In an acute angled triangle PQR, angle PQR = 45°. Prove that (PR)sq.= (PQ)sq. +(OR)sq. -4ar(triangle PQR)
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construct PN perpendicular to QR
in triangle PNQ, angle Q=45
so QN=PN
PQ^2+QR^2-4*areaPQR=PN^2+QN^2+(QN+NR)2-4*0.5*QR*PN
=2PN^2+PN^2+NR^2+2*PN*NR-2*QR*PN
=2PN^2+PR^2+2PN.NR-2(PN+NR)*PN
=2PN^2-2PN^2+2PN*NR-2PN*NR+PR^2
in triangle PNQ, angle Q=45
so QN=PN
PQ^2+QR^2-4*areaPQR=PN^2+QN^2+(QN+NR)2-4*0.5*QR*PN
=2PN^2+PN^2+NR^2+2*PN*NR-2*QR*PN
=2PN^2+PR^2+2PN.NR-2(PN+NR)*PN
=2PN^2-2PN^2+2PN*NR-2PN*NR+PR^2
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