see attachment and explain it

see attachment and explain it

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1 Answers

Nishant Vora IIT Patna
askIITians Faculty 2467 Points
8 years ago
Hello Student, Please find the solution

Consider triangle FAC and EBC
\angle FAC = \angle EBC = 90
\angle FCA = \angle ECB (vertically opposite angles)
Hence by AA traingle FAC and EBC are similar
so \frac{AF}{BE} = \frac{AC}{BC}........... (Eqn 1)

Now consider Triangle OAF and OBD
\angle FOA = \angle DOB
\angle OAF = \angle OBD = 90
So again by AA similarity \Delta OAF = \Delta OBD

\frac{OA}{OB} = \frac{AF}{BD} .........eqn 2

From (1) and (2) and also BD=BE
Therefore, \frac{AC}{BC} = \frac{OA}{OB}

=>\frac{OC-OA}{OB-OC} = \frac{OA}{OB}

=> \frac{OC-OA}{OA} = \frac{OB-OC}{OB}

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

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