4. Let $A B C D$ be a quadrilateral with $B C=C D=D A=1, \angle D A B=135^{\circ}$ and $\angle A B C=75^{\circ}$. Find $A B$.
(1 mark)
$A B C D$ 是四邊形，其中 $B C=C D=D A=1 、 \angle D A B=135^{\circ} 、 \angle A B C=75^{\circ}$ 。求 $A B$ 。
(1 分)
4. $\frac{\sqrt{6}-\sqrt{2}}{2}$
4. We claim that $A C=1$. If $A C>1$, then by considering $\triangle D A C$ we see that $\angle D A C$ must be smaller than $60^{\circ}$ (recall that in a triangle, the side opposite a larger angle is longer), while $\angle C A B$ must be smaller than $75^{\circ}$ by considering $\triangle A B C$. It leads to the contradiction $\angle D A B<135^{\circ}$. In a similar way, we can get a contradiction when $A C<1$, thereby establishing the claim. Now $\triangle C A B$ is isosceles with $C A=C B=1$ and base angles $75^{\circ}$. Therefore
$$
\begin{aligned}
A B & =2 \cos 75^{\circ} \\
& =2 \cos \left(45^{\circ}+30^{\circ}\right) \\
& =2\left(\cos 45^{\circ} \cos 30^{\circ}-\sin 45^{\circ} \sin 30^{\circ}\right) \\
& =\frac{\sqrt{6}-\sqrt{2}}{2}
\end{aligned}
$$