For each $n \geq 2$. Let $\alpha+\beta < \frac{\pi}{2}$, we denote by $\triangle(\alpha,\beta)$ a triangle with angles $\alpha,\beta$. We write $\triangle_1 = \triangle_2+\triangle_3$ if $\triangle_1$ can be cut into two triangles similar to $\triangle_2,\triangle_3$ respectively. Let us take small enough $\alpha << 1$. Our triangles are $\triangle(\alpha,\alpha),\triangle(\alpha,2\alpha),\ldots,\triangle(\alpha,n\alpha)$. Using one simple cut we can easily get $\triangle(\alpha,k\alpha) = \triangle(\alpha,(k+1)\alpha)+\triangle(\alpha,k\alpha)$ and $\triangle(\alpha,k\alpha) = \triangle(\alpha,(k-1)\alpha)+\triangle(\alpha,k\alpha)$, for each $k\leq n$. Using this formulas recursively we get $\triangle(\alpha,k\alpha) = \triangle(\alpha,\alpha)+\triangle(\alpha,2\alpha)+\cdots+\triangle(\alpha,n\alpha)$, for each $1 \leq k \leq n$.