## Task A-1.3.

Given is a triangle $A B C$ where $\varangle B A C=45^{\circ},|A B|=4,|A C|=3 \sqrt{2}$. Let $\overline{A D}$ and $\overline{B E}$ be the altitudes of this triangle. The perpendicular to $\overline{A B}$ through point $E$ intersects the segment $\overline{A D}$ at point $P$.

Determine $|E P|$.
## First Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_76afda38443ee3ddf2f1g-03.jpg?height=982&width=1265&top_left_y=294&top_left_x=401)

In the right triangle $A B E$, we have $\varangle B A E=45^{\circ}$, so we can conclude that it is an isosceles right triangle. From $|A B|=4$, we conclude

$$
|A E|=|B E|=2 \sqrt{2}
$$

It follows that $|E C|=|A C|-|A E|=3 \sqrt{2}-\sqrt{2}=\sqrt{2}$.

Let $H$ be the orthocenter of triangle $A B C$. Note that $\varangle C H E=\varangle B A C=45^{\circ}$, because these are angles with perpendicular legs. Therefore, triangle $C E H$ is also an isosceles right triangle, so $|E H|=|E C|=\sqrt{2}$.

We can also conclude that $\varangle P E H=\varangle B A C$, $\varangle E H P=\varangle A C B$, and $\varangle E P H=\varangle C B A$, because these are pairs of angles with perpendicular legs. According to the AA similarity criterion, we can conclude that triangles $E P H$ and $A B C$ are similar. Therefore,

$$
\frac{|E P|}{|E H|}=\frac{|A B|}{|A C|}
$$

from which we have

$$
|E P|=\frac{|A B|}{|A C|}|E H|=\frac{4}{3 \sqrt{2}} \cdot \sqrt{2}=\frac{4}{3}
$$