Question
Prove that the angle in a segment shorter than a semicircle is greater than a right angle.
✓
Answer
Given: $\angle\text{ACB}$ is an angle in minor segment.
To prove: $\angle\text{ACB}>90^\circ$
Proof: By degree measure theorem Reflex
$\angle\text{AOB}=2\angle\text{ACB}$ And reflex
$\angle\text{AOB}>180^\circ$ Then, $2\angle\text{ACB}>180^\circ$
$\Rightarrow\angle\text{ACB}>\frac{180^\circ}{2}$
$\Rightarrow\angle\text{ACB}>90^\circ$
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