Question

Suppose AB and CD are equal chords of a circle and a line parallel to the tangent at A intersects the chords at D and E. Prove that AD = AE.

Answer 1

$$\begin{gathered} \mathrm{QR}\parallel \mathrm{DE}\mathrm{and}\mathrm{OA}\perp \mathrm{QR}\mathrm{at}\mathrm{A}. \\ \mathrm{Hence},\angle \mathrm{DPA}=\angle \mathrm{PAR}=90°(\mathrm{Alternate}\mathrm{angles}) \\ \mathrm{i}.\mathrm{e}.,\mathrm{OP}\perp \mathrm{DE} \\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{a}\mathrm{perpendicular}\mathrm{line}\mathrm{from}\mathrm{the}\mathrm{centre}\mathrm{to}\mathrm{the}\mathrm{chord}\mathrm{bisects}\mathrm{the}\mathrm{chord}. \\ \mathrm{So},\mathrm{PD}=\mathrm{PE}...\left(\mathrm{i}\right) \\ \mathrm{In}∆\mathrm{APD}\mathrm{and}∆\mathrm{APE},\mathrm{we}\mathrm{have}: \\ \mathrm{PD}=\mathrm{PE}\left[\mathrm{From}\mathrm{equation}\left(\mathrm{i}\right)\right] \\ \mathrm{AP}=\mathrm{AP}\left(\mathrm{Common}\right) \\ \angle \mathrm{APD}=\angle \mathrm{APE}=90° \\ \mathrm{i}.\mathrm{e}.,∆\mathrm{APD}\cong ∆\mathrm{APE}(\mathrm{By}\mathrm{SAS}\mathrm{congruency}) \\ \mathrm{Hence},\mathrm{AD}=\mathrm{AE}\left(\mathrm{CPCT}\right) \end{gathered}$$