Question

Two congruent circles with centers $\mathrm{O}$ and $\mathrm{O}\text{'}$ intersect at two points $\mathrm{A}$ and $\mathrm{B}$. Then $\angle \mathrm{AOB}=\angle \mathrm{AO}\text{'}\mathrm{B}$. Write True or False and justify your answer in each of the following.

• A. True
• B. False

Answer 1

The correct option is A True

Chord of a circle:

In a circle, if the endpoints of a line segment touch, it is called a chord of the circle.

Congruent circles:

If the radii of two or more circles are the same, it is a congruent circle.

Side-angle-side property:

In a triangle, any two sides and the angle between them are congruent is called the side-angle-side property.

Corresponding parts of congruent triangles property:

If two or more triangles are congruent to each other, then the corresponding angles and the sides of the triangles are also congruent to each other.

Proving $\mathbf{\angle }\mathbf{AOB}\mathbf{=}\mathbf{\angle }\mathbf{AO}\mathbf{\text{'}}\mathbf{B}$:

The center of the two circles is $\mathrm{O}$ and $\mathrm{O}\text{'}$.

Joining $\mathrm{AB},\mathrm{OA},\mathrm{OB},\mathrm{O}\text{'}\mathrm{A}$and $\mathrm{BO}\text{'}$ it forms two triangles $△\mathrm{AOB}$ and $△\mathrm{AO}\text{'}\mathrm{B}$.

Since the two circles are congruent the radii of the two circles is same.

In $△\mathrm{AOB}$ and $△\mathrm{AO}\text{'}\mathrm{B}$

$\mathrm{OA}={\mathrm{AO}}^{\text{'}}$ and $\mathrm{OB}=\mathrm{BO}\text{'}$

$\mathrm{AO}=\mathrm{OB}$ and $\mathrm{AO}\text{'}=\mathrm{O}\text{'}\mathrm{B}$

$\mathrm{AB}$ is a common chord between the two circles.

$△\mathrm{AOB}\cong △\mathrm{AO}\text{'}\mathrm{B}$ (By using the side-angle-side property)

$\angle \mathrm{AOB}=\angle \mathrm{AO}\text{'}\mathrm{B}$ (By using corresponding parts of congruent triangles property)

Hence, the given statement is true.