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Using Coordinate Geometry to Prove Triangles Similar

Use coordinate geometry to prove ΔABC ~ ΔADE.

$If\ BC||DE,\ then\ \frac{AB}{BD}=\frac{AC}{CE}.$

$(x_{1},\ y_{1})\rightarrow (0,\ 3)$

$(x_{2},\ y_{2})\rightarrow (6,\ 0)$

$\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{0-3}{6-0}=\frac{-3}{6}=-\frac{1}{2}$

$(x_{1},\ y_{1})\rightarrow (0,\ 5)$

$(x_{2},\ y_{2})\rightarrow (10,\ 0)$

$\frac{y_{2}-y{1}}{x_{2}-x_{1}}=\frac{0-5}{10-0}=\frac{-5}{10}=-\frac{1}{2}$

$\frac{3}{2}=\frac{6}{4}$

$4(3)=2(6)$

$12=12$