Chapter 10.3, Problem 37HP

(a)

To determine

To construct: $\overline{CD}$ is passes through the center.

Explanation of Solution

Given information: The centre of the circle is not on $\overline{CD}$ .

Construction:

  

• Draw a circle with chord $\overline{AB}$ .
• Use compass to construct arcs.
• Join the intersection point to get the perpendicular bisector of $\overline{AB}$ .
• Bisector intersects the circle in $C,D$ .

  

• Perform the same operation and get $\overline{FG}$ .
• $\overline{FG}$ is the intersection point of $\overline{CD}$ .
• The intersection point termed as O.

  

Assume that the center X is not on $\overline{CD}$ .

We have drawn the radii $\overline{XA},\overline{XB}$ .

Since $\overline{CD}$ is perpendicular bisector of $\overline{AB}$ ,

  $\overline{AE}\cong \overline{BE}$

By side-side-side congruence postulate,

  $\Delta XEA\cong \Delta XEB$

Corresponding elements are congruent in congruent triangles,

  $\angle XEA\cong \angle XEB$

  $\begin{array}{l}\angle XEA+\angle XEB=180\\ 2\angle XEA=180\\ \angle XEA=90\\ \therefore \overline{XE}\perp \overline{AB}\end{array}$

Therefore, the assumption is false. The centre of the circle lies on $\overline{CD}$ .

(b)

To determine

To prove: $O$ is the center of circle.

Explanation of Solution

Given information: The following figure is given.

  

Proof:

  

Point $X$ is on $\overline{CD}$ and $C,D$ belongs to the circle. So $\overline{CD}$ is a diameter. As line $FG$ bisects $\overline{CD}$ at point O. So O is the midpoint of a diameter. It means Point O is Point $X$ . Therefore, the centre of the circle is $O$ .