China Girls' Mathematical Olympiad

Without loss of generality, we may assume that $\angle C>90^{\circ }$. Since $\angle A+\angle B<90^{\circ }$, we may assume without loss of generality that $\angle A<45^{\circ }$.

We can then construct a semicircle $\omega$ with $AB$ as its diameter such that point $C$ lies inside $\omega$. Let $O$ be the center of $\omega$. Then $AO=BO$. Construct rays $AT$ and $OE$ such that $\angle BAT=\angle BOE=45^{\circ }$ with $E$ lying on $\omega$. Let $l$ be the line tangent to $\omega$ at $E$, and let $l$ meet rays $AT$ and $AB$ at $F$ and $D$ respectively. It is not difficult to see that triangle $AFD$ is an isosceles right-angled triangle, with $\angle F=90^{\circ }$, that covers triangle $ABC$.

It suffices to show that $AD<\sqrt{2}+1$. Note that $$AD=AO+OD=AO+\sqrt{2}OE=(\sqrt{2}+1)AO.$$ Applying the sine rule to triangle $ABC$ gives $AB=2\sin C<2$, or $AO<1$. It follows that $AD<\sqrt{2}+1$, as desired.