This is an old problem from Igor Sharygin (exercise 3 in his Kvant article "Теоремы Чевы и Менелая"). It is the starting point of a theory about some triangle centers which I later named "Sharygin points". Here is a solution of the problem: Let the tangents to the circumcircle of triangle ABC at the points A, B, C meet the lines BC, CA, AB at the points X, Y, Z, respectively. It is well-known that these points X, Y, Z are collinear. Now, we will prove that A" = X, B" = Y and C" = Z; this will clearly solve the problem. Since the point B" lies on the perpendicular bisector of the segment BB', the triangle BB"B' is isosceles. Thus, < B"BB' = < B"B'B. Hence, < B"BA = < B"BB' - < ABB' = < B"B'B - < ABB' = < AB'B - < ABB' = (180 - < ABB' - < BAB') - < ABB' = 180 - 2 < ABB' - < BAB' = 180 - 2 B/2 - A = 180 - B - A = C = < BCA. On the other hand, since the line BY is tangent to the circumcircle of triangle ABC, we have < YBA = < BCA (by the secant-tangent angle theorem). Together with < B"BA = < BCA, this implies < YBA = < B"BA, and hence the point Y lies on the line BB". Since we also know that both points Y and B" lie on the line CA, it follows that B" = Y. Similarly, C" = Z and A" = X. This completes the solution. Darij

Consider that $\angle CBA > \angle BCA$ now we have $\angle A''AA'=\angle A''A'A=\angle BAC/2+\angle ACB$ $\longrightarrow \angle A''AB=\angle ACB$ , so $\triangle A''AB$ and $\triangle A''AC$ are similiar and we have : $\frac{A''B}{A''A}=\frac{A''A}{A''C}=\frac{AB}{AC}\longrightarrow$ $\frac{A''B}{A''C}=\frac{AB^{2}}{AC^{2}}$ . similiary we get that : $\frac{B''C}{B''A}=\frac{BC^{2}}{BA^{2}}$ and $\frac{C''A}{C''B}=\frac{CA^{2}}{CB^{2}}$ . now we can check the Menelaus' theorem easily !

This nice, easy and old problem belongs to the clasical synthetical geometry. Proof. Denote $X\in AA\cap BC$. Observe that $\{\begin{array}{c}m(\widehat{XAA'})=m(\widehat{XAB})+m(\widehat{BAA'})=C+\frac{A}{2}\\\ m(\widehat{XA'A})=m(\widehat{A'AC})+m(\widehat{A'CA})=\frac{A}{2}+C\end{array}$ $\implies$ $\widehat{XAA'}\equiv\widehat{XA'A}$ $\implies$ $\triangle AXA'$ is isosceles. In conclusion $X\equiv A''$ a.s.o.

Well i tried an ugly vectorial proof. Sorry for eventual mistakes, i am not into the style.

Approach by mihai miculita: Let $ A_1; B_1; C_1$ the intersections of external bisectors of angles $ A; B; C$ with the opposite sides. The circles of diametres $ [A'A_1], [B'B_1], [C'C_1]$ is the Appollonius circles of $ \triangle{ABC}$ and the points $ A", B", C"$ is the Appolonian circles centres $ \Rightarrow$ the points $ A", B", C"$ coliniar!

Also, it has been posted as An interesting line of a triangle, by nayel. Kostas Vittas.

Sorry for reviving, but... Let the $A$-tangent to $(ABC)$ meet $\overleftrightarrow{BC}$ in $A^{*}$, then $\measuredangle A^{*}A'A=\measuredangle A'AC+\measuredangle A'CA=\measuredangle BAA'+\measuredangle BAA^{*}=\measuredangle A^{*}AA'$, so $\Delta A^{*}AA'$ is $A^{*}$-isosceles thus perpendicular bisector of $\overline{AA'}$ passes through $A^{*}$. This shows that $A^{*}=A''$. As we know, $A^{*}$, $B^{*}$ and $C^{*}$ form the Lemoine's line of $\Delta ABC$, so $A''$, $B''$ and $C''$ are colinear.

Note that $\angle A''AB = \angle A''AA' - \angle A/2 = 180^{\circ}-\angle B - \angle A/2 - \angle A/2 = \angle C$ so $A''A$ is tangent to $(ABC)$ at $A$. Let $X,Y,Z$ be the points on $BC,CA,AB$ respectively such that $AX,BY,CZ$ are symmedians. Let $T$ be the intersection of the tangents to $(ABC)$ at $B,C$. Then $A'' \in BC$ the polar of $T$ so $T$ is in the polar of $A''$. Similarly $A$ lies in the polar of $A''$ because $A''A$ is tangent to $(ABC)$. Therefore $AT$ is the polar of $A''$ and it is well known that $AT$ is the symmedian so $X$ is also in the polar of $A''$. Thus by a well known lemma we have $(B,C;X,A'') = -1$. We know that $AX,BY,CZ$ are concurrent at the symmedian point of $\triangle ABC$ so $YZ \cap BC = A_1$ satisfies $(B,C;X,A_1) = -1$. Therefore $A'' = YZ \cap BC$ and similarly $B'' = XZ \cap AC, C'' = XY \cap AB$. Since $AX,BY,CZ$ concur we know that triangles $ABC$ and $XYZ$ are perspective meaning $AB \cap XY, BC \cap YZ, CA \cap XZ = A'',B'',C''$ are collinear as desired.

See P5 from here