In the given figure, two circles intersect each other at points S and R. Their common tangent PQ touches the circle at points P, Q. Prove that, ∠ PRQ + ∠ PSQ = 180°
Chapter 3 – Circle – Text Book Solution
Problem Set 3 | Q 22 | Page 89
In the given figure, two circles intersect each other at points S and R. Their common tangent PQ touches the circle at points P, Q.
Prove that, ∠ PRQ + ∠ PSQ = 180°
It is given that two circles intersect each other at points S and R.
Join RS.
The angle between a tangent of a circle and a chord drawn from the point of contact is congruent to the angle inscribed in the arc opposite to the arc intercepted by that angle.
PQ is the tangent to the smaller circle and PR is the chord.
∴ ∠RPQ = ∠PSR .....(1)
Also, PQ is the tangent to the bigger circle and RQ is the chord.
∴ ∠RQP = ∠QSR .....(2)
Adding (1) and (2), we get
∠RPQ + ∠RQP = ∠PSR + ∠QSR
⇒ ∠RPQ + ∠RQP = ∠PSQ .....(3)
In ∆PRQ,
∠RPQ + ∠RQP + ∠PRQ = 180º .....(4) (Angle sum property)
From (3) and (4), we get
∠PSQ + ∠PRQ = 180º
Hence proved.
Explanation:-
We are given two circles that intersect each other at points S and R. We need to prove that the sum of opposite angles of a cyclic quadrilateral is 180°.
To prove this, we draw a tangent PQ to the smaller circle and a chord PR. By the angle between a tangent of a circle and a chord drawn from the point of contact, we know that the angle ∠RPQ is congruent to the angle inscribed in the arc opposite to the arc intercepted by that angle, which is ∠PSR. Hence, we have ∠RPQ = ∠PSR.
Similarly, we draw a tangent PQ to the bigger circle and a chord RQ. By the same angle between a tangent and a chord property, we have ∠RQP = ∠QSR.
Adding equations (1) and (2), we get: ∠RPQ + ∠RQP = ∠PSR + ∠QSR ∠RPQ + ∠RQP = ∠PSQ ……(3)
Next, we consider the triangle PRQ. By the angle sum property of triangles, we know that: ∠RPQ + ∠RQP + ∠PRQ = 180° ……(4)
Substituting equation (3) in (4), we get: ∠PSQ + ∠PRQ = 180°
Therefore, we have proved that the sum of opposite angles of a cyclic quadrilateral is 180°.
Chapter 3 – Circle – Text Book Solution
Problem Set 3 | Q 22 | Page 89
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