Question

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

Answer 1

In $\Delta OAF$ and $\Delta OBD$,

$\angle O=\angle O$ ...[Common angle]

$\angle OAE=\angle OBD$ ...[Each ${90}^o$]

$\therefore \Delta OAF\sim \Delta OBD$ ...[AA similarity criterion]

$\Rightarrow \frac{OA}{OB}=\frac{FA}{DB}$ ...(i) [CSST]

Similarly,

$\Delta FAC\sim \Delta EBC$ ...[AA similarity criterion]

$\Rightarrow \frac{FA}{EB}=\frac{AC}{BC}$ ...[CSST]

But $EB=DB$ ...[Given]

$\therefore \frac{FA}{DB}=\frac{AC}{BC}$ ...(ii)

From (i) and (ii), we get

$\frac{OA}{OB}=\frac{AC}{BC}$

$\Rightarrow \frac{OA}{OB}=\frac{OC-OA}{OB-OC}$ ... [$\because AC=OC-OA$ and $BC=OB-OC$]

$\Rightarrow OA\times (OB-OC)=OB\times (OC-OA)$

$\therefore OA\times OB-OA\times OC=OB\times OC-OB\times OA$

$\Rightarrow OA\times OB+OA\times OB=OA\times OC+OB\times OC$

$\Rightarrow (OA+OB)\times OC=2OA\times OB$

Dividing both sides by $OA\times OB\times OC$ , we get

$\frac{(OA+OB)\times OC}{OA\times OB\times OC}=\frac{2OA\times OB}{OA\times OB\times OC}$

$\Rightarrow \frac{OA}{OA\times OB}+\frac{OB}{OA\times OB}=\frac{2}{OC}$

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

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

Hence proved