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Question
Prove that the angle in a segment greater than a semi-circle is less than a right angle.
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Solution

\[\text{ To prove } : \angle ABC \text{ is an acute angle } \]
\[\text{ Proof } : \]
\[\text{ AD being the diameter of the given circle } \]
\[ \Rightarrow \angle ACD = 90° \left[ \text{ Angle in a semicircle is a right angle } \right]\]
\[\text{ Now, in } \bigtriangleup ACD, \angle ACD = 90° \text{ which means that } \angle ADC \text{ is an acute angle } . . . . . . \left( 1 \right)\]
\[\text{ Again, } \angle ABC = \angle ADC \left[ \text{ Angle in a same segment are always equal } \right]\]
\[ \Rightarrow \angle ABC \text{ is also an acute angle } . \left[ \text{ Using } \left( 1 \right) \right]\]
\[\text{ Hence proved } \]
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