triangle pqr is right angled at …

Given that a right angle {tex}\triangle PQR{/tex} in which {tex}\angle P Q R = 90 ^ { \circ }{/tex}And S and T are the points of trisection of QR. Thus,  QS = ST = TR
To Prove: 8PT² = 3PR² + 5PS².
Proof: Let QS = ST = TR = x units
Then, QS = x units, QT = 2x units and QR = 3x units. 
From right angle triangles PQS, PQT and PQR where {tex}\angle Q= 90^°{/tex} in all, by using Pythagoras Theorem, we have
PS² = PQ² + QS² ... (i) 
PT² = PQ² + QT²... (ii) 
and PR² = PQ² + QR² ​​​​​​... (iii)
Multiplying (iii) by 3, (i)  by 5 and (ii) by 8
Adding thus obtained (iii)  and (i) and Subtracting (ii) from it, we get
{tex}\therefore{/tex} 3PR² + 5PS² - 8PT²
=3(PQ² + QR²) + 5(PQ² + QS²) - 8(PQ² + QT²)
= 3QR² + 5QS² - 8QT²
{tex}= 3 \times ( 3 x ) ^ { 2 } + 5 ( x ) ^ { 2 } - 8 \times ( 2 x ) ^ { 2 }{/tex} [{tex}\because{/tex} QR = 3x, QS = x and QT = 2x]
(27x² + 5x² - 32x² ) = 0
Thus, 3PR² + 5PS² - 8PT² = 0
Hence, 8PT² = 3PR² + 5PS²
Hence Proved