Asked by ashish854 | 2nd Feb, 2010, 07:13: PM
Draw EF parallel to AB passing through O.
In right triangles OEA, OFC
OA2 = OE2 + AE2 and
OC2 = OF2+ CF2
OA2+OC2 = OE2 +AE2+ OF2+CF2……..(A)
In right triangles OFB & ODE
OB2 =OF2 +FB2.
OD2 =OE2 + DE2.
Adding the above equations
OB2+OD2=OF2+FB2 + OE2+DE2 ………. (B)
From the figure DE =CF; AE = BF
We can say both the LHS can be equated.
OA2 + OC2 = OB2 + OD2
Answered by | 3rd Feb, 2010, 04:29: PM
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