Question

In the given fig PQR is a triangle where vertices R lies on the circumference of the semicircle with diameter PQ and centre O.
PRS and QS are the lone segments.
Prove that PR×PS+QT×QR=PQ^2

Answer 1

All Vertex lie on Circumference in Cyclic Quadrilateral

Step-by-step explanation:

In a Cyclic Quadrilateral, Opposite angles sum up 180°

PRTQ is a cyclic quadrilateral with $\angle RTS = \angle RPQ$

In triangle PSQ and triangle TSR

$\frac {PS}{TS} = \frac {QS}{RS}$

$\frac {PS}{QS-QT} = \frac {QS}{PS-PR}$

$PS^2 - PR.PS = QS^2 - QT \times QS$

$QS^2 - PS^2 = QT \times QS - PR \times PS$ """(1)

$PQ^2 = PR^2 + QR^2$

= $PR^2 + (QS^2 - RS^2)$

= $PR^2 + QS^2 - (PS-PR)^2$

= $PR^2 + QS^2 - PS^2 - PR^2 + 2PR \times PS$

= $QS^2 - PS^2 + 2PR \times PS$ """(2)

From (1) & (2) we get :

$PQ^2 = QT \times QS - PR \times PS + 2PR \times PS$

= $PR \times PS + QT \times QS$ Proved !

Answer 2

Answer:

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