Question

In the fig given below $OB$ is the perpendicular bisector of the line segment $DE,FA\perp OB$ and $FE$ intersects $OB$ at the point $C$. Prove that $\frac{1}{OA}\frac{1}{OB}=\frac{2}{OC}$.

Answer 1

$OB$ is $\perp$er bisector of line segment $DE,FA$ $\perp$er to $OB$ and $FE$ intersects $OB$ at the point $C$ as shown in figure

Now $△OAF$ and $△ODB$

$\angle OAF=\angle OBD={90}^o$ (because $OB$ $\perp$er bisector of DE,So $OB\perp AF$ and $OB\perp DE$)

$\angle FOA=\angle DOB$ (common angle)

From $A-A$, $△OAF\cong △ODB$

Similarly $△AFC$ and $△BEC$

$\angle FCA=\angle BCE,\angle FAC=\angle CBE={90}^o$

from $A-A$, $△AFC\cong △BEC$

So $AF/BE=AC/CB=FC/CE$

We know taht $DE=BE$ ($\perp$ er bisector of $DE$ is $OB$)

$AF/DB=AC/CB=FC/CE$

As we have $OA/OB=AF/DB=OF/OD$

$OA/OB=AC/CB=(OC-OA)/(OB-OC)$

$OA/OB=(OC-OA)/(OB-OC)$

$OA(OB-OC)=(OC-OA)OB$

$OA.OB-OA.OC=OB.OC-OA.OB$

$2OAOB=OB.OC+OA.OC$

Divide by $OA.OB.OC$

$\frac{2OAOB}{OAOBOC}=\frac{OB.OC}{OAOBOC}+\frac{OA.OC}{OAOBOC}$

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

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