Question

In the given figure, PQR is a triangle where vertices R lies on the circumference of the semi-circle with diameter
PQ and centre O. PRS and QS are the line segments. Prove that PR x PS + QT x QS PQ?.

Answer 1

In a Cyclic Quadrilateral, Opposite angles sum up 180°

Step-by-step explanation:

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 !