Question

in triangle PQR,S and t trisect the side QR of triange PQR right angled at Q.prove that 8PT^2=3PR^2+5PS^2

Answer 1

Let 
QS = a
Thus
QT = 2a
and QR = 3a

In triangle PQR
PR^2 = PQ^2 + QR^2
PR^2 = PQ^2 + 3a^2
PR^2 = PQ^2 + 9a^2         Eqn 1

In triangle PQT
PT^2 = PQ^2 + 4a^2          Eqn 2

In Triangle PQS
PS^2 = PQ^2 + a^2            Eqn 3

Thus 
3 PR^2 + 5 PS^2 = 3 PQ^2 + 27a^2 + 5 PQ^2 + 5a^2
                            = 8 [ PQ^2 + 4a^2 ]
                             = 8 PT^2
Hence proved

Answer 2

Here's your answer my friend


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