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प्रश्न
If the two sides of a pair of opposite sides of a cyclic quadrilateral are equal, prove that its diagonals are equal.
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उत्तर
To prove: AC = BD
Proof: We know that equal chords subtend equal at the centre of circle and the angle subtended by a chord at the centre is twice the angle subtended by it at remaining part of the circle.
\[\angle AOD = \angle BOC \left( \text{ O is the centre of the circle } \right)\]
\[\angle AOD = 2\angle ACD \]
\[\text{ and } \angle BOC = 2\angle BDC\]
\[\text{ Since, } \angle AOD = \angle BOC\]
\[ \Rightarrow \angle ACD = \angle BDC . . . . . \left( 1 \right) \]
\[\angle ACB = \angle ADB . . . . . \left( 2 \right) \left( \text{ Angle in the same segment are equal } \right)\]
\[\text{ Adding } \left( 1 \right) \text{ and } \left( 2 \right)\]
\[\angle BCD = \angle ADC . . . . . \left( 3 \right)\]
\[\text{ In } \bigtriangleup ACD \text{ and } \bigtriangleup BDC\]
\[CD = CD \left( \text{ common } \right)\]
\[\angle BCD = \angle ADC \left[ \text{ Using } \left( 3 \right) \right]\]
\[AD = BC \left( given \right)\]
\[\text{ Hence } , \bigtriangleup ACD \cong BDC \left( \text{ SAS congruency criterion } \right)\]
\[ \therefore AC = BD \left( \text{ cpct } \right)\]
Hence Proved
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