證明:設(shè)?$\frac {AB}{A'B'}=\frac {BC}{B'C'}=k$?
則?$AB=kA'B',$??$BC=kB'C'$?
∵?$∠C=∠C'=90°$?
∴?$AC2= AB2- BC2,$??$A'C'2= A'B'2- B'C'2$?
∴?$AC2=k2(A'B'2-B'C'2)= k2A'C'2$?
∴?$AC= kA'C'$?
∴?$\frac {AB}{A'B'}=\frac {AC}{A'C'}=\frac {BC}{B'C'}$?
∴?$Rt△ABC∽Rt△A'B'C'$?