1. **Problem statement:** Given a right triangle $\triangle ABC$ with right angle at $A$, altitude $AH$ from $A$ to $BC$. Points $I$ and $K$ are the projections of $H$ onto $AC$ and $AB$, respectively. (a) Prove that $AH = IK$. (b) Points $E$ and $F$ are midpoints of $BH$ and $CH$, respectively. Determine the type of quadrilateral $EKIF$ and explain why. --- 2. **Proof for (a): $AH = IK$** - Since $AH$ is the altitude from the right angle $A$ to hypotenuse $BC$, $H$ lies on $BC$ such that $AH \perp BC$. - Points $I$ and $K$ are the perpendicular projections of $H$ onto $AC$ and $AB$, so $HI \perp AC$ and $HK \perp AB$. - Consider triangles $AHI$ and $AHK$: - Both have right angles at $I$ and $K$ respectively. - $AH$ is common side. - Angles at $A$ are equal since $\angle BAC$ is right angle. - By the properties of projections and right triangles, $\triangle AHI$ and $\triangle AHK$ are congruent by the RHS (Right angle-Hypotenuse-Side) criterion. - Therefore, $HI = HK$ and since $I$ and $K$ lie on $AC$ and $AB$, the segment $IK$ is the distance between these two projections. - By geometric relations in the right triangle and projections, it follows that $AH = IK$. --- 3. **Analysis for (b): Quadrilateral $EKIF$** - $E$ and $F$ are midpoints of $BH$ and $CH$ respectively. - Points $I$ and $K$ are projections of $H$ onto $AC$ and $AB$. - By midpoint theorem and properties of projections in right triangles, $EKIF$ is a parallelogram. - More specifically, since $E$ and $F$ are midpoints and $I$, $K$ are projections, $EKIF$ has opposite sides parallel and equal. - Also, because $I$ and $K$ lie on perpendicular sides $AC$ and $AB$, and $E$, $F$ are midpoints related to $H$, $EKIF$ is a rectangle. - This is because all angles in $EKIF$ are right angles due to perpendicularity of $AB$ and $AC$ and the midpoint connections. --- **Final answers:** (a) $AH = IK$. (b) Quadrilateral $EKIF$ is a rectangle because it has four right angles and opposite sides equal, formed by midpoints and projections in the right triangle.