1. Projection and Perpendicularity:
- Let $H$ be the projection of $A$ onto the plane $BCD$.
- Let $X$, $Y$, and $Z$ be the projections of $H$ onto lines $CD$, $BD$, and $BC$, respectively.
- Since $AH$ is perpendicular to the plane $BCD$, $AH$ is perpendicular to $BC$. Similarly, $AZ$ is perpendicular to $BC$, implying $BC$ is perpendicular to the plane $AHZ$ and thus to $HZ$.
- Similarly, $BD$ is perpendicular to $HY$ and $CD$ is perpendicular to $HX$.

2. Orthocenters and Collinearity:
- Denote the orthocenters of triangles $ACD$, $ABD$, and $ABC$ by $H_b$, $H_c$, and $H_d$, respectively.
- $H_b$ lies on line $AX$, $H_c$ on line $AY$, and $H_d$ on line $AZ$.
- Since no face of the pyramid is a right triangle, $H_b$, $H_c$, and $H_d$ do not coincide at $A$, $X$, $Y$, or $Z$.

3. Plane and Line Intersection:
- Consider the plane $AXY$. Both $H_b$ and $H_c$ lie on this plane, so line $H_bH_c$ lies in the plane $AXY$.
- Since $H_b$, $H_c$, and $H_d$ are collinear, $H_d$ also lies on this plane. Thus, line $AH_d$ lies in the plane $AXY$.
- Point $Z$ lies on both planes $AXY$ and $BCD$, so $Z$ lies on the intersection line of these planes, which is line $XY$.
- Therefore, $X$, $Y$, and $Z$ are collinear.

4. Simson Line and Circumcircle:
- Since $X$, $Y$, and $Z$ are the projections of $H$ onto lines $CD$, $BD$, and $BC$, respectively, by the Simson Theorem, $H$ lies on the circumcircle of triangle $BCD$.

5. Circumcenter of the Pyramid:
- Consider a point $O$ on the plane passing through the midpoints of $AB$, $AC$, and $AD$, with its projection onto the plane $BCD$ being the circumcenter $O_A$ of triangle $BCD$.
- Let $R$ be the radius of the circumcircle of triangle $BCD$.
- The distances $OB$, $OC$, and $OD$ are all equal to $\sqrt{|OO_A|^2 + R^2}$.
- The distance $OA$ is given by:
$$|OA| = \sqrt{|HO_A|^2 + (|AH| - |OO_A|)^2} = \sqrt{|HO_A|^2 + |OO_A|^2} = \sqrt{|OO_A|^2 + R^2}$$
- Since $O$ lies on the plane passing through the midpoints of $AB$, $AC$, and $AD$, we have $|AH| = 2|OO_A|$.

6. Conclusion:
- Therefore, $O$ is equidistant from $A$, $B$, $C$, and $D$, making $O$ the circumcenter of the pyramid.
- Hence, $O$ lies on the plane passing through the midpoints of $AB$, $AC$, and $AD$.

$\blacksquare$