In this video we will learn about properties related to chords.
A chord is a point joining two points on the circumference of a circle.
Equal chords are equidistant from the center
To illustrate this we take a circle with center O. AB and CD are two chords on this circle which are equal in length. E is a point on chord AB while F is a point on chord CD. We have to prove that OE = OF. We join the points A, B and C, D to the center O, thus forming ΔAOB and ΔCOD. To prove OE = OF we need to prove ΔAOB ≅ ΔCOD. On observing carefully we will find that AO, BO, CO and DO are all radius of the circle. Thus, AO = CO and BO = DO. It is already given that, AB = CD. So, ΔAOB ≅ ΔCOD by the Side-Side-Side Congruency rule. Thus, OE = OF since in congruent triangles, the angles and the altitudes are of equal measure.
Conversely, if two chords are equidistant from the center, they must be equal.
Equal chords subtend equal angles at the center
We say that an angle is subtended at the center when it is angled at the center of the circle. For example, we have taken a circle with center as O. AB is a chord on this circle and subtends angle AOB at the centre. We know that equal chords are equidistant from the centre. Using the example in the previous property we can say that the angles of ΔAOB and ΔCOD will also be equal to each other. Thus, ∠AOB = ∠COD.
Note that in the previous example, arc AB will be equal to arc CD and the area between the arc and the chords will also be equal due to the congruency of triangles.
Conversely, if the angles subtended at the center are equal, the chords which subtend them must be equal.
Bisection of a chord
A perpendicular drawn from the center of the circle to a chord bisects the chord. To illustrate this we take a circle with center as O. AB is a chord and OD is a perpendicular drawn on AB. Points A and B are connected to O thus forming ΔAOD and ΔDOB. To prove that OD bisects AB, we need to prove ΔAOD ≅ ΔDOB.
In ΔAOD and ΔDOB,
AO = OB (radius of the circle)
OD = OD (OD is a common line)
∠ADO = ∠BDO ( angle formed by perpendicular bisector)
Thus, ΔAOD ≅ ΔDOB by the RHS rule. Having proven them congruent we can say that their third side will also be equal to each other.
Therefore, AD = DB.
Conversely if a line passing through the center bisects the chord of a circle, it must be a perpendicular to it.