$D$ is the midpoint of $AA'$ and $M,N$ are the orthogonal projections of $C,B$ onto $AA'.$ From $DA_1 \parallel CM \parallel BN,$ we get $\frac{_{A_1B}}{^{A_1C}}=\frac{_{DN}}{^{DM}} \ (\star).$ On the other hand, from $|AB^2-A'B^2|=|AN^2-A'N^2|$ and $|AC^2-A'C^2|=|AM^2-A'M^2|$ we get $DN= \frac{|AB^2-A'B^2|}{2 \cdot AA'} \ , \ DM= \frac{|AC^2-A'C^2|}{2 \cdot AA'}$ Substituting $DN,DM$ into $(\star)$ yields $\frac{A_1B}{A_1C}=\frac{|AB^2-A'B^2|}{|AC^2-A'C^2|}$ Multiplying the cyclic expressions together gives $\frac{A_1B}{A_1C} \cdot \frac{B_1C}{B_1A} \cdot \frac{C_1A}{C_1B}=\frac{|AB^2-A'B^2|}{|AC^2-A'C^2|} \cdot \frac{|BC^2-B'C^2|}{|BA^2-B'A^2|} \cdot \frac{| CA^2-C'A^2|}{|CB^2-C'B^2|}=1$ By Menelaus' theorem in $\triangle ABC,$ we conclude that $A_1,B_1,C_1$ are collinear.

Let $A_0$ be the midpoint of $AA'$ and let the perpendiculars from $B$ and $C$ to the line $AA'$ meet it at points $B_A,C_A$, respectively. Notice that $\frac{BA_1}{CA_1}=\frac{A_0B_A}{A_0C_A}$ since we are taking projections along $AA'$, the $\cos \angle (AA',BC)$ factor cancels out. Notice that $AB^2-AB'^2=AB_A^2-A'B_A^2=2AA'\cdot A_0B_A$ and $AC^2-AC'^2=AC_A^2-A'C_A^2=2AA'\cdot A_0C_A$. Dividing the two relations yields that $\frac{BA_1}{CA_1}=\frac{A_0B_A}{A_0C_A}=\frac{AB^2-A'B^2}{AC^2-A'C^2}$. Hence, multiplying similar relations gives that $\frac{BA_1}{A_1C}\cdot \frac{CB_1}{B_1A}\cdot \frac{AC_1}{C_1A}=-1$ and by the Menelaus' theorem, the points $A_1,B_1,C_1$ are collinear. Note:- This is a rather standard solution and I would like to know a more "synthetic" solution. Specifically, an idea that didn't yield much but is potentially helpful is that the circles with center $A_1$ and radius $A_1A=A_1A_0$ and analogous ones for $B,C$ are coaxial. Any proof along these lines?