Question

in right angle triangle PRQ, M is midpoint of QR, and Angle PRQ is 90°. Prove that PQ^2 =4PM^2 - 3PR^2

Answer 1

In ∆ PQR  , we apply Pythagoras theorem , and get

QR2  =  PR2  + PQ2                                  --------------------- ( 1 )

And

In ∆ PMR  , we apply Pythagoras theorem , and get

MR2 =  PR2 + PM2     
                         
MR2 =  PR2 + ( PQ2 )2                                           ( As given PM  = PQ2 )

Taking L.C.M.  we get

MR2 =  4PR2  + PQ24      

4MR2 =  4PR2 + PQ2                                                 --------------------- ( 2 )

Now we subtract equation 1 from equation 2 , we get

4MR2 - QR2 =  3PR2 


 QR2 =  4MR2 - 3PR2                                     ( Hence proved )