Q7 of 62 Page 193

A point O in the interior of a rectangle ABCD is joined to each of the vertices A, B, C and D. Prove that OA2 + OC2 = OB2 + OD2


Draw EF || AB || DC passing through O.


Also AB = EF = DC


∴ ABCD and CDEF are rectangles


Now by Pythagoras theorem,


BO2 = BF2 + OF2 …(1)


And,


OD2 = OE2 + ED2 …(2)


Adding (1) and (2),


BO2 + OD2 = BF2 + OF2 + OE2 + ED2


⇒ BO2 + OD2 = AE2 + OF2 + OE2 + CF2


⇒ BO2 + OD2 = AE2 + OE2 + OF2 + CF2


⇒ BO2 + OD2 = AO2 + OC2


Hence, Proved


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