Question

In fig 4.42 P is the centre of a circle E is the point on the diameter RS. Two chords AB and CD intersect the diameter in E such that $\angle$AEP $\cong$ $\angle$DEP shows that seg AB $\cong$ seg CD.

Answer 1

$$\begin{gathered} \mathrm{Draw}\mathrm{OM}\perp \mathrm{AB}\mathrm{and}\mathrm{ON}\perp \mathrm{CD}. \\ \mathrm{Now},\mathrm{in}△\mathrm{ONE}\mathrm{and}△\mathrm{OME}, \\ \angle \mathrm{ONE}=\angle \mathrm{OME}\left(\text{b}\mathrm{oth}\mathrm{equal}\mathrm{to}90°\right) \\ \angle \mathrm{NEO}=\angle \mathrm{MEO}\left(\text{g}\mathrm{iven}\right) \\ \mathrm{OE}=\mathrm{OE}\left(\mathrm{common}\mathrm{side}\right) \\ \therefore △\mathrm{ONE}\cong △\mathrm{OME}\left(\mathrm{by}\mathrm{AAS}\mathrm{congruency}\mathrm{rule}\right) \\ \therefore \mathrm{ON}=\mathrm{OM}\left(\mathrm{by}\mathrm{cpct}\right) \\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{two}\mathrm{chords}\mathrm{are}\mathrm{equal}\mathrm{if}\mathrm{they}\mathrm{are}\mathrm{equidistan}\text{t}\mathrm{from}\mathrm{the}\mathrm{centre}. \\ \therefore \mathrm{seg}\mathrm{AB}\hspace{0.17em}=\mathrm{seg}\mathrm{CD} \end{gathered}$$