Answer with explanation:
To prove two triangles are similar using SSS Similarity Criterion we need to prove that sides of triangles are Proportional.
Sides of two triangles can be obtained using Distance formula.
Coordinates of ΔABC are= A(0,0) , B(6,3) and C(-3,3)
Coordinates of ΔFGH are=F(-3,-2),G(-1,-3) and H(-4, -3)
[tex]AB=\sqrt{(6-0)^2+(3-0)^2}\\\\=\sqrt{36+9}\\\\=\sqrt{45}\\\\=5\sqrt{3}\\\\BC=\sqrt{(6+3)^2+(3-3)^2}\\\\BC=9\\\\AC=\sqrt{(-3-0)^2+(3-0)^2}\\\\AC=\sqrt{9+9}\\\\=\sqrt{18}\\\\=3\sqrt{2}\\\\FG=\sqrt{(-3+1)^2+(-2+3)^2}\\\\FG=\sqrt{4+1}\\\\FG=\sqrt{5}\\\\GH=\sqrt{(-1+4)^2+(-3+3)^2}\\\\GH=3\\\\HF=\sqrt{(-4+3)^2+(-3+2)^2}\\\\HF=\sqrt{1+1}\\\\HF=\sqrt{2}[/tex]
Ratio of Corresponding sides are
[tex]1.\rightarrow \frac{AB}{FG}=\frac{3\sqrt{5}}{\sqrt{5}}\\\\=3\\\\2.\rightarrow \frac{AC}{FH}=\frac{3\sqrt{2}}{\sqrt{2}}\\\\=3\\\\3.\rightarrow \frac{CB}{HG}=\frac{9}{3}\\\\=3\\\\\rightarrow \frac{AB}{FG}=\frac{AC}{FH}=\frac{CB}{HG}[/tex]
As corresponding sides are proportional, so trinagles are Similar.
⇒ ΔABC ≅ ΔFGH--------[SSS]
Hence proved.
