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In the given figure, OB is perpendicular bisector of the line segment DE, $F A ⊥ O B$ and, FE intersects OB in C. Is the given statement true or false, $1 O A + 1 O B = 2 O C$ .

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False

True

answer is B.

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Detailed Solution

It is given that OB⊥DE, FA⊥OB and FE intersects OB in C.
We shall first prove the similarities of ΔOAF, ΔODB, ΔAFC, and ΔBEC.
Question Image

Now considering, ΔOAF and ΔODB

∠ O A F = ∠ O B D = 900

(OB⊥AF and OB⊥DE)
∠FOA=∠DOB (common angle)
Hence,

ΔOAF \approx ΔODB

by RHS similarity criterion.
Hence, we can write that,

\frac{OA}{OB} = \frac{AF}{DB} = \frac{OF}{OD}

Similarly, for △AFC and △BEC
∠FCA=∠BCE
∠FAC=∠CBE=90⁰
So, with AA similarity, △AFC≅△BEC

So, A F B E = A C C B = F C C E … (1)

We know that DB=BE (Perpendicular bisector of DE is OB).
So, from (1)

\frac{AF}{DB} = \frac{AC}{CB} = \frac{FC}{C}

Also we have, \frac{OA}{OB} = \frac{AF}{DB} = \frac{OF}{OD}

O A O B = A C C B = O C − O A O B − O C

→ O A O B − O C = O C − O A O B

→ O A . O B − O A . O C = O B . O C − O A . O B .

\Rightarrow 2 OAOB = OB . OC + OA . OC

Dividing by, OA . OB . OC

\frac{2 OAOB}{(OA . OB . OC)} = \frac{(OA . OC)}{OA . OB . OC} + \frac{(OB . OC)}{(OA . OB . OC)}

\Rightarrow \frac{1}{OA} + \frac{1}{OB} = \frac{2}{OC}

Hence, from the given conditions it is derived that

1 O A + 1 O B = 2 O C

.
Therefore, the given statement is true i.e.,

\frac{1}{OA} + \frac{1}{OB} = \frac{2}{OC}

.
Hence, the correct option is 2.

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