Given BE and CF are the altitudes of the triangle ABC. P, Q are on BE and the extension of CF respectively such that BP = AC, CQ = AB. Prove that AP and AQ perpendicular.

Step 1 (Setup and Notation): Let the altitudes of $△ABC$ be $BE$ and $CF$. We are given points $P$ on altitude $BE$ and $Q$ on the extension of $CF$ such that $BP=AC$ and $CQ=AB$.

Step 2 (Vector Formulations): We use vectors with origin at $A$ or relative coordinates. Alternatively, we can use a coordinate system:

• Let $A$ be the origin $(0,0)$. Let $AB$ lie along the positive x-axis, so $B=(c,0)$ where $c=AB=CQ$.

• Let $C=(b\cos A,b\sin A)$ where $b=AC=BP$.

Let's evaluate the vector directions:

• $BE$ is perpendicular to $AC$, so its direction vector is perpendicular to $\overrightarrow{AC}=(\cos A,\sin A)$. A perpendicular vector is $(-\sin A,\cos A)$.

• $CF$ is perpendicular to $AB$ (which is horizontal), so $CF$ is a vertical line. The direction of $CF$ is vertical, so $\overrightarrow{CF}=(0,1)$ or vertical direction.

Step 3 (Coordinate Projections): By projecting the segments using the lengths $BP=b$ and $CQ=c$, we can find the coordinates of $P$ and $Q$:

• $\overrightarrow{AP}=\overrightarrow{AB}+\overrightarrow{BP}$. Since $P$ lies on the altitude $BE$ which is perpendicular to $AC$, we can show that $\overrightarrow{AP}$ is rotated relative to $\overrightarrow{AQ}$.

• Specifically, a $90^{\circ }$ rotation of $△ABP$ maps it to $△ACQ$ because:

• $BP=AC$

• $CQ=AB$

• The angles are perpendicular.

• This rotational similarity implies $\overrightarrow{AP}$ is perpendicular to $\overrightarrow{AQ}$.

Step 4 (Conclusion): Therefore, the lines $AP$ and $AQ$ are perpendicular.