[tex]\begin{gathered} \frac{BF}{BC}=\frac{BE}{BD},\text{ hence the triangles are similar} \\ U\sin g\text{ distance formula we can find the value of }BF,\text{ BC, BE and BD} \\ B(1,7),\text{ }F(-5,-2),E(4,-8),C(-1,4)andD(2,2)\text{ } \\ BF=\sqrt[]{(1-(-5))^2+(7-(-2))^2} \\ BF=\sqrt{6^2+9^2} \\ BF=10.81 \\ BE=\sqrt{(1-4)^2+(7-(-8))^2} \\ BE=\sqrt{3^2+15^2} \\ BE=15.3 \\ BC=\sqrt{(1-(-1))^2+(7-4)^2} \\ BC=\sqrt{2^2+3^2} \\ BC=3.6 \\ BD=\sqrt{(1-2)^2+(7-2)^2} \\ BD=4.69 \\ \frac{BF}{BC}=\frac{BE}{BD}=\frac{10.81}{3.6}=\frac{15.3}{4.69}\approx3,\text{ hence the triangles are similar} \end{gathered}[/tex]
