Square inscribed in a right triangle problem

Let A be a point on a fixed semicircle with diameter BC. MNPQ is a square such that $M \in AB, N \in AC, P \in BC, Q \in BC$. Let D be the intersection of BN and CM and E be the center of the square. Prove that as A varies, DE always passes through a fixed point.

The fixed point is the midpoint of the semicircle. Any suggestion ?

Edit : I have proved that AE is the angle bisector of angle BAC and AD passes through the midpoint of BC.

I think that DE also passes through the feet of the altitude to BC. May be harmonic bundle is useful. (D,H,E,J) = -1 ?

Expanding upon my comment ...

In fact, the semicircle and even half the elements of $\triangle ABC$ are irrelevant to the result. All that really matters is that $\square MNPQ$ is a square ---any square (barring degeneracies)--- such that $\overline{MN}\parallel\overline{BC}$. The construction's fixed point, which you have keenly observed is the midpoint of the semicircle, is, more simply, the center of a square erected upon $\overline{BC}$. (Below, we resolve the ambiguity of which of two candidate squares is meant.)

The parallelism condition guarantees $\triangle DBC\sim\triangle DNM$, so that $|DB|/|DN|=|DC|/|DM|$, making $D$ the center of a dilation/homothety that carries $N$ to $B$ (which my figure also denotes $N'$) and carries $M$ to $C=M'$. Necessarily, the dilated images $P'$ and $Q'$ of $P$ and $Q$ complete square $\square M'N'P'Q'$ as the dilated image of $\square MNPQ$. (Note: The fact that the squares have opposite orientations resolves the ambiguity mentioned above.) Thus also, the dilation carries center $E$ of one square to center $F=E'$ of the other; since a point and its dilated image are collinear with the center of dilation, we are done. $\square$

Here is an easy coordinate geometric proof:

Let the radius of semicircle be $1$ and centre be $(0,0)$ without loss of generality, and the side length of the square be $a$. Then coordinates of $B$ are $(-1,0)$, coordinates of $C$ are $(1,0)$ and that of $M$ are $(b,a)$ for some $b$. Coordinates of $N$ are $(b+a,a)$ and coordinates of $E$ are $\left(b+\frac{a}{2},\frac{a}{2}\right)$. Equation of $BN$ is $$\frac{y}{x+1}=\frac{a}{a+b+1}$$ and equation of $CM$ is $$\frac{y}{x-1}=\frac{a}{b-1}$$ Solving these to find the coordinates of $D$, we get that the coordinates of $D$ are $\left(\frac{a+2b}{a+2},\frac{2a}{a+2}\right)$.

Now, we wish to show that $F(0,1),D$ and $E$ are collinear, so what we wish is as follows: $$\text{Slope of EF = Slope of ED}$$ $$\iff \frac{\frac{a}{2}-1}{\frac{a+2b}{2}}=\frac{\frac{a}{2}-\frac{2a}{a+2}}{\frac{a+2b}{2}-\frac{a+2b}{a+2}}$$ $$\iff a-2 = \frac{a^2+2a-4a}{a+2-2}$$ which is indeed true.