Question

Show that ∆ ABC with vertices A (–2, 0), B (0, 2) and C (2, 0) is similar to ∆ DEF with vertices D (–4, 0), F (4, 0) and E (0, 4) ?

Answer 1

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

\[AB = \sqrt{\left( - 2 - 0 \right)^2 + \left( 0 - 2 \right)^2} =\sqrt{8} = 2\sqrt{2}\]

\[BC = \sqrt{\left( 0 - 2 \right)^2 + \left( 2 - 0 \right)^2} = \sqrt{8} = 2\sqrt{2}\]

\[AC = \sqrt{\left( - 2 - 2 \right)^2 + \left( 0 - 0 \right)^2} = \sqrt{16} = 4\]

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

\[DE = \sqrt{\left( - 4 - 0 \right)^2 + \left( 0 - 4 \right)^2} = \sqrt{32} = 4\sqrt{2}\]
\[EF = \sqrt{\left( 0 - 4 \right)^2 + \left( 4 - 0 \right)^2} = \sqrt{32} = 4\sqrt{2}\]
\[DF = \sqrt{\left( - 4 - 4 \right)^2 + \left( 0 - 0 \right)^2} = \sqrt{64} = 8\]

For ΔABC and ΔDEF to be similar, their corresponding sides should be proportional.

Now,

\[\frac{AB}{DE} = \frac{2\sqrt{2}}{4\sqrt{2}} = \frac{1}{2}\]

\[\frac{BC}{EF} = \frac{2\sqrt{2}}{4\sqrt{2}} = \frac{1}{2}\]

\[\frac{AC}{DF} = \frac{4}{8} = \frac{1}{2}\]

\[So, \frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF}\]

Thus, ΔABC is similar to ΔDEF.