prove that

Asked by ashish854 | 2nd Feb, 2010, 07:13: PM

Expert Answer:

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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