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In figure, $OB$ is perpendicular bisector of line segment $DE$, $FA$ perpendicular to $OB$ and $FE$ intersects $OB$ at the point $C$, then find the sum $\displaystyle \frac{OC}{OA}+\frac{OC}{OB}$

I have proved the similarity of the triangles FAC~EBC and DBO~FAO please help after that.

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Some Guy
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Mohit Kumar
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2 Answers2

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From $\triangle FAC \sim \triangle EBC$:

$$\frac {BE}{AF} = \frac {BC}{AC}$$

From $\triangle DBO \sim \triangle FAO$:

$$\frac {DB}{FA} = \frac {OB}{OA}$$

Since $OB$ is the perpendicular bisector of $DE$, we have $BD = BE$. Therefore:

$$\frac {BO - CO}{CO-AO} = \frac{BC}{AC} = \frac {BE}{AF} = \frac {DB}{FA} = \frac {OB}{OA}$$

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

$$2 \times OA \times OB = OA \times OC + OB \times OC$$

Thus

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

player3236
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Observe that, $$\frac {AC}{BC}=\frac {AF}{EB}=\frac {AF}{BD}=\frac {OA}{OB}$$ and thereafter $$\frac {AC}{OA}=\frac {BC}{OB}$$.

Therefore, $$\frac {OC}{OA} + \frac {OC}{OB}=\frac {OA+AC}{OA}+\frac {OB-BC}{OB}=2+\frac {AC}{OA}-\frac {BC}{OB}=\boxed {2}$$

Limestone
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