# 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.

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