Question

Are you really intelligent.Can you answer this.If you answer this you will get the answer yes.

Answer 1

Here, QM is angle bisector of ∠PQR.

In any triangle, the bisector of an angle divides the opposite side in the ratio of the sides of the angle.

∴ PQ : QR = PM : MR

$\frac{PQ}{QR} = \frac{PM}{MR} \\ \\ \therefore \frac{PM^2}{MR^2} = \frac{PQ^2}{QR^2} \\ \\ PQ^2 = PS^2 + QS^2 \\ \\ QR^2 = SR^2 + QS^2 \\ \\ \therefore \frac{PM^2}{MR^2} = \frac{PS^2 + QS^2}{SR^2 + QS^2}$

Consider ΔQPS and ΔQRS.

∠PSQ = ∠RSQ = 90°

Let ∠QPS = x°

∠PQS = 90 - x°

∠RQS = x°

∠QRS = 90 - x°

∠QPS = ∠SQR = x°

∠PQS = SRQ = 90 - x°

∴ ΔQPS and ΔQRS are similar.

$\therefore \frac{PS}{QS} = \frac{QS}{SR} \\ \\ \therefore PS \times SR = QS^2 \\ \\ \\ \therefore \frac{PM^2}{MR^2} = \frac{PS^2 + (PS \times SR)}{SR^2 + (PS \times SR)} \\ \\ \frac{PM^2}{MR^2} = \frac{(PS \times PS) + (PS \times SR)}{(SR \times SR) + (PS \times SR)} \\ \\ \frac{PM^2}{MR^2} = \frac{PS(PS + SR)}{SR(SR + PS)} \\ \\ \frac{PM^2}{MR^2} = \frac{PS \times PR}{SR \times PR} \\ \\ \frac{PM^2}{MR^2} = \frac{PS}{SR}$

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