## Task B-1.7.

Let the length of the side of square $A B C D$ be $a$. Point $M$ is the midpoint of side $\overline{B C}$. Point $X$ is the intersection of line $M D$ and the perpendicular line drawn from vertex $A$ to line $M D$. Calculate the perimeter of triangle $A B X$ in terms of $a$.
## First Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_4b7629e40479ae2c2357g-06.jpg?height=460&width=848&top_left_y=1712&top_left_x=547)

Sketch

Let point $E$ be the intersection of lines $A B$ and $D M$. Triangles $C D M$ and $B E M$ are congruent because $|C M|=|B M|$, $\angle C M D=\angle B M E$ (vertical angles), and they are right triangles.

It follows that $|B E|=|C D|=a$.

In the right triangle $A E X$, the length $\overline{B X}$ is the median from vertex $X$ to the hypotenuse $A E$, so $|B X|=\frac{1}{2}|A E|=a$.

By the Pythagorean theorem, $|D E|=\sqrt{|A D|^{2}+|A E|^{2}}=a \sqrt{5}$.

The area of the right triangle $A E D$ is $P_{A E D}=\frac{a \cdot 2 a}{2}=a^{2}$.

The same area can be expressed using the hypotenuse $|D E|$ and the height $|A X|$:

$$
P_{A E D}=\frac{|D E| \cdot|A X|}{2}
$$

Equating the expressions for the area of triangle $A E D$, we get

$$
a^{2}=\frac{|A X| \cdot a \sqrt{5}}{2}
$$

or $|A X|=\frac{2 \sqrt{5}}{5} a$.

The perimeter of triangle $A B X$ is finally:

$$
o=2 a+\frac{2 \sqrt{5}}{5} a=2 a\left(1+\frac{\sqrt{5}}{5}\right)
$$