# Show that ΔABC with vertices A(–2, 0), B(0, 2) and C(2, 0) is similar to ΔDEF with vertices D(–4, 0), F(4, 0) and E(0, 4).

Question:

Show that ΔABC with vertices A(–2, 0), B(0, 2) and C(2, 0) is similar to ΔDEF with vertices D(–4, 0), F(4, 0) and E(0, 4).

Solution:

In ΔABC, the coordinates of the vertices are A(–2, 0), B(0,2), C(2,0).

$\mathrm{AB}=\sqrt{(0+2)^{2}+(2-0)^{2}}=\sqrt{8}=2 \sqrt{2}$

$\mathrm{CB}=\sqrt{(0-2)^{2}+(2-0)^{2}}=\sqrt{8}=2 \sqrt{2}$

$\mathrm{AC}=\sqrt{(2+2)^{2}+(0-0)^{2}}=4$

In ΔDEF, the coordinates of the vertices are D(–4, 0), E(4, 0), F(0, 4).

$\mathrm{DF}=\sqrt{(4+4)^{2}+(0-0)^{2}}=8$

$\mathrm{FE}=\sqrt{(0-4)^{2}+(4-0)^{2}}=4 \sqrt{2}$

$\mathrm{DE}=\sqrt{(0+4)^{2}+(4-0)^{2}}=4 \sqrt{2}$

Now, for ΔABC and ΔDEF to be similar, the corresponding sides should be proportional.

So, $\frac{\mathrm{AC}}{\mathrm{DF}}=\frac{\mathrm{BC}}{\mathrm{FE}}=\frac{\mathrm{AB}}{\mathrm{DE}}$

$\Rightarrow \frac{4}{8}=\frac{2 \sqrt{2}}{4 \sqrt{2}}=\frac{2 \sqrt{2}}{4 \sqrt{2}}$

$\Rightarrow \frac{1}{2}=\frac{1}{2}=\frac{1}{2}$

Since, the corresponding sides are proportional

Therefore, given two triangles are similar.