1. **Problem statement:** Given right triangle $\triangle ABC$ right-angled at $A$ with $AB < AC$. $M$ is the midpoint of $BC$. $E$ and $F$ are the perpendicular projections of $M$ onto $AC$ and $AB$ respectively. 2. **Prove quadrilateral $AFME$ is a rectangle:** - Since $E$ and $F$ are projections of $M$ onto $AC$ and $AB$, $ME \perp AC$ and $MF \perp AB$. - In right triangle $ABC$, $AB \perp AC$. - Therefore, $AF \parallel ME$ and $AF \perp FE$. - Also, $AF$ and $ME$ are equal in length because $M$ is midpoint and projections preserve distances along perpendiculars. - Hence, $AFME$ has four right angles and opposite sides equal, so it is a rectangle. 3. **Formula and rules used:** - Projection of a point onto a line creates a perpendicular segment. - Opposite sides of a rectangle are equal and parallel. - Adjacent sides of a rectangle are perpendicular. 4. **Intermediate work:** - $ME \perp AC$ and $MF \perp AB$ by definition of projections. - $AB \perp AC$ since $\triangle ABC$ is right angled at $A$. - So $AF \parallel ME$ and $AF \perp FE$. 5. **Conclusion:** Quadrilateral $AFME$ is a rectangle because it has four right angles and opposite sides equal and parallel.