Q8 of 27 Page 289

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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