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Question

In the given figure seg PS is the median of ∆PQR and PT ⊥ QR. Prove that,

(1) PR2=PS2+QR×ST+QR22

(2) PQ2=PS2-QR×ST+QR22

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Solution

According to Pythagoras theorem, in ∆PTQ

PQ2=PT2+QT2 ...1

In ∆PTS

PS2=PT2+TS2 ...2

In ∆PTR

PR2=PT2+RT2 ...3

In ∆PQR, point S is the midpoint of side QR.

QS=SR=12QR ...4

PQ2+PR2=2PS2+2QS2 by Apollonius theoremPR2=2PS2+2QS2-PQ2PR2=PS2+PS2+QS2+QS2-PQ2PR2=PS2+PT2+TS2+QS2+QR22-PT2+QT2 From 1, 2 and 4PR2=PS2+QR22+PT2+TS2+QS2-PT2-QT2PR2=PS2+QR22+QS2+TS2-QT2PR2=PS2+QR22+QS2+TS+QTTS-QTPR2=PS2+QR22+QS2+QSTS-QTPR2=PS2+QR22+QS2+QS×TS-QS×QTPR2=PS2+QR22+QS×TS+QS2-QS×QTPR2=PS2+QR22+QS×TS+QSQS-QTPR2=PS2+QR22+QS×TS+QS×TSPR2=PS2+QR22+2QS×TSPR2=PS2+QR22+QR×TS

Hence, PR2=PS2+QR×ST+QR22.

Now,

PQ2+PR2=2PS2+2QS2 by Apollonius theoremPQ2=2PS2+2QS2-PR2PQ2=PS2+PS2+QS2+QS2-PR2PQ2=PS2+PT2+TS2+QS2+QR22-PT2+RT2 From 2, 3 and 4PQ2=PS2+QR22+PT2+TS2+QS2-PT2-RT2PQ2=PS2+QR22+QS2-RT2-TS2PQ2=PS2+QR22+QS2-TS+RTRT-TSPQ2=PS2+QR22+QS2-TS+RTRSPQ2=PS2+QR22+QS2-TS+RTQSPQ2=PS2+QR22+QS2-QS×TS-QS×RTPQ2=PS2+QR22-QS×TS+QS2-QS×RTPQ2=PS2+QR22-QS×TS-QSRT-QSPQ2=PS2+QR22-QS×TS-QSRT-SRPQ2=PS2+QR22-QS×TS-QS×TSPQ2=PS2+QR22-2QS×TSPQ2=PS2+QR22-QR×TS

Hence, PQ2=PS2-QR×ST+QR22.

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