In ΔRST, ∠S is right angle. The mid- points of tow sides RS and ST are X and Y respectively; let us prove that, RY² + XT² = 5XY²

Given: ΔRST is a right angled triangle with ∠S as right angle.

X is midpoint of RS.

Y is midpoint of ST.

To prove: RY² + XT² = 5XY²

Proof:

In ΔRSY,

RY² = RS² + SY² [∵ H² = P² + B²]

⇒ RY² = (2XS)² + SY² [∵ X is midpoint of RS]

⇒ RY² = 4XS² + SY² ……………… (1)

In ΔXST,

XT² = XS² + ST² [∵ H² = P² + B²]

⇒ XT² = XS² + (2SY)² [∵ Y is midpoint of ST]

⇒ XT² = XS² + 4SY² ……………… (2)

In ΔXSY,

XY² = XS² + SY² …………… (3) [∵ H² = P² + B²]

Now, adding eqn. (1) and (2) gives,

RY² + XT² = 4XS² + SY² + XS² + 4SY²

⇒ RY² + XT² = 5XS² + 5SY²

⇒ RY² + XT² = 5(XS² + SY²)

Substituting from (3) gives,

RY² + XT² = 5XY²

Hence proved.