Chapter 10.6, Problem 33PPS

To determine

Prove that $m\angle CAB=\frac{1}{2}marcCDA$ and $m\angle CAE=\frac{1}{2}marcCA$

Explanation of Solution

Given :

Tangent $\overleftrightarrow{AB}$ and secant $\overrightarrow{AC}$ , $\angle CAB$ is obtuse.

Calculation:

Let the center of the circle be O and diameter FA.

Since the radius is perpendicular to the tangent, so , $\overline{OA}\perp \overline{AB}$

So, $m\angle FAE=90°$ and $marcFCA=180°\mathrm{..............}[semi-circle]$

By angle and arc addition postulate,

$\begin{array}{l}m\angle FAE=m\angle FAC+m\angle CAE\\ \Rightarrow 90°=m\angle FAC+m\angle CAE\mathrm{..............}(1)\\ marcFCA=marcFC+marcCA\\ \Rightarrow 180°=marcFC+marcCA\\ \Rightarrow 90°=\frac{1}{2}marcFC+\frac{1}{2}marcCA\mathrm{................}[\text{divide each side by 2]}\\ \Rightarrow m\angle FAC+m\angle CAE=\frac{1}{2}marcFC+\frac{1}{2}marcCA\mathrm{.....}[\text{from (1)]}\mathrm{..........}\text{(2)}\end{array}$

By Inscribed Angle Theorem, since the angle FAC is inscribed by arc FC , so,

$m\angle FAC=\frac{1}{2}marcFC$

Substitute in equation (2) :

$\begin{array}{l}m\angle FAC+m\angle CAE=\frac{1}{2}marcFC+\frac{1}{2}marcCA\\ \Rightarrow \frac{1}{2}marcFC+m\angle CAE=\frac{1}{2}marcFC+\frac{1}{2}marcCA\\ \Rightarrow m\angle CAE=\frac{1}{2}marcCA\end{array}$

Since the radius is perpendicular to the tangent, so , $\overline{OA}\perp \overline{AB}$

So, $m\angle FAB=90°$ and $marcFDA=180°\mathrm{..............}[semi-circle]$

By angle and arc addition postulate,

$\begin{array}{l}m\angle CAB=m\angle CAF+m\angle FAB\\ \Rightarrow m\angle CAB=m\angle CAF+90°\\ marcCDA=marcCF+marcFDA\\ \Rightarrow marcCDA=marcCF+180°\end{array}$

By Inscribed Angle Theorem, since the angle CAF is inscribed by arc CF , so,

$m\angle CAF=\frac{1}{2}marcCF$

$\begin{array}{l}m\angle CAB=m\angle CAF+90°\\ \Rightarrow m\angle CAB=\frac{1}{2}marcCF+90°\mathrm{...............}(3)\end{array}$

$\begin{array}{l}marcCDA=marcCF+180°\\ \Rightarrow \frac{1}{2}marcCDA=\frac{1}{2}marcCF+90°\mathrm{...........}[\text{divide each side by 2]}\\ \Rightarrow \frac{1}{2}marcCDA=m\angle CAB\mathrm{........................}[\text{from equation (3)]}\end{array}$

Hence proved.