Question

Prove that equal chords of a circle are equidistant from the centre.

Answer 1

Given: A circle (O,r) in which chord AB = chord CD, OL ⊥ AB and OM ⊥ CD.
To Prove: OL = OM
Construction: Join OA and OC.

Proof: We know that the perpendicular from the centre of a circle to a chord bisects the chord.
∴ $\mathrm{AL}=\frac{1}{2}\mathrm{AB}\mathrm{and}\mathrm{CM}=\frac{1}{2}\mathrm{CD}$
Now, AB = CD
$$\begin{gathered} \Rightarrow \frac{1}{2}\mathrm{AB}=\frac{1}{2}\mathrm{CD} \\ \Rightarrow \mathrm{AL}=\mathrm{CM} \end{gathered}$$ ...(i)
Now, in the right angled ΔOLA and ΔOMC, we have:
AL = CM [From (i)]
OA = OC [Each equal to r]
∴ ΔOLA ≅ ΔOMC [By RHS congruency]
i.e., OL = OM
Hence, AB and CD are equidistant from O.