4. The diagram shows a square $P Q R S$ with sides of length 2 . The point $T$ is the midpoint of $R S$, and $U$ lies on $Q R$ so that $\angle S P T=\angle T P U$.
What is the length of $U R$ ?
Method 1
Let $F$ be the point on $P U$ so that $\angle T F P=90^{\circ}$ and join $T$ to $F$ and $U$, as shown. Then triangles $P T S$ and $P T F$ are congruent (AAS), so that $T F=1$.

Hence triangles $T U R$ and $T U F$ are congruent (RHS), so that $\angle R T U=\angle U T F$.

Now the four angles at $T$ are angles on the straight line $R T S$, so they add up to $180^{\circ}$. From the fact that $\angle T S P=90^{\circ}$ (because $P Q R S$ is a square), we have
$$
\left(90^{\circ}-\angle S P T\right)+\left(90^{\circ}-\angle T P F\right)+\angle R T U+\angle U T F=180^{\circ} \text {. }
$$

It follows that $\angle R T U=\angle S P T$.
Therefore triangles $R T U$ and $S P T$ are similar (AA), so that
$$
\frac{U R}{R T}=\frac{T S}{S P}=\frac{1}{2} .
$$

Thus $U R=\frac{1}{2}$.

Method 2
Let $\angle S P T$ and $\angle T P U$ both be equal to $\theta$ and join $T$ to $Q$ and $U$, as shown. Then triangles $T S P$ and $T R Q$ are congruent (SAS), so that $\angle T Q R$ is also equal to $\theta$.

From the converse of 'angles in the same segment' it follows that quadrilateral $U T P Q$ is cyclic. Hence, from 'opposite angles of a cyclic quadrilateral', we know that $\angle U T P$ is $90^{\circ}$, because $\angle P Q R$ is $90^{\circ}$.

Now the four angles at $T$ are angles on the straight line $R T S$, so they add up to $180^{\circ}$. It follows that $\angle R T U$ is equal to $\theta$.

Therefore triangles $R T U$ and SPT are similar (AA), so that
$$
\frac{U R}{R T}=\frac{T S}{S P}=\frac{1}{2} .
$$

Thus $U R=\frac{1}{2}$.