Question

Prove that congruent chords of a circle are equidistant from the centers.

Answer 1

Consider a circle with centre O.
It has two equal chords AB and CD.
Draw perpendiculars OL and OM from centre O on the chords AB and CD, respectively,
Join OB and OC.
We have to prove that OL = OM.
Proof:
$$\begin{gathered} \mathrm{In}\mathrm{right}△\mathrm{OLB}\mathrm{and}△\mathrm{OMC}, \\ \mathrm{LB}\hspace{0.17em}=\mathrm{MC}\left(s\mathrm{ince}\mathrm{AB}\hspace{0.17em}=\mathrm{CD}\mathrm{and}\mathrm{perpendicular}\mathrm{from}\mathrm{centre}\mathrm{bisects}\mathrm{the}\mathrm{respective}\mathrm{chord}\right) \\ \angle \mathrm{OLB}=\angle \mathrm{OMC}\left(b\mathrm{oth}\mathrm{equal}\mathrm{to}90°\right) \\ \mathrm{OB}=\mathrm{OC}\left(\mathrm{radii}\mathrm{of}\mathrm{the}\mathrm{same}\mathrm{circle}\right) \\ \therefore △\mathrm{OLB}\cong △\mathrm{OMC}\left(\text{by}\mathrm{RHS}\mathrm{congruency}\mathrm{rule}\right) \\ \therefore \mathrm{OL}=\mathrm{OM}\left(\text{by}\mathrm{CPCT}\right) \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$