Angles Subtended by a Chord at a Point

If in a circle AB is the chord and is making  \angleACB at any point of the circle then this is the angle subtended by the chord AB at a point C.

 \angleAOB is the angle subtended by chord AB at point O i.e. at the centre and  \angleADB is also the angle subtended by AB at point D on the circle.

Theorem 1: Any two equal chords of a circle subtend equal angles at the centre.

Here in the circle, the two chords are given and PQ = RS with centre O.

So OP = OS = OQ = OR (all are radii of the circle)

\mathrm{\triangle POQ\cong\triangle SOR}

\mathrm{\angle POQ=\angle SOR}

This shows that the angles subtended by equal chords to the centre are also equal.

Theorem 2: If the angles made by the chords of a circle at the centre are equal, then the chords must be equal.

Theorem 3: If we draw a perpendicular from the centre of a circle to any chord then it bisects the chord.

If we draw a perpendicular from the centre to the chord of the circle then it will bisect the chord. And the bisector will make 90° angle to the chord.

Theorem 4: The line which is drawn from the centre of a circle to bisect a chord must be perpendicular to the chord.

If we draw a line OB from the centre of the circle O to the midpoint of the chord AC i.e. B, then OB is the perpendicular to the chord AB.

If we join OA and OC, then

In ∆OBA and ∆OBC,

AB = BC (B is the midpoint of AC)

OA = OC (Both are the radii of the same circle)

OB = OB (same side)

Hence, \mathrm{\triangle OBA\cong\triangle OBC} (both are congruent by SSS congruence rule)

\mathrm{\Rightarrow \angle OBA=\angle OBC} (respective angles of congruent triangles)

\mathrm{\angle OBA+\angle OBC=\angle ABC=180^o} [Linear pair]

\mathrm{\angle OBC+\angle OBC=180^o}  [Since \mathrm{\angle OBA=\angle OBC} ]

\mathrm{2\times\angle OBC=180^o}

\mathrm{\angle OBC=90^o}

\mathrm{\angle OBC=\angle OBA=90^o}

\mathrm{\therefore OB\bot AC}

Theorem 5: There is one and only one circle passing through three given non-collinear points.

In this figure, we have three non-collinear points A, B and C. Let us join AB and BC and then make the perpendicular bisector of both so that RS and PQ the perpendicular bisector of AB and BC respectively meet each other at Point O.

Now take the O as centre and OA as the radius draw the circle which passes through the three points A, B and C.

This circle is known as Circumcircle. Its centre and radius are known as the Circumcenter and Circumradius.

Theorem 6: Two equal chords of a circle are at equal distance from the centre.

AB and CD are the two equal chords in the circle. If we draw the perpendicular bisector of these chords then the line segment from the centre to the chord is the distance of the chord from the centre.

If the chords are of equal size then their distance from the centre will also be equal.

Theorem 7: Chords at equal distance from the centre of a circle are also equal in length. This is the reverse of the above theorem which says that if the distance between the centre and the chords are equal then they must be of equal length.

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