In an acute △ PQR ,\angle PQR=45^° .Prove that (PQ)^{2 }=(PR)^{2 }+(QR)^{2 }- 4× area(△ PQR

In an acute △ PQR ,\angle PQR=45^° .Prove that (PQ)^{2 }=(PR)^{2 }+

1.	In an acute △ PQR ,\angle PQR=45^° .Prove that (PQ)^{2 }=(PR)^{2 }+(QR)^{2 }- 4× area(△ PQR
Answer» In an acute △ PQR ,\angle PQR=45^° .Prove that (PQ)^{2 }=(PR)^{2 }+(QR)^{2 }- 4× area(△ PQR