1. Problem statement: Given a right triangle $\triangle ABC$ with right angle at $A$ and $AB < AC$, and altitude $AH$. Points $E$ and $F$ are the feet of perpendiculars from $H$ to $AB$ and $AC$ respectively. We need to prove that quadrilateral $AEHF$ is a rectangle. 2. Recall that a quadrilateral is a rectangle if it has four right angles. Since $E$ and $F$ are projections of $H$ onto $AB$ and $AC$, $HE \perp AB$ and $HF \perp AC$ by definition. 3. Since $AH$ is the altitude, $AH \perp BC$. Also, $AB \perp AC$ at $A$ because $\triangle ABC$ is right angled at $A$. 4. Consider the quadrilateral $AEHF$. We know: - $HE \perp AB$ so angle at $E$ is $90^\circ$. - $HF \perp AC$ so angle at $F$ is $90^\circ$. - $AB \perp AC$ so angle at $A$ is $90^\circ$. 5. Since three angles are right angles, the fourth angle at $H$ must also be $90^\circ$ (sum of angles in quadrilateral is $360^\circ$). 6. Therefore, $AEHF$ has four right angles and is a rectangle. Final answer: Quadrilateral $AEHF$ is a rectangle.