The perpendicular from the centre of a circle to a chord bisects the chord.
A circle has a radius of 5 cm. The chord EF is 7 cm.
How far is the midpoint of the chord from the centre of the circle?
The two radii OE and OF are the hypotenuses of 2 right-angled triangles.
The distance FM is half of the length of the chord. (The perpendicular from the centre of a circle to a chord bisects the chord.)
FM = 3.5 cm
Use Pythagoras’ theorem to calculate the length OM.
OF2 = FM2 + OM2
52 = 3.52 + OM2
OM2 = 52 – 3.52
OM2 = 12.75
OM = 3.6 cm (to 1 decimal place)
In the diagram below, AB is the chord of a circle with centre O.
OM is the perpendicular from the centre to the chord.
Look at triangles OAM and OBM.
The hypotenuses (OA and OB) are the same, as they are both the radius of the circle. OM is common to both triangles. OMA and OMB are both right angles.
Triangles OAM and OBM are congruent (RHS), so it follows that AM = MB.
Therefore, M is the midpoint of AB, and the chord has been bisected.