Question

In the adjoining figure, O is the centre of a circle. Chord CD is parallel to diameter AB. If ∠ABC = 25°, calculate ∠CED.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{angle}\mathrm{subtended}\mathrm{by}\mathrm{an}\mathrm{arc}\mathrm{of}\mathrm{a}\mathrm{circle}\mathrm{at}\mathrm{the}\mathrm{centre}\mathrm{is}\mathrm{double} \\ \mathrm{the}\mathrm{angle}\mathrm{subtended}\mathrm{by}\mathrm{it}\mathrm{at}\mathrm{any}\mathrm{point}\mathrm{on}\mathrm{the}\mathrm{remaining}\mathrm{part}\mathrm{of}\mathrm{the}\mathrm{circle}. \\ \mathrm{So},\angle AOC=2\angle ABC \\ \Rightarrow \angle AOC=2\times 25°=50° \\ \angle AOC=\angle DCO(\text{A}\mathrm{lternate}\mathrm{angles}) \\ \text{Now},\text{s}\mathrm{ince}OD=OC,∆DOC\mathrm{is}\mathrm{an}\mathrm{isoceles}\mathrm{triangle}. \\ \mathrm{In}∆DOC, \\ \angle DOC+\angle ODC+\angle DCO=180°(\text{A}\mathrm{ngle}\mathrm{sum}\mathrm{property}) \\ \Rightarrow \angle DOC=180°-\angle ODC-\angle ODC \\ \Rightarrow \angle DOC=180°-50°-50°(\angle ODC=\angle DCO) \\ \Rightarrow \angle DOC=80° \end{gathered}$$
and
$$\begin{gathered} \angle CED=\frac{1}{2}\angle COD \\ =\frac{1}{2}\times 80° \\ =40° \end{gathered}$$