It is easy to see that $M$ is incenter of $\Delta PQL$. Using easy angle chase we can paraphrase the problem as follows: Let $ABC$ be a triangle. Points $D$ and $E$ are constructed s.t. $\Delta DBA \sim \Delta ABC $ (in that order, namely $D\mapsto A$ ect.) and $\Delta ECA \sim \Delta ABC$. Suppose $X$ is a point s.t. $\Delta XED \sim \Delta ABC$. Let $P$ and $Q$ be points on $AB$ and $AC$ s.t. $DA=DP$ and $EA=EQ$. Next, $Y$ be the center of spiral similarity that sends $BC$ to $PQ$. Then $X$, $A$, $Y$ are collinear. We do this by complex number. By similarity we can get $d=\frac{a^2+bc-2ab}{c-b}$ and $e=\frac{a^2+bc-2ac}{b-c}$. Also it is not hard to see that $p=b+d-ab\overline{d}$, $q=c+e-ac\overline{e}$. Spiral similarity center is given $y=\frac{bq-pc}{b+q-p-c}$. After some calculation we get $y=\frac{a(b+c)-2bc}{2a-b-c}$. Now we employ a trick. It is not hard to see $XD||AB$ and $XE||AC$. From these two we get pair of equations: \begin{align} x+\overline{x}ab=d+ab \overline{d} \\ x+\overline{x}ac=e+ac \overline{e} \end{align}Collinearity of $x,y,a$ is equivalent with \begin{align*} x+\overline{x}ay=a+y && \text{(3)} \end{align*}Now, without calculating $x$, lets show that from $(1)$, $(2)$ equation $(3)$ follows. We do this by considering following: $$(1)* \frac{a-c}{2a-b-c}+(2)* \frac{a-b}{2a-b-c}= $$LHS trivially matches, not lets show that RHS matches. $$(a-c)(d+ab \overline{d})+(a-b)(e+ac \overline{e})=a(2a-b-c)+a(b+c)-2bc$$Notice that $\overline{d}=\frac{e}{a^2}$ and by using this we get: $$(a-c)\left(d+\frac{eb}{a}\right)+(a-b)\left(e+\frac{cd}{a}\right)=2(a^2-bc)$$$$a(d+e)-\frac{bce}{a}-\frac{bcd}{a}=2(a^2-bc)$$$$\left (d+e \right)\left(a-\frac{bc}{a}\right)=2(a^2-bc)$$$$\left(\frac{a^2-bc}{a}\right)\left(\frac{2ab-2ac}{b-c}\right)=2(a^2-bc)$$Which is correct and we are done.