1. Problem: Let $A B C D$ be a convex quadrilateral such that $A B+B C=2021$ and $A D=C D$. We are also given that
$$
\angle A B C=\angle C D A=90^{\circ}
$$

Determine the length of the diagonal $B D$.
Solution: Since $A D=D C$ and $\angle A D C=90^{\circ}$, we can fit four copies of quadrilateral $A B C D$ around vertex $D$ as shown in the diagram.
The outer shape is a quadrilateral because $\angle D A B+\angle B C D=180^{\circ}$. Moreover it is a rectangle because $\angle A B C=90^{\circ}$. In fact it is a square with side-length 2021 because of rotational symmetry and $A B+B C=2021$. Also $D$ is the centre of the square because it is the centre of the rotational symmetry. So $B D$ is the distance from a vertex to the centre of the square, which is half the length of the diagonal of the square. Thus
$$
B D=\frac{1}{2}(2021 \sqrt{2})=\frac{2021}{\sqrt{2}} \text {. }
$$

Alternative Solution:
First let $x=A D=D C$ and $a=B D$ and $y=A B$ and $z=B C$. Now initially we can apply Pythagoras in triangles $C D A$ and $A B C$ to get $x^{2}+x^{2}=A C^{2}$ and $y^{2}+z^{2}=A C^{2}$ respectively. Putting this together gives us
$$
x^{2}=\frac{y^{2}+z^{2}}{2} \text {. }
$$

Now note that the opposite angles $\angle A B C$ and $\angle C D A$ (in quad $A B C D)$ are supplimentary. Therefore $A B C D$ is a cyclic quadrilateral. Equal chords subtend equal arcs (and chords $A D=D C$ are equal) so $\angle A B D=\angle D B C$. Furthermore, since $\angle A B C$ is a right angle, this means that $\angle A B D=\angle D B C=45^{\circ}$.

For any three points $P, Q$ and $R$, let $|P Q R|$ denote the area of triangle $P Q R$. Now consider the total area of quadrilateral $A B C D$ calculated in two ways:
$$
|A B C|+|C D A|=|A B D|+|D B C| \text {. }
$$

We calculate the areas of the right-angled triangles using the $\triangle=\frac{b h}{2}$ formula, and we calculate the area of the $45^{\circ}$-angled triangles using the $\triangle=\frac{1}{2} a b \sin C$ formula.
$$
\frac{y z}{2}+\frac{x^{2}}{2}=\frac{1}{2} a y \sin \left(45^{\circ}\right)+\frac{1}{2} a z \sin \left(45^{\circ}\right)
$$

At this point we can substitute $x^{2}=\frac{1}{2}\left(y^{2}+z^{2}\right)$ into this equation, and rearrange:
$$
\begin{aligned}
\frac{y z}{2}+\frac{x^{2}}{2} & =\frac{a y \sin \left(45^{\circ}\right)}{2}+\frac{a z \sin \left(45^{\circ}\right)}{2} \\
\frac{y z}{2}+\frac{y^{2}+z^{2}}{4} & =\frac{a y}{2 \sqrt{2}}+\frac{a z}{2 \sqrt{2}} \\
\frac{2 y z+y^{2}+z^{2}}{4} & =\frac{a y+a z}{2 \sqrt{2}} \\
\frac{(y+z)^{2}}{4} & =\frac{a(y+z)}{2 \sqrt{2}} \\
\frac{y+z}{\sqrt{2}} & =a .
\end{aligned}
$$

Finally since $y+z=2021$ this gives our final answer of $B D=a=\frac{2021}{\sqrt{2}}$.