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Theorems on Midpoints and Perpendicular Bisectors of Chords:
Theorem: The line joining the centre of a circle and the midpoint of a chord of the circle is perpendicular to the chord.
Explanation:
The theorem states that if \(O\) is the centre and \(R\) is the mid-point of the chord \(PQ\), the line joining the centre \(O\) and the mid-point \(R\) is perpendicular to the chord \(PQ\).
Proof of the theorem:
Consider a circle with centre \(O\) and a chord \(PQ\) as follows:
Draw a line \(OR\) joining the centre and the mid-point \(R\) of the chord \(PQ\).
Join the points \(OP\) and \(OQ\).
Here, \(OP\) and \(OQ\) are radius. So, they are equal.
Since \(R\) is the mid-point, the sides \(PR = RQ\).
Also, the side \(OR\) is common to the triangle \(ORP\) and \(ORQ\).
Therefore, by the SSS rule (If three sides of one triangle are equal to the three sides of another triangle, then the two triangles are congruent), the triangles \(ORP\) and \(ORQ\) are congruent.
By the congruency the corresponding parts of the triangles are equal. So, \(\angle ORP = \angle ORQ\).
It is observed that \(\angle ORP\) and \(\angle ORQ\) are linear pairs.
This implies:
\(\angle ORP\) \(+\) \(\angle ORQ\) \(=\) \(180^{\circ}\)
\(2 \angle ORP\) \(=\) \(180^{\circ}\)
\(\angle ORP\) \(=\) \(\frac{180^{\circ}}{2}\)
\(=\) \(90^{\circ}\)
Since \(\angle ORP = \angle ORQ = 90^{\circ}\) we say that \(OR \perp PQ\).
Therefore, the line drawn through the centre of the circle to bisect a chord is perpendicular to the chord.
Theorem: The perpendicular from the centre of a circle to a chord of a circle bisects the chord.
Explanation:
The theorem states that if \(O\) is the centre and \(AB\) is the chord, then the perpendicular \(OC\) from the centre \(O\) to the chord \(AB\) bisects the chord \(AB\) (i.e.) \(CA = CB\).
Proof of the theorem:
Consider a circle with centre \(O\) and a chord \(AB\).
Let \(OC\) be drawn perpendicular from the centre of a circle to the chord \(AB\).
Join the points \(OA\) and \(OB\).
Consider the \(\Delta OAC\) and \(\Delta OBC\).
The sides \(OA\) and \(OB\) are radii of the circle. So, they are equal.
That is, \(OA\) \(=\) \(OB\).
Also, the side \(OC\) is common to the triangle \(OCA\) and \(OCB\).
Given, \(OC \perp AB\).
So, \(\angle OCA = \angle OCB = 90^{\circ}\).
Therefore, \(\Delta OAC\) \(≅\) \(\Delta OBC\) [By RHS congruence]
By the congruency the corresponding parts of the triangles are equal.
So, \(CA = CB\).
This implies, \(C\) is the midpoint of the chord \(AB\).
Therefore, perpendicular from the centre of a circle to a chord of the circle bisects the chord.
Let \(OC\) be drawn perpendicular from the centre of a circle to the chord \(AB\).
Join the points \(OA\) and \(OB\).
Consider the \(\Delta OAC\) and \(\Delta OBC\).
The sides \(OA\) and \(OB\) are radii of the circle. So, they are equal.
That is, \(OA\) \(=\) \(OB\).
Also, the side \(OC\) is common to the triangle \(OCA\) and \(OCB\).
Given, \(OC \perp AB\).
So, \(\angle OCA = \angle OCB = 90^{\circ}\).
Therefore, \(\Delta OAC\) \(≅\) \(\Delta OBC\) [By RHS congruence]
By the congruency the corresponding parts of the triangles are equal.
So, \(CA = CB\).
This implies, \(C\) is the midpoint of the chord \(AB\).
Therefore, perpendicular from the centre of a circle to a chord of the circle bisects the chord.
Distance of Chords from the Centre:
Theorem: Chords of a circle having the same length are all at the same distance from the centre of the circle.
Proof:
\(CH\) is perpendicular to \(RS\).
\(CE = CH\)
To prove: \(PQ = RS\)
Proof:
Consider a circle with centre \(O\) and two equal chords \(PQ\) and \(RS\).
Join the endpoints of the chords \(PQ\) and \(RS\) with the centre.
Draw a line \(OA\) perpendicular to \(RS\) and \(OB\) perpendicular to \(PQ\).
Method - 1:
Consider the triangles \(POQ\) and \(SOR\).
Here, \(OP\) and \(OR\) are radii of the circle, so \(OP = OR\).
Similarly, \(OQ\) and \(OS\) are radii of the circle, so \(OQ = OS\).
Given that, \(PQ = RS\).
So, by SSS (Side-Side-Side) congruence, the \(\Delta POQ ≅ \Delta SOR\)
Thus, the altitudes \(OA\) and \(OB\) are congruent and \(OA = OB\).
Therefore, the chords \(PQ\) and \(RS\) are at equal distance from the centre \(O\).
Method - 2:
Two right triangles \(OAR\) and \(OBQ\) are obtained where \(\angle OAR = \angle OBQ = 90^{\circ}\) since \(OA \perp RS\) and \(OB \perp PQ\).
Here, \(OR\) and \(OQ\) are radius. So, they are equal.
According to the theorem, the perpendicular from the centre of the circle bisects the chord. So, \(RA = BQ\) since the chords are equal.
Therefore, by the RHS rule(In two right-angled triangles, if the length of the hypotenuse and one side of one triangle is equal to the length of the hypotenuse and corresponding side of the other triangle, then the two triangles are congruent), the triangles are congruent.
This implies that the sides \(OA = OB\).
Therefore, the chords \(PQ\) and \(RS\) are at equal distance from the centre \(O\).
Theorem: Chords of a circle that are equidistant from the centre have equal length.
Given:
\(CE\) is perpendicular to \(PQ\).
\(CE\) is perpendicular to \(PQ\).
\(CH\) is perpendicular to \(RS\).
\(CE = CH\)
To prove: \(PQ = RS\)
Proof:
Consider the right angled triangles \(CEP\) and \(CHR\).
\(CP = CR\) [Radii of the circle]
\(CE = CH\) [Given]
Also, \(\angle CEP = \angle CHR = 90^{\circ}\)
So, \(\Delta CEP ≅ \Delta CHR\) [By RHS congruence]
Thus, \(PE = HR\) ......(1) [By CPCT]
By the theorem, the perpendicular from the centre to a chord bisects the chord.
\(CP = CR\) [Radii of the circle]
\(CE = CH\) [Given]
Also, \(\angle CEP = \angle CHR = 90^{\circ}\)
So, \(\Delta CEP ≅ \Delta CHR\) [By RHS congruence]
Thus, \(PE = HR\) ......(1) [By CPCT]
By the theorem, the perpendicular from the centre to a chord bisects the chord.
We have, \(PE = EQ\) and \(HR = HS\).
Therefore, \(PQ = 2PE\) and \(RS = 2HR\)
From equation (1), we get \(PQ = RS\)
Hence, proved.
Therefore, \(PQ = 2PE\) and \(RS = 2HR\)
From equation (1), we get \(PQ = RS\)
Hence, proved.
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