Chapter 9.4, Problem 6WE

To determine

To Find: The arc CD for the given circle.

Answer to Problem 6WE

The measure of arc CD is 45°.

Explanation of Solution

Given information:

  

  $BC=CD\text{ and}\angle BOD={90}^{\circ }$

In the figure, circle contains a diameter AD , two chords BC and CD , point O which is a center point of circle.

OB is a radius and it is a perpendicular distance to diameter AD .

Measure of $\angle BOD={90}^{\circ }$ .

If radius OC bisects angle BOD in two equal parts.

Then $m\angle BOC=m\angle COD$

If measure of BOD is 90° then measure of $m\angle BOC=m\angle COD$ will be

  $\begin{array}{l}\angle BOD=2\angle COD\hfill \\ 90°=2\angle COD\hfill \end{array}$

Divide both sides of equation $90°=2\angle COD$ by 2 .

  $\begin{array}{l}\frac{90°}{2}=\frac{2\angle COD}{2}\\ \begin{array}{l}45°=\angle COD\hfill \\ \angle COD=45°\hfill \end{array}\end{array}$

Hence, $m\angle BOC=m\angle COD$ =45°

If $\angle COD=45°$ then $arcCD=45°$ because according to theorem, in same circle congruent central angles have a congruent arc.

Therefore, measure of $arcCD\text{ is }45°$ .