## Task A-1.2.

Given is a right triangle $A B C$ with a right angle at vertex $C$ and side lengths $|A B|=$ $26,|B C|=24$. A semicircle is inscribed in it with the diameter on side $\overline{B C}$ which contains point $C$. The semicircle touches side $\overline{A B}$. What is the radius of the inscribed semicircle?
## First Solution.

Let $S$ be the center of the given semicircle, and $D$ the point where the semicircle touches the hypotenuse $\overline{A B}$. Let the desired radius be $|C S|=|D S|=r$.

![](https://cdn.mathpix.com/cropped/2024_05_30_a897c7e1109e2574c8d7g-02.jpg?height=400&width=811&top_left_y=654&top_left_x=560)

By the Pythagorean theorem, $|A B|^{2}=|A C|^{2}+|B C|^{2}$, or $26^{2}=|A C|^{2}+24^{2}$, from which it follows that $|A C|=10$.

Since $|C S|=|D S|$, $\varangle A C S=\varangle A D S=90^{\circ}$, and side $\overline{A S}$ is common, triangles $A C S$ and $A D S$ are congruent by the S-S-K theorem.

It follows that $|A D|=|A C|=10$.

Now, $|B D|=|A B|-|A D|=26-10=16$.

Applying the Pythagorean theorem to triangle $B D S$, we have $|B S|^{2}=16^{2}+r^{2}$.

We also have $|B S|=|B C|-|C S|=24-r$, so:

$$
16^{2}+r^{2}=(24-r)^{2}
$$

Rearranging gives $256+r^{2}=576-48 r+r^{2}$, or $48 r=576-256=320$.

Finally, we have $r=\frac{320}{48}=\frac{20}{3}$.

Note: The student does not need to prove the mentioned congruence; instead, they can write that $\overline{A D}$ and $\overline{A C}$ are of the same length because they are segments of tangents from the same point to the circle.