclude this section with an extraordinary theorem due to Ptolemy of Alexandria, an astronomer and mathematician. Ptolemy used this theorem to develop a table of values of sines and cosines.
- Theorem 7.13:
Ptolemy's Theorem
If $ABCD$ is a quadrilateral inscribed in a circle, then the product of the lengths of the diagonals is equal to the sum of the products of the lengths of the two pairs of opposite sides. That is:$$AC\cdot BD=AB\cdot CD+BC\cdot AD.$$
The proof is outlined in the exercises.