A straight line through point P of a triangle PQR intersects the side QR at the point S and the circumcircle of the triangle PQR at the point T. lf S is not the centre of the circumcircle, then which of the following is true?
Using properties of secant, $$PS \times ST = QS \times SR$$ -------------Eqn(I)
Also, for two numbers, $$PS$$ and $$ST$$, we know that harmonic mean is less than geometric mean.
=> $$\frac{2}{\frac{1}{PS} + \frac{1}{ST}} < \sqrt{PS \times ST}$$
=> $$\frac{1}{PS} + \frac{1}{ST} > \frac{2}{\sqrt{PS \times ST}}$$
Using Eqn(I)
=> $$\frac{1}{PS} + \frac{1}{ST} > \frac{2}{\sqrt{QS \times SR}}$$ ------Eqn(II)
Also, for two numbers, $$QS$$ and $$SR$$, geometric mean is less than arithmetic mean.
=> $$\sqrt{QS \times SR} < \frac{QS + SR}{2}$$
=> $$\frac{1}{\sqrt{QS \times SR}} > \frac{2}{QR}$$Â $$(\because QS + SR = QR)$$
Multiplying both sides by $$2$$
=> $$\frac{2}{\sqrt{QS \times SR}} > \frac{4}{QR}$$ ----------Eqn(III)
From eqn(II) and (III)
$$\therefore \frac{1}{PS} + \frac{1}{ST} > \frac{4}{QR}$$