Two circles are drawn with side PQ , PR of a triangle PQR as diameter. the circle intersects at a point S . prove that S lies on QR ?
Answers
If the sides of triangle PQR, PQ and PR are diameters of the two circle intersecting at point S then it is proved that S lies on QR.
Step-by-step explanation:
Referring to the figure attached below, let’s make certain assumptions:
- Let “O” be the centre of the circle with diameter PQ and “ O’ ” be the centre of the circle with diameter PR.
- Both the circles are given to be intersecting at point S.
- Let’s join points P and S.
We know that the angle subtended by the diameter of the circle on any point of the circle is 90°.
∴ ∠PSQ = ∠PSR = 90° …… [since it is given that PQ and PR are the diameters of circles O and O’]
Now, let add the angle PSQ and angle PSR which will give us
∠PSQ + ∠PSR = 90° + 90° = 180°
i.e., the ∠PSR and ∠PSQ are linear pair of angles
∴ We can say that QSR is a straight line
Thus, point S lies on side QR of triangle PQR.
Hence proved
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Join PS. Proof: QAP is a diameter.
∴ ∠QSP = 90° (angle in the semi circle) Similarly, ABR is a diameter. ∠PSR – 90° (angle in the semicircle) ∠QSR = ∠QSP + ∠RSP = 90 + 90 ∠QSR = 180°.
∴ ∠QSR is straight angle.
∴ QSR is a straight line.
∴ Point ‘S’ is on third side QR of ∆PQR