Question

In an acute angled triangle PQR,angle PQR=45 degree.Prove that (PR)2=(PQ)2+(QR)2-4ar(trianglePQR).

Answer 1

construct PN perpendicular to QR
in triangle PNQ, angle Q=45
so QN=PN
RHS=PQ²+QR²-4*areaPQR
=PN²+QN²+(QN+NR)²-4*0.5*QR*PN
=2PN²+PN²+NR²+2*PN*NR-2*QR*PN
=2PN²+PR²+2PN.NR-2(PN+NR)*PN
=2PN²-2PN²+2PN*NR-2PN*NR+PR²
=PR²=LHS
proved

Answer 2

in ∆ pqr, 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
=PR^2