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A point OI in the interior of a rectangle ABCD is joined with each of the vertices A, B, C and D. Prove that OB^{2} + OD^{2 }= OC^{2} + OA^{2}

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#### Solution

Let ABCD be the given rectangle and let O be a point within it.

Join OA, OB, OC and OD.

Through O, draw EOF || AB. Then, ABFE is a rectangle.

In right triangles ΔOEA and ΔOFC, we have

OA^{2} = OE^{2} + AE^{2} and OC^{2} = OF^{2 }+ CF^{2}

⇒ OA^{2} + OC^{2} = (OE^{2 }+ AE^{2}) + (OF^{2} + CF^{2})

⇒ OA^{2 }+ OC^{2} = OE^{2} + OF^{2} + AE^{2} + CF^{2} ......(i)

Now, in right triangles OFB and ODE, we have

OB^{2} = OF^{2} + FB^{2} and OD^{2 }= OE^{2} + DE^{2}

⇒ OB^{2} + OD^{2 }= (OF^{2} + FB^{2}) + (OE^{2 }+ DE^{2})

⇒ OB^{2 }+ OD2 = OE^{2} + OF^{2} + DE^{2} + BF^{2}

⇒ OB^{2} + OD^{2 }= OE^{2} + OF^{2} + CF^{2} + AE^{2} [∵ DE = CF and AE = BF]....(ii)

From (i) and (ii), we get

OA^{2} + OC^{2 }= OB^{2} + OD^{2}.

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