10. We use the following lemma. Lemma. Let $A B C$ be a triangle and $X \in A B$ such that $\overrightarrow{A X}: \overrightarrow{X B}=m: n$. Then $(m+n) \cot \angle C X B=n \cot A-m \cot B$ and $m \cot \angle A C X=$ $(n+m) \cot C+n \cot A$. Proof. Let $C D$ be the altitude from $C$ and $h$ its length. Then using oriented segments we have $A X=A D+D X=h \cot A-h \cot \angle C X B$ and $B X=B D+D X=h \cot B+h \cot \angle C X B$. The first formula in the lemma now follows from $n \cdot A X=m \cdot B X$. The second formula immediately follows from the first part applied to the triangle $A C X$ and the point $X^{\prime} \in A C$ such that $X X^{\prime} \| B C$. Let us set $\cot A=x, \cot B=y$, and $\cot C=z$. Applying the second formula in the lemma to $\triangle A B C$ and the point $X$, we obtain $4 \cot \angle A C X=$ $9 z+5 x$. Applying the first formula in the lemma to $\triangle C X Z$ and the point $Y$ and using $\angle X Y Z=45^{\circ}$ and $\cot \angle C X Z=-y$, we obtain $3 \cot \angle X Y Z=$ $\cot \angle A C X-2 \cot \angle C X Z=\frac{9 z+5 x}{4}+2 y \Rightarrow 5 x+8 y+9 z=12$. We now use the well-known relation for cotangents of a triangle $x y+y z+$ $x z=1$ to get $9=9(x+y) z+9 x y=(x+y)(12-5 x-8 z)+9 x y=9 \Rightarrow$ $(4 y+x-3)^{2}+9(x-1)^{2}=0 \Rightarrow x=1, y=\frac{1}{2}, z=\frac{1}{3}$. It follows that $x, y$, and $z$ have fixed values, and hence all triangles $T$ in $\Sigma$ are similar, with their smallest angle $A$ having cotangent 1 and thus being equal to $\angle A=45^{\circ}$.