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Question
A point OI in the interior of a rectangle ABCD is joined with each of the vertices A, B, C and D. Prove that OB2 + OD2 = OC2 + OA2
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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
OA2 = OE2 + AE2 and OC2 = OF2 + CF2
⇒ OA2 + OC2 = (OE2 + AE2) + (OF2 + CF2)
⇒ OA2 + OC2 = OE2 + OF2 + AE2 + CF2 ......(i)
Now, in right triangles OFB and ODE, we have
OB2 = OF2 + FB2 and OD2 = OE2 + DE2
⇒ OB2 + OD2 = (OF2 + FB2) + (OE2 + DE2)
⇒ OB2 + OD2 = OE2 + OF2 + DE2 + BF2
⇒ OB2 + OD2 = OE2 + OF2 + CF2 + AE2 [∵ DE = CF and AE = BF]....(ii)
From (i) and (ii), we get
OA2 + OC2 = OB2 + OD2.
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