In figure 3.100, 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^° Figure: P is on the first circle and Q on the second; R is the intersection point nearer to the tangent PQ and S the farther one. Segments PR, QR, PS, QS and RS are drawn.

In figure 3.100, 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°

In figure 3.100, two circles intersect each other at points S and R. Their common tangent PQ touches the circle at points P, Q. Prove that, \(\angle PRQ + \angle PSQ = 180^\circ\)

Figure: P is on the first circle and Q on the second; R is the intersection point nearer to the tangent PQ and S the farther one. Segments PR, QR, PS, QS and RS are drawn.

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Answer: By the tangent–secant angle theorem \(\angle RPQ = \angle PSR\) and \(\angle RQP = \angle QSR\); adding and using the angle sum of \(\triangle PRQ\) gives the result.

In the first circle, \(\angle RPQ\) is the angle between tangent PQ and chord PR, so it equals the inscribed angle on chord PR in the alternate segment:

\[\angle RPQ = \angle PSR\quad \text{(I)}\]

Similarly in the second circle,

\[\angle RQP = \angle QSR\quad \text{(II)}\]

Adding (I) and (II):

\[\angle RPQ + \angle RQP = \angle PSR + \angle QSR = \angle PSQ\]

In \(\triangle PRQ\),

\[\angle PRQ + \angle RPQ + \angle RQP = 180^\circ\] \[\therefore \angle PRQ + \angle PSQ = 180^\circ\]

Hence proved.

Doubtly AI · 24 Oct 2026 AI-verified · Oct 2026 Report

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In figure 3.100, 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°

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