## Task A-1.7. (10 points)

A right triangle $A B C$ with legs of length $15 \mathrm{~cm}$ and $20 \mathrm{~cm}$ and a right angle at vertex $B$ is congruent to triangle $B D E$ with a right angle at vertex $D$. Point $C$ lies on the segment $\overline{B D}$, and points $A$ and $E$ are on the same side of line $B D$.

a) Determine the distance between points $A$ and $E$.

b) Calculate the area of the intersection of triangles $A B C$ and $B D E$.
## First Solution.

Given $|A B|=|B D|=20, \quad|B C|=|D E|=15$, the lengths of the hypotenuses of the given right triangles are $|A C|=|B E|=25$.

(1 point)

Let $S$ be the intersection of segments $\overline{A C}$ and $\overline{B E}$, and let $\varangle B A C=\alpha$.

![](https://cdn.mathpix.com/cropped/2024_05_30_5dbbe9edfad2d8808b54g-04.jpg?height=480&width=485&top_left_y=2053&top_left_x=777)

Then $\varangle E B D=\alpha$, so

$$
\varangle A B E=\varangle A B C-\varangle E B D=90^{\circ}-\alpha.
$$

Therefore, $\varangle A S B=90^{\circ}$, i.e., $A C \perp B E$.

(1 point)

In this solution, we use the Pythagorean theorem.

Let $|A S|=x,|C S|=y$, and $|B S|=v$.

We have $x+y=25$.

Moreover, from the right triangles $A B S$ and $B C S$, we get

$$
x^{2}+v^{2}=20^{2}, \quad y^{2}+v^{2}=15^{2}
$$

Subtracting these equations, we get: $x^{2}-y^{2}=(x+y)(x-y)=400-225=175$,

so $x-y=175 / 25=7$.

Thus, $x=16, y=9$.

Therefore, $v=12$.

Now we can answer both questions.

a) The distance $|A E|$ will be obtained from the right triangle $A S E$. We know $|A S|=16$ and $|S E|=|B E|-|B S|=25-12=13$.

Thus, $|A E|=\sqrt{16^{2}+13^{2}}=\sqrt{425}=5 \sqrt{17} \mathrm{~cm}$.

b) The area of triangle $B S C$ is $P=\frac{1}{2}|B S| \cdot|S C|=54 \mathrm{~cm}^{2}$.

## Variations of the First Solution.

To determine the lengths $x$, $y$, and $v$, Euclid's theorem or similarity can be used. In each case, the correct calculation of these lengths is worth 3 points.

## Euclid's Theorem.

We have

$$
x=|A S|=\frac{|A B|^{2}}{|A C|}=\frac{20^{2}}{25}=16, \quad y=|C S|=\frac{|B C|^{2}}{|A C|}=\frac{15^{2}}{25}=9
$$

and finally $v=\sqrt{x y}=12$.

## Similarity of Triangles.

Triangles $A B C$ and $B S C$ are similar because they have the same angles

so $|B S|:|B C|=|A B|:|A C|$.

Thus, $v=|B S|=\frac{|A B| \cdot|B C|}{|A C|}=12$,

$y=\sqrt{15^{2}-12^{2}}=9, x=25-9=16$.