Q. 145.0( 2 Votes )

# In ΔRST, ∠S is right angle. The mid- points of tow sides RS and ST are X and Y respectively; let us prove that, RY^{2} + XT^{2} = 5XY^{2}

Answer :

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

X is midpoint of RS.

Y is midpoint of ST.

To prove: RY^{2} + XT^{2} = 5XY^{2}

Proof:

In ΔRSY,

RY^{2} = RS^{2} + SY^{2} [∵ H^{2} = P^{2} + B^{2}]

⇒ RY^{2} = (2XS)^{2} + SY^{2} [∵ X is midpoint of RS]

⇒ RY^{2} = 4XS^{2} + SY^{2} ……………… (1)

In ΔXST,

XT^{2} = XS^{2} + ST^{2} [∵ H^{2} = P^{2} + B^{2}]

⇒ XT^{2} = XS^{2} + (2SY)^{2} [∵ Y is midpoint of ST]

⇒ XT^{2} = XS^{2} + 4SY^{2} ……………… (2)

In ΔXSY,

XY^{2} = XS^{2} + SY^{2} …………… (3) [∵ H^{2} = P^{2} + B^{2}]

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

RY^{2} + XT^{2} = 4XS^{2} + SY^{2} + XS^{2} + 4SY^{2}

⇒ RY^{2} + XT^{2} = 5XS^{2} + 5SY^{2}

⇒ RY^{2} + XT^{2} = 5(XS^{2} + SY^{2})

Substituting from (3) gives,

RY^{2} + XT^{2} = 5XY^{2}

Hence proved.

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