Question

Prove that, “Angles inscribed in the same arc are congruent.”

Hint:

All angles subtending the same arc (or equal arcs) in a circle are equal in measure.

Answer 1

Given: A circle with center $O$, and points $A$, $B$, and $C$ lying on the circle such that $\angle ACB$ and $\angle ADB$ are inscribed in the same arc $AB$. 

To Prove: $\angle ACB = \angle ADB$ 
Proof: 
Step 1: Join $O$ to $A$, $B$, $C$, and $D$. Then, $\triangle OAC$, $\triangle OBC$, $\triangle OAD$, and $\triangle OBD$ are formed. 
Step 2: $\angle AOB$ is the central angle subtending arc $AB$, and $\angle ACB$, $\angle ADB$ are inscribed angles subtending the same arc $AB$. 
Step 3: By the property of a circle, the measure of an inscribed angle is half the measure of the central angle subtending the same arc. 
\[ \angle ACB = \frac{1}{2}\angle AOB \quad \text{and} \quad \angle ADB = \frac{1}{2}\angle AOB \] Step 4: Therefore, \[ \angle ACB = \angle ADB \] Hence proved. 
Result: Angles inscribed in the same arc are congruent.