Question

In Figure, `A B C` is a right triangle right angled at `B` and points `Da n dE` trisect `B C` . Prove that `8A E^2=3A C^2+5A D^2dot`

Answer 1

Let QS= ST = TR=x
Now by pythagoras theorem
`PR^(2) = PQ^(2)+QR^(2)=PQ^(2)+(3x)^(2)`
` 3PR^(2)= 3PQ^(2) +27x^(2)`
Also , `PS^(2)=PQ^(2)+QS^(2)`
(by pythagoras theorem)
`= PQ^(2)+ x^(2)`
`Rightarrow " " 5PS^(2) = 5PQ^(2) +5x^(2)`
Adding (1) and (2) ,we get
`R.H.S= 3PR^(2)+5PS^(2)=8PQ^(2)+32x^(2)=8(PQ^(2)+4x^(2))= 8[PQ^(2)+(2x)^(2)]`
` = 8 ( PQ^(2) + QT^(2)) = 8 PT^(2)` ( by Pyhagores theorem)
L.H.S. Hence proved.