Question

use a text box to explain your proof that an inscribed angle is half the measure of its intercepted arc. case 3: the diameter lies in the exterior of the inscribed angle

Explanation:

To prove an inscribed angle (\(\angle QSR\)) is half its intercepted arc (\(\overset{\frown}{QR}\)) for Case 3 (diameter outside the angle):

1. Let the center be \(O\), draw radii \(OS\), \(OQ\), \(OR\). Let \(\angle QOR\) be the central angle (intercepts \(\overset{\frown}{QR}\)), and \(\angle QSR\) be the inscribed angle. Let \(\angle QOS = x\) (central angle for \(\overset{\frown}{QS}\)) and \(\angle ROS = y\) (central angle for \(\overset{\frown}{RS}\)).
2. By the central angle theorem, \(m\angle QOR = m\angle QOS - m\angle ROS = x - y\) (since the diameter/center line is outside \(\angle QSR\), the larger central angle minus the smaller).
3. In \(\triangle OSQ\), \(OS = OQ\) (radii), so \(\triangle OSQ\) is isosceles. Thus, \(\angle OQS=\angle OSQ\). The exterior angle \(\angle QOS = \angle OQS + \angle OSQ = 2\angle OSQ\), so \(\angle OSQ=\frac{x}{2}\).
4. Similarly, in \(\triangle OSR\), \(OS = OR\) (radii), so \(\triangle OSR\) is isosceles. \(\angle ORS=\angle OSR\), and exterior angle \(\angle ROS = \angle ORS + \angle OSR = 2\angle OSR\), so \(\angle OSR=\frac{y}{2}\).
5. Now, \(\angle QSR=\angle OSQ - \angle OSR\) (since \(\angle OSQ\) is larger, as \(x > y\) here) \(=\frac{x}{2}-\frac{y}{2}=\frac{x - y}{2}\).
6. But \(m\angle QOR = x - y\) (central angle for \(\overset{\frown}{QR}\)), so \(m\angle QSR=\frac{1}{2}m\angle QOR\). Since the measure of a central angle equals its intercepted arc, \(m\angle QOR = m\overset{\frown}{QR}\). Thus, \(m\angle QSR=\frac{1}{2}m\overset{\frown}{QR}\), proving the inscribed angle is half its intercepted arc.

Answer:

An inscribed angle is half the measure of its intercepted arc (proven via central angles, isosceles triangles, and exterior angle theorem, as detailed above).