+5265 Given: A(2, 4), B(-4, 2), C(4, -2), D(-1, 3), E(3, 1)
Prove: Triangle ABC is similar to Triangle ADE
+2498 what you need to find is distance from one point to another (sides of triangle):
\(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\ \text{ABC}_{triangle}\\ AB=\sqrt {(-6)^2 + (-2)^2}= 6.324555\\ AC=\sqrt {(4 - 2)^2 + (-2 - 4)^2}= 6.324555\\ BC=\sqrt {(4 - (-4))^2 + (-2 - 2)^2}=8.944272 \)
\(ADE_{triangle}\\ AD=\sqrt {(-1 - 2)^2 + (3 - 4)^2}=3.162278\\ AE=\sqrt {(3 - 2)^2 + (1 - 4)^2}=3.162278\\ DE=\sqrt {(3 - (-1))^2 + (1 - 3)^2}=4.472136\)
SO:
\(\frac{AB}{AD}=\frac{AC}{AE}=\frac{BC}{DE}=1.999999683772268\)
+2498 what you need to find is distance from one point to another (sides of triangle):
\(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\ \text{ABC}_{triangle}\\ AB=\sqrt {(-6)^2 + (-2)^2}= 6.324555\\ AC=\sqrt {(4 - 2)^2 + (-2 - 4)^2}= 6.324555\\ BC=\sqrt {(4 - (-4))^2 + (-2 - 2)^2}=8.944272 \)
\(ADE_{triangle}\\ AD=\sqrt {(-1 - 2)^2 + (3 - 4)^2}=3.162278\\ AE=\sqrt {(3 - 2)^2 + (1 - 4)^2}=3.162278\\ DE=\sqrt {(3 - (-1))^2 + (1 - 3)^2}=4.472136\)
SO:
\(\frac{AB}{AD}=\frac{AC}{AE}=\frac{BC}{DE}=1.999999683772268\)
+118608 Solveit,
Your answer is very good but you would have been better off to stick to exact values.
AD =sqrt (10)
AB = 2*sqrt(10)
so AB/AD = 2
I am sure that this would be true for the other ratios that you looked at as well.
So the triangles are similar AND the dimensions of the bigger one are exactly twice that of the smaller one :))
