euclidean-geometry
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.
It is easier to prove the following equivalent version:
Let $N$ be the point of intersection between the circumcircle of $ABC$ (centered at $O$) and the angle bisector from $A$. Then $N$ lies on the perpendicular bisector of $BC$.
Proof: since $\widehat{BAN}=\widehat{NAC}$, the $BN$ and $NC$ arcs in the circumcircle of $ABC$ have the same length, since they are subtended by equal angles at $O$. It follows that the segments $BN$ and $NC$ have the same length, and since $OB=OC=R$, the line through $O$ and $N$ is the perpendicular bisector of $BC$.
Reflect $A$ across $D$ to a new point $Q$. Then $\triangle QDE\cong \triangle ECB$ (sas).
Since $$\angle QEC = \angle DEB $$ we see that quadrilateral $ACEQ$ is cyclic, so $$\angle QCE =\angle QAE = \angle DAE = \angle DNE$$ so $QC||DN$. Now since $D$ halves $AQ$ the line $DI$ is a middle line for triangle $QAC$ and thus it halves $AC$.
Best Answer
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
$$\frac {OC}{OA} + \frac {OC}{OB} = \frac {OB \times OC + OA \times OC}{OA \times OB} = 2$$