Chapter 9.5, Problem 13CE

To determine

To prove:Case III of theorem 9.7 that states that the measure of an inscribed angle is equal to half the measure of its intercepted arc.

Answer to Problem 13CE

Below proof shows the case III of theorem 9.7, in which point O lies outside $\angle ABC$ .

Explanation of Solution

Giveninformation: In below given figure,a circle is given in which $\angle ABC$ is inscribed so that point O lies outside this angle.

  

Concept used:Use following properties of a triangle,

(i) In an isosceles triangle, angles opposite to its equal opposite sides are also equal.

(ii) In a triangle, an exterior angle is the sum of its two opposite interior angles.

To prove: $m\angle ABC=\frac{1}{2}m(arcAC)$

Key steps of proof of case III:

1. Draw radius $\overline{OA}$ . Join points A,B and C from center O . Let $m\angle ABC=x$ .
2. Using above concept (i), $m\angle BAO=x$ .
3. Using above concept (ii), $m\angle AOC=2x$ .
4. As an arc subtends angle at the center, equal to its own angle measure, so, angle measure of arc AC, corresponding to angle $m\angle AOC$ is also $2x$degree.
5. By substitution property, $m\angle ABC=\frac{1}{2}m(\text{arc }AC)$ .

Conclusion: Thus,It proofs that in a circle, even if point O lies outside $\angle ABC$ ,the measure of an inscribed angle is equal to half the measure of its intercepted arc.