A chord PQ of a circle is parallel to the tangent drawn at a point R of the circle. Prove that R bisects the arc PRQ
Solution:
Given, a chord PQ of a circle is parallel to the tangent drawn at a point R of the circle.
We have to prove that R bisects the arc PRQ.
From the figure,
MN is the tangent to the circle at the point R.
PQ is the chord
Given, PQ || MN
The alternate interior angles are equal.
i.e., ∠MRP = ∠RPQ
∠1 = ∠2 ------------------ (a)
We know that the angle between the tangent and the chord of a circle is equal to the angle made by the chord in the alternate segment.
So, ∠MRP = ∠RQP
∠1 = ∠3 ----------------- (b)
From (a) and (b),
∠2 = ∠3
∠MRP = ∠RQP
We know that the sides equal to the opposite angles are equal.
So, PR = QR.
This implies that R bisect PQ.
Therefore, R bisects the arc PRQ.
✦ Try This: In the figure, PQ is a chord of a circle with centre O and PT is the tangent. If angle QPT = 60°, find angle PRQ.
Given, PQ is a chord of the circle with centre O.
PT is the tangent to the circle.
Also, ∠QPT = 60°
We have to find the measure of the angle PRQ.
Consider a point X on the tangent PT.
∠QPT + ∠QPX = 180°
60° + ∠QPX = 180°
∠QPX = 180° - 60°
∠QPX = 120°
The alternate segment theorem, also known as the tangent-chord theorem states that in any circle, the angle between a chord and a tangent through one of the end points of the chord is equal to the angle in the alternate segment.
∠QPX = ∠PRQ
Therefore, ∠PRQ = 120°
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 10
NCERT Exemplar Class 10 Maths Exercise 9.3 Problem 8
A chord PQ of a circle is parallel to the tangent drawn at a point R of the circle. Prove that R bisects the arc PRQ
Summary:
A chord PQ of a circle is parallel to the tangent drawn at a point R of the circle. It is proven that R bisects the arc PRQ
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