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In an equilateral triangle PQR, prove that PS^{2} = 3(QS)^{2}.

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#### Solution

**Given:** PS is the altitude of ΔPQR.

In ΔPSQ and ΔPSR,

∠PSQ ≅ ∠PSR ......[Each angle is equal to 90°]

PS ≅ SP ......[Common side]

PQ ≅ PR ......[Sides of an equilateral triangle]

By R.H.S. criterion of congruence,

ΔPSQ ≅ ΔPSR

∴ QS ≅ SR ......[C.S.C.T.]

Now, QS + SR = QR

QS + QS = QR .......[∵ SR = QS]

2QS = QR

QS = `(QR)/2` ......(i)

In right-angled triangle PQS, by Pythagoras theorem,

PS^{2} + QS^{2} = PQ^{2}

PS^{2} + QS^{2} = QR^{2} ......[∵ PQ = QR]

PS^{2} = QR^{2} – QS^{2}

ps^{2} = (2QS)^{2} – QS^{2} ......[∵ QR = 2QS]

PS^{2} = 4QS^{2} – QS^{2}

PS^{2} = 3QS^{2}

Hence proved.

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