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.

$$\begin{gathered} \angle AOD=\angle BOC\left(O\mathrm{is}\mathrm{the}\mathrm{centre}\mathrm{of}\mathrm{the}\mathrm{circle}\right) \\ \angle AOD=2\angle ACD \\ \mathrm{and}\angle BOC=2\angle BDC \\ \mathrm{Since},\angle AOD=\angle BOC \\ \Rightarrow \angle ACD=\angle BDC.....\left(1\right) \\ \angle ACB=\angle ADB.....\left(2\right)\left(\mathrm{Angle}\mathrm{in}\mathrm{the}\mathrm{same}\mathrm{segment}\mathrm{are}\mathrm{equal}\right) \\ \mathrm{Adding}\left(1\right)\mathrm{and}\left(2\right) \\ \angle BCD=\angle ADC.....\left(3\right) \\ \mathrm{In}△ACD\mathrm{and}△BDC \\ CD=CD\left(\mathrm{common}\right) \\ \angle BCD=\angle ADC\left[\mathrm{Using}\left(3\right)\right] \\ AD=BC\left(\mathrm{given}\right) \\ \mathrm{Hence},△ACD\cong BDC\left(\mathrm{SAS}\mathrm{congruency}\mathrm{criterion}\right) \\ \therefore AC=BD\left(\mathrm{cpct}\right) \\ \mathrm{Hence}\mathrm{Proved} \end{gathered}$$