Q. 8

# If two diagonals of a quadrilateral ABCD intersect each other perpendicularly, then let us prove that, AB2 + CD2 = BC2 + DA2

Given: AC BD

To prove: AB2 + CD2 = BC2 + DA2

Proof:

By applying Pythagoras theorem in ΔAOB, we get,

AB2 = AO2 + BO2 ……………… (1)

By applying Pythagoras theorem in ΔBOC, we get,

BC2 = BO2 + CO2 ……………… (2)

By applying Pythagoras theorem in ΔCOD, we get,

CD2 = CO2 + DO2 ……………… (3)

By applying Pythagoras theorem in ΔDOA, we get,

DA2 = AO2 + DO2 ……………… (4)

By adding eqn. (1) and (3), we get,

AB2 + CD2 = AO2 + BO2 + CO2 + DO2

AB2 + CD2 = (BO2 + CO2) + (AO2 + DO2)

Substituting from eqn. (2) and (4) gives,

AB2 + CD2 = BC2 + DA2

Hence proved.

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