The tangent at $A$ to the circumcircle $ABC$ meets the line $BC$ at the point $A''$; the points $B''$ and $C''$ are defined similarly. The points $A''$, $B''$ and $C''$ are collinear on Lemoine's line. We shall prove that the lines $AA'$, $BB'$ and $CC'$ are the polars of the points $A''$, $B''$ and $C''$, respectively, relative to the nine-point circle $\gamma$, so they are indeed concurrent. Clearly, it is sufficient to prove that that $AA'$ is the polar of $A''$ with respect to $\gamma$.

Let $A_1, B_1$ and $C_1$ be the perpendicular feet dropped from $A, B$ and $C$, respectively, on the lines $BC, CA$ and $AB$, respectively. Let further $A_2$ be the midpoint of the side $BC$, and let $A_3$ be the midpoint of the segment joining $A$ to the orthocenter of the triangle $ABC$. It is easily seen that the line $A_2A_3$ is the perpendicular bisector of the segment $B_1C_1$, so it is perpendicular to the tangent at $A$ to the circumcircle $ABC$. Consequently, $A_3$ is the orthocenter of the triangle $AA''A_2$, so the lines $AA_2$ and $A''A_3$ are perpendicular; it is easily seen that they meet at some point on $\gamma$, so $A''$ lies on the polar of $A$ with respect to $\gamma$. Finally, $A''$ lies on $BC$, which is the polar of $A'$ with respect to $\gamma$, so $A''$ is the pole of $AA'$ with respect to $\gamma$.