Question

In the given figure, ABCD is a cyclic quadrilateral with O as the centre of the circle. If ∠BOD = 136°, find ∠BCD.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{angle}\mathrm{subtended}\mathrm{by}\mathrm{an}\mathrm{arc}\mathrm{of}\mathrm{a}\mathrm{circle}\mathrm{at}\mathrm{the}\mathrm{centre}\mathrm{is} \\ \mathrm{double}\mathrm{the}\mathrm{angle}\mathrm{subtended}\mathrm{by}\mathrm{it}\mathrm{at}\mathrm{any}\mathrm{point}\text{in}\mathrm{the}\mathrm{remaining}\mathrm{part}\mathrm{of}\mathrm{the}\mathrm{circle}. \\ \mathrm{So},\angle BAD=\frac{1}{2}\angle BOD \\ \Rightarrow \angle BAD=\frac{1}{2}\times 136°=68° \end{gathered}$$
$$\begin{gathered} \angle BCD+\angle BAD=180°(ABCD\mathrm{is}\mathrm{a}\mathrm{cyclic}\mathrm{quadrilateral}) \\ \Rightarrow \angle BCD=180°-\angle BAD \\ =180°-68° \\ =112° \end{gathered}$$