parmenides51 wrote: In a right triangle $ABC$ ($AC = 90^o$) it is known that $AC = 4$ cm, $BC = 3$ cm. Which angle ? $AC = 90^o$

$\angle C=90^o$

Let $A(4,0),B(0,3),C(0,0)$. Point $A_{1}(4,a)$, then the line $BB_{1}$ has equation $y=\frac{a}{4}x+3$. Point $B_{1}(b,\frac{ab}{4}+3)$, then the line $CC_{1}$ has equation $y=\frac{\frac{ab}{4}+3-a}{b-4} \cdot x$. Point $C_{1}$. With the two conditions, we find $a=8\sqrt{2}$ and $b=\frac{4-2\sqrt{2}}{3}$. $CA_{1}B_{1}C_{1}$ is a rectangle with $CC_{1}=1$ and $B_{1}C_{1}=12$.