018. Rotation

Let $P$ be any point in the square $ABCD$, and let the perpendiculars drawn from points $A$, $B$, $C$, and $D$ to the lines $BP$, $CP$, $DP$, and $AP$ be $l_1$, $l_2$, $l_3$, and $l_4$, respectively. Prove that the four lines $l_1$, $l_2$, $l_3$, and $l_4$ intersect at one point.

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Let the center of square $ABCD$ be $O$. Let $R$ be a 90-degree counterclockwise rotation centered at O. Then, by $R$, each point and line are transformed as follows. $$\begin{aligned} A \xrightarrow{R} B, \qquad B \xrightarrow{R} C, &\qquad C \xrightarrow{R} D, \qquad D \xrightarrow{R} A \\\\ l_1 \xrightarrow{R} \overrightarrow{BP}, \qquad l_2 \xrightarrow{R} \overrightarrow{CP}, &\qquad l_3 \xrightarrow{R} \overrightarrow{DP}, \qquad l_4 \xrightarrow{R} \overrightarrow{AP} \end{aligned}$$

Since the four straight lines $BP$, $CP$, $DP$, and $AP$ share a point $P$, let $Q$ be the point where $P$ is transformed by $R^{-1}$, then this $Q$ lies on the four lines $l_1$, $l_2$, $l_3$, and $l_4$.

Briefly, when drawing only the relationship between points $A$ and $B$ and lines $BP$, $CP$, $l_1$, $l_2$, the similarity of triangles by rotation is discovered.