Question

In the given figure , OB is perpendicular bisector of the line segment DE, FA is perpendicular to OB and FE intersects OB at C.
Prove that 1/OA + 1/OB = 2/OC .

Answer 1

we have to use similar triangles! 
we can claim TRIAngle CAF ~ TRIangle CBE, and make this proportion: 
FA/OA = BE/BC AND also another similar triangles: TRI OAF ~ TRI OBD, which gives us: FA/OA = BD/OB 
but we know BC = OB - OC and AC = OB - OA - BC = OB - OA - OB + OC = OC - OA 
subbing back in, we get: OA/OB = (OC - OA)/(OB - OC) OA(OB - OC) = OB(OC - OA) OAOB - OAOC = OAOC - OAOB 
2OAOB = OAOC + OBOC 
now if we divide through by OAOBOC, we get what you're looking for: 
2/OC = 1/OB = 1/OA hope it helps you

Answer 2

Answer:

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Answer is in attachment

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