Question

If the two sides of a pair of opposite sides of a cyclic quadrilateral are equal, prove that its diagonals are equal.

 

Answer 1

         

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