Question

a triangle PQR is right angeled at Q and points S and T trisect side QR.prove that 8PT^2= 3PR^2+5PS^2.

Answer 1

Let,QS=x
QS=ST=TR=x
QT=2x
QR=3x
 In ΔPQT, ∠Q=90°
PT²=PQ²+QT²
      =PQ²+4x²  -------------- 1
In ΔPQR, ∠Q=90°
PR²=PQ²+QR²
      =PQ²+9x²
In ΔPQS, ∠Q=90°
PS²=PQ²+QS²
      =PQ²+x²
From eq.1
8PT²=8PQ²+32x²--------------------- eq.2
[by multiplying eq.1 by 8]
3PR²+5PS²=3(PQ²+9x²)+5(PQ²+x²)
                  =3PQ²+27x²+5PQ²+5x²
                  =8PQ²+32x²
                  =8(PQ²+4x²)
                  =8PT²       [from eq.2]

Answer 2

Here's your answer my friend