Chapter 9.6, Problem 26WE

To determine

To prove: that $m\angle 1=\frac{1}{2}\left(x°-y°\right)$ .

Explanation of Solution

Consider the given circle with center at $0,AB$ is a secant to the circle and $BC$ is tangent to the circle.

Assume that measure of arc $AC$ is $x°$ and arc $CD$ is $y°$ .

  

Also it is given that $BC$ is tangent and $AB$ is secant.

Prove that $\angle ABC=\frac{1}{2}\left(x°-y°\right)$ as follows:

Proof:

Draw Chord   $BC$

$m\angle 3=m\angle 1+m\angle 2$

$m\angle 1=m\angle 3-m\angle 2$

$m\angle 3=\frac{1}{2}x°$

$m\angle 2=\frac{1}{2}y°$

$m\angle 1=\frac{1}{2}x°-\frac{1}{2}y°$

Therefore, it can be concluded that

  $\begin{array}{c}m\angle 1=\frac{1}{2}x°-\frac{1}{2}y°\\ m\angle 1=\frac{1}{2}\left(x°-y°\right)\end{array}$

Hence, proved.