9.2: Inscribed Angles
Learning Objectives
- Find the measure of inscribed angles and the arcs they intercept.
Inscribed Angle, Intercepted Arc
An inscribed angle is an angle whose vertex is on the circle and whose sides contain chords of the circle.
- The __________________________ of an inscribed angle is on the circle.
- The sides of an inscribed angle are ________________________ of the circle.
An inscribed angle is said to intercept an arc of the circle. In the diagram above, the intercepted arc is shown between the two sides of the angle.
- The sides of an inscribed angle intercept an _________________ of the circle.
We will prove shortly that the measure of an inscribed angle is half of the measure of the arc it intercepts.
Notice that the vertex of the inscribed angle can be anywhere on the circumference of the circle; it does not need to be diametrically opposite the intercepted arc.
Measure of Inscribed Angle
The measure of a central angle is twice the measure of the inscribed angle that intercepts the same arc.
Proof
\begin{align*}\angle COB\end{align*}
We draw the diameter of the circle through points \begin{align*}A, O,\end{align*}
We see that \begin{align*}\Delta AOC\end{align*}
From this we can conclude that \begin{align*}m \angle ACO = x^\circ\end{align*}
Similarly, we can conclude that \begin{align*}m \angle ABO = y^\circ\end{align*}
We use the property that the sum of angles inside a triangle equals \begin{align*}180^\circ\end{align*}
Then, \begin{align*}m \angle COD = 180^\circ -m \angle AOC = 180^\circ -(180^\circ -2x^\circ) = 2x^\circ\end{align*}
Therefore: \begin{align*}m \angle COB = 2x^\circ + 2y^\circ = 2(x^\circ + y^\circ) = 2(m \angle CAB)\end{align*}
This proves that the measure of a central angle is two times the measure of the inscribed angle that intercepts the same arc.
Another way of looking at this is that the measure of the inscribed angle is half the measure of the __________________________ angle that intercepts the same arc.
Do you remember what a corollary is?
A corollary is a statement that follows a theorem.
If we know that a theorem is true, we can often make other statements that are true. These are called corollaries.
- A corollary is a statement that follows a ___________________________.
Inscribed Angle Corollaries
The theorem we just proved has several corollaries, which you can prove on your own:
- Inscribed angles intercepting the same arc are congruent.
- Opposite angles of an inscribed quadrilateral are supplementary.
- An angle inscribed in a semicircle is a right angle.
- An inscribed right angle intercepts a semicircle.
- A _________________________ angle is inscribed in a semicircle.
- Opposite angles of an inscribed quadrilateral are __________________________.
Example 1
Find the angle marked \begin{align*}x\end{align*}
\begin{align*}m \angle EOB\end{align*}
So \begin{align*}m \angle OAB = 35^\circ\end{align*}
This means that the angle marked \begin{align*}x = 180^\circ - 35^\circ = 145^\circ\end{align*}
Example 2
Find the angle marked \begin{align*}x\end{align*}
We know that an arc has the same measure as the central angle that intercepts it and double the measure of the inscribed angle that intercepts it.
- An arc has the _______________ measure as the central angle that intercepts it.
- An arc has ______________ the measure of the inscribed angle that intercepts it.
Therefore,
\begin{align*}m \widehat{ADC} & = 2 \cdot 95^\circ = 190^\circ\\
\text{so} \quad m \widehat{ABC} & = 360^\circ - 190^\circ = 170^\circ\\
\text{and} \quad m \widehat{AB} & = 2 \cdot 35^\circ = 70^\circ\\
\text{so} \quad m \widehat{BC} & = 170^\circ - 70^\circ = 100^\circ\end{align*}
\begin{align*}m \angle BDC\end{align*}
\begin{align*}m \angle BDC & = \frac{1}{2} ( m \widehat{BC})\\ m \angle BDC & = \frac{1}{2} ( \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}) = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\end{align*}
Therefore, \begin{align*}x = 50^\circ\end{align*}.
Example 3
Find the angles marked \begin{align*}x\end{align*} and \begin{align*}y\end{align*} in the circle:
First we use \begin{align*}\Delta ABD\end{align*} to find the measure of angle \begin{align*}x\end{align*}, since all three angles in the triangle add up to \begin{align*}180^\circ\end{align*}:
\begin{align*}x + 15^\circ + 32^\circ & = 180^\circ\\ x + 47^\circ & = 180^\circ\\ x & = 133^\circ\end{align*}
So \begin{align*}m \angle CBD = 180^\circ - 133^\circ = 47^\circ\end{align*} because it is a linear pair with \begin{align*}\angle ABD\end{align*} or \begin{align*}x\end{align*}.
\begin{align*}m \angle BCE = m \angle BDE\end{align*} because they are both inscribed angles and both intercept the same arc, \begin{align*}\widehat{EB}\end{align*}. This makes \begin{align*}m \angle BCE = 15^\circ\end{align*}.
Using \begin{align*}\Delta BFC\end{align*} to find the measure of angle \begin{align*}y\end{align*}:
\begin{align*}y + 47^\circ + 15^\circ & = 180^\circ\\ y + 62^\circ & = 180^\circ\\ y & = 118^\circ\end{align*}
Therefore, \begin{align*}x = 133^\circ\end{align*} and \begin{align*}y = 118^\circ\end{align*}.
Reading Check:
1. What was the most important thing you learned about inscribed angles in this lesson?
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2. How does a central angle relate to the inscribed angle that shares the same intercepted arc?
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