Need some help with the following problem.
Problem: In $\triangle ABC$ the midpoints of $BC$, $AC$, $AB$ are $L, M,$ and $N$ respectively, and the points on the circumcircle opposite to $A, B,$ and $C$ are $A', B',$ and $C'$ respectively. Prove that the lines $A'L$, $B'M$, $C'N$ are concurrent at the orthocenter.
Kind of at a loss on how to start, though I've tried using power of a point and contradiction without much effect.
On
The triangle $LMN$ is the image of $ABC$ by the homothety $h_1$ centered on the gravity center of the triangle $ABC$ (which you should add to the picture ; let's call it $G$) and scaling $-1/2$.
The triangle $A'B'C'$ is the image of $ABC$ by the homothety $h_2$ centered on $Y$ (the center of the circumscribed circle of $ABC$) and scaling $-1$.
Hence $LMN$ is the image of $A'B'C'$ by $h_1 \circ h_2^{-1}$, a composition of two homotheties of scaling $-1$ and $-1/2$, which has to be an homothety of scaling $-1 \times -1/2 = 1/2$. In particular, the lines $(A'L), (B'M)$ and $(C'N)$ all pass through the center $X$ of this homothety.
But what is the center of this homothety ? $X$ is fixed by $h_1 \circ h_2^{-1}$, so put $X' = h_2^{-1}(X)$, so that $X = h_1(X')$. Then we must have $\vec{YX'} = - \vec{YX}$ and $\vec{GX} = - \frac 1 2 \vec{GX'}$.
Hence $\vec{GX} = - \frac 1 2 (\vec {GY} - \vec {YX})$, and so $\frac 3 2 \vec{GX} = \vec{YX}$ : $X$ is the barycenter of $(G,3)(Y,-2)$
Then it is a well-known fact about triangles that this point is the orthocenter of the triangle.
$AXCB'$ is a parallelogram with center $M$. This easy to prove fact (use $AB' \perp AB$) and its equivalents for the other two sides show that:
the dilation of the plane by a factor of $2$, centered at $X$, carries $L,M$ and $N$ to $A', B',$ and $C'$.
Really that dilation carries the nine-point circle of $ABC$ to the circumcircle. Six points on the circumcircle are drawn in the diagram, and the other three are the reflections of $X$ in the sides of the triangle.