Question for class 10

Question

In the above figure, $OB$ is the perpendicular bisector of the line segment $DF$, $FA\perp OB$ and $FE$ intersects $OB$ at the point $C$. Prove that $\displaystyle \frac{1}{OA}+\frac{1}{OB}=\frac{2}{OC}$.

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Triangle similarities to be used.

Answer 1

Let's write out some of the information that is outright obvious.

1. It can be seen that the triangles $DOB$ and $FOA$ are similar. Therefore, $\angle DOB = \angle FOA$ and $\dfrac{OA}{OB} = \dfrac{AF}{BD}$.
2. It is also said that the line $OB$ bisects the line $DE$. We can deduce then that $BD$ = $BE$.
3. It can be shown that $\angle ACF = \angle BCE$ because lines $FA$ and $DE$ are parallel and the triangles $ACF$ and $BCE$ are similar. Therefore, $\dfrac{FA}{DE} = \dfrac{AC}{CB}$.
4. $AC = OC - OA$
5. $CB = OB - OC$

With all this information, you can easily prove that $\dfrac{1}{OA} + \dfrac{1}{OB} = \dfrac{2}{OC}$