Question

In the given figure, $OB$ bisects $DE$. Then, which of the following is true?

• A. $\frac{1}{OA}+\frac{1}{OC}=\frac{1}{OB}$
• B. $\frac{1}{OA}+\frac{1}{OB}=\frac{1}{OC}$
• C. $\frac{1}{OA}+\frac{1}{OB}=\frac{2}{OC}$
• D. $\frac{1}{OA}+\frac{1}{OC}=\frac{2}{OB}$

Answer 1

The correct option is C $\frac{1}{OA}+\frac{1}{OB}=\frac{2}{OC}$

In $\Delta AOF$ and $\Delta BOD$,
$\angle O=\angle O$ (Common angle)
$\angle A=\angle B={90}^{\circ }$
Hence, by $AA$ similarity, $\Delta AOF\sim \Delta BOD$.
Thus, $\frac{OA}{OB}=\frac{FA}{DB}$ --- (i)

In $\Delta FAC$ and $\Delta EBC$,
$\angle A=\angle B={90}^{\circ }$
$\angle FCA=\angle ECB$ (Vertically opposite angles)
Hence, by $AA$ similarity, $\Delta FAC\sim \Delta EBC$.
Thus, $\frac{FA}{EB}=\frac{AC}{BC}$
But $EB=DB$ ($B$ is the midpoint of $DE)$
Hence, $\frac{FA}{DB}=\frac{AC}{BC}$ --- (ii)
From (i) andd (ii),
$\frac{AC}{BC}=\frac{OA}{OB}$
$\Rightarrow \frac{(OC-OA)}{(OB-OC)}=\frac{OA}{OB}$
$\Rightarrow (OB+OA)\left(OC\right)=2\left(OA\right)\left(OB\right)$
$\Rightarrow \frac{1}{OA}+\frac{1}{OB}=\frac{2}{OC}$