Problem
GEO-B1-M05-P018 Distances From the Centre to Equal Chords
#18
★★★☆☆ Level 3 of 5
In a circle with centre \(O\), chords \(AB\) and \(CD\) are equal. Perpendiculars from \(O\) to these chords meet them at \(M\) and \(N\). Prove that \(OM=ON\).
The perpendicular from the centre to a chord bisects the chord.
Since \(OM\perp AB\), point \(M\) is the midpoint of \(AB\). Similarly, \(N\) is the midpoint of \(CD\). From \(AB=CD\), we get \(AM=CN\). In right triangles \(OMA\) and \(ONC\), hypotenuses \(OA\) and \(OC\) are equal as radii, and legs \(AM\) and \(CN\) are equal. Thus the triangles are congruent, so \(OM=ON\).
A slightly longer problem: it requires two chord facts at once.