In an isosceles right triangle $ABC$, on the extension of the hypotenuse $AB$ beyond point $B$, a segment $BD$ equal to $BC$ is laid out, and point $D$ is connected to $C$. Find the sides of triangle $ADC$, if the leg $BC=a$.

#
From the right triangle $A B C$, we find that

$$
A B=\frac{B C}{\cos 45^{\circ}}=a \sqrt{2}, A D=A B+B D=a \sqrt{2}+a=a(\sqrt{2}+1)
$$

By the Law of Cosines in triangle $C B D$, we find that

$$
\begin{gathered}
C D^{2}=B C^{2}+B D^{2}-2 B C \cdot B D \cos \angle C B D= \\
=a^{2}+a^{2}-2 a^{2} \cos \left(180^{\circ}-45^{\circ}\right)=2 a^{2}+a^{2} \sqrt{2}=a^{2}(2+\sqrt{2})
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_845c09db4774ba28be39g-19.jpg?height=497&width=532&top_left_y=1515&top_left_x=769)

## Answer

Apply the Law of Cosines.