Question

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Answer 1

Answer:

★Answer★

Given :-

• OB is perpendicular bisector of DE. FA ⊥ OB and FE intersects OB at the point C.

To prove :-

$\bf \: \dfrac{1}{OA} + \dfrac{1}{OB} = \dfrac{2}{OC}$

Proof :-

OB is perpendicular bisector of DE.

⟹ DB = BE ...........[1]

& OB ⊥ DE...........[2]

In ∆OAF & ∆OBD

/_AOF = /_BOD [COMMON]

/_OAF = /_OBD [Each 90°]

⟹ ∆AOF ~ ∆BOD [AA similarity]

$\bf\implies \:\dfrac{OA}{OB} = \dfrac{AF}{DB} \: \: \: (cpst)$

As DB = BE, So above can be rewritten as

$\bf\implies \:\dfrac{OA}{OB} = \dfrac{AF}{BE} ..........(3)$

Now, In ∆ACF & ∆BCE

/_ACF = /_BCE [Vertically opposite angles]

/_FAC = /_CBE [Each 90°]

⟹ ∆ACF ~ ∆BCE [AA Similarity]

$\bf\implies \:\dfrac{AF}{BE} = \dfrac{AC}{BC} .....(4)$

On equating (3) and (4), we get

$\bf\implies \:\dfrac{OA}{OB} = \dfrac{AC}{BC}$

$\bf\implies \:\dfrac{OA}{OB} = \dfrac{OC - OA}{OB - OC}$

$\bf\implies \:OA \times OB - OA \times OC = OB \times OC - OA \times OB$

$\bf\implies \:2OA \times OB = OB \times OC + OA \times OC$

Now, divide both sides by OA × OB × OC, we get

$\bf\implies \:\dfrac{2}{OC} = \dfrac{1}{OA} + \dfrac{1}{OB}$

$\bf\implies \:Hence, Proved$