Trivial problem: It's well known that $B'B^{\#},C'C^{\#}$ pass throw the Lemoine’s point so $K$ is lemoine's point so $AK$ is $A$-symmedian line and we are done. (Lemoine’s point is the intersection of three symmedian lines)

Yeah I'm curious as to why a well-known lemma was inserted as a problem here. Anyways, let's prove the lemma. Lemma. Let $\triangle ABC$ be a triangle and let $M$ be the midpoint of $\overline{BC}.$ Let $H$ be the projection of $A$ onto $BC$ and let $N$ be the midpoint of $\overline{AH}.$ Then $MN$ passes through the symmedian point $K.$ Proof. Let $\triangle T_aT_bT_c$ be the tangential triangle of $\triangle ABC.$ Denote $X \equiv AT_a \cap BC$ and $Y \equiv AB \cap T_aT_b.$ It is well-known (Lemma 1) that $AT_a, BT_b, CT_c$ concur at $K.$ Therefore, the division $(Y, C; T_a, T_b)$ is harmonic. It follows that \[-1 = B(Y, C; T_a, T_b) = (A, X; T_a, K) = M(A, X; T_a, K) = (A, H; \infty, MK \cap AH).\] Therefore $MK$ bisects $\overline{AH}$ and the desired result follows. $\blacksquare$ By the lemma, we see that $B'B^{\#} \cap C'C^{\#}$ is the symmedian point, which implies the result.

Dear Mathlinkers, for the Lemoine's point you can see http://jl.ayme.pagesperso-orange.fr/Docs/The%20cross-cevian%20point.pdf p. 20 - 21 based on http://jl.ayme.pagesperso-orange.fr/Docs/An%20Another%20Unlikely%20Concurrence.pdf Sincerely Jean-Louis

We rephrase the question as notation is bad. In triangle $ABC$ let $A',B',C'$ be midpoints of sides and $A_1,B_1,C_1$ be feet of altitudes. Let $X,Y$ be the midpoints of $BB_1,CC_1$ respectively. Then $L=B'X \cap C'Y$. Prove that $L$ lies on $A$ symmedian. Notice that $X,Y$ are isogonal in angle $A$ (trivial) and by the forgotten isogonality Lemma, we get $AL,A(B'Y \cap C'X)$ as isogonal in $A$. Now, $B'Y$ is simply parallel from $B'$ to $AB$ meaning that the intersection is in fact $A'$. The result holds.

We prove that the midpoint of the $A$-altitude, the symmedian point, and the midpoint of $BC$ are collinear, and this clearly solves the problem as it implies that the concurrence point is the symmedian point. We use barycentric coordinates with respect to $ABC$. The midpoint of the altitude is \[(1/2,0,0)+\frac{1}{2}(0,S_C/a^2,S_B/a^2)=(a^2:S_C:S_B)\]where we are using Conway's notation. Thus, all we want to show is that \[\begin{vmatrix}0 & 1 &1\\ a^2 & b^2 & c^2 \\ a^2 & S_C & S_B\end{vmatrix}=0,\]which is an easy computation.

It's well - known that $K$ is symmedian point of $\triangle$ $ABC$, so: $AK$ is $A$ - symmedian of $\triangle$ $ABC$ or $\widehat{LAB}$ = $\widehat{CAA'}$

We use complex numbers on $a$, $b$, $c$ lying on the unit circle centered at $O=0$. $a^{'}=\frac{1}{2}(b+c)$, $b^{'}=\frac{1}{2}(c+a)$, $c^{'}=\frac{1}{2}(a+b)$, $b^{*}=\frac{1}{2}(a+b+c-\frac{ca}{b})$, $c^{*}=\frac{1}{2}(a+b+c-\frac{ab}{c})$, $b^{\#}=\frac{3b^2+ab+bc-ca}{4b}$, $c^{\#}=\frac{3c^2+bc+ca-ab}{4c}$. $k \stackrel{def}{=} \overline{c^{'}c^{\#}}\cap \overline{b^{'}b^{\#}}=\frac{ (\overline{c^{'}}c^{\#}-\overline{c^{\#}}c^{'})(b^{'}-b^{\#}) - (\overline{b^{'}}b^{\#}-\overline{b^{\#}}b^{'})(c^{'}-c^{\#}) }{ (\overline{c^{'}-c^{\#}})(b^{'}-b^{\#}) - (\overline{b^{'}-b^{\#}})(c^{'}-c^{\#})}=\frac{2(\sum_{cyc}^{}(a^2b^2-a^2bc))}{\sum_{sym}^{}(a^2b)-6abc}$, $\overline{k}=\frac{2(a^2+b^2+c^2-ab-bc-ca)}{\sum_{sym}^{}(a^2b)-6abc}$, $1-a\overline{k}=\frac{(a-b)(a-c)(b+c-2a)}{\sum_{sym}^{}(a^2b)-6abc}$, $k-a=\frac{(a-b)(a-c)(2bc-ab-ca)}{\sum_{sym}^{}(a^2b)-6abc}$. Let $\varphi_{1}=\measuredangle BAL=\measuredangle BAK$. Then $\frac{b-a}{k-a}=e^{2i\varphi_1} \big( \overline{\frac{b-a}{k-a}} \big)$, whence $e^{2i\varphi_1}=b\frac{1-a\overline{k}}{k-a}=b\frac{b+c-2a}{2bc-ab-ca}$. Let $\varphi_2=\measuredangle A^{'}AC$. Then $\frac{a^{'}-a}{c-a}=e^{2i\varphi_2}\big( \overline{\frac{a^{'}-a}{c-a}}\big)$, whence $e^{2i\varphi_2}\frac{1}{c}\frac{a^{'}-a}{1-a\overline{a^{'}}}=b\frac{b+c-2a}{2bc-ab-ca}$. Thus $e^{2i\varphi_1}=e^{2i\varphi_2} \implies \varphi_1=\varphi_2$ as needed. $\square$