Question

in the graph below, triangle abc is similar to triangle dec
which statements are true? select all that apply.
a. slope of \\( \overline { e c } = \\) slope of \\( \overline { c b } \\)
b. slope of \\( \overline { d e } = \\) slope of \\( \overline { a c } \\)
c. \\( \frac { d e } { c d } = \frac { b c } { a c } \\)
d. \\( \frac { c d } { d e } = \frac { c a } { a b } \\)

Explanation:

Step1: Recall the slope formula

The slope formula is \(m = \frac{y_2 - y_1}{x_2 - x_1}\). For similar triangles, the ratios of corresponding sides are equal. Also, since the line \(EB\) is a straight - line, the slope between any two points on the line is the same.
Let's assume the coordinates: \(A=(0,3)\), \(B=(3,3)\), \(C=(0,2)\), \(D=(0,1)\), \(E = (- 3,1)\)

• For the slope of \(\overline{EC}\): Using the formula \(m=\frac{y_2 - y_1}{x_2 - x_1}\), with \(E(-3,1)\) and \(C(0,2)\), \(m_{EC}=\frac{2 - 1}{0+3}=\frac{1}{3}\)
• For the slope of \(\overline{CB}\): Using the formula \(m=\frac{y_2 - y_1}{x_2 - x_1}\), with \(C(0,2)\) and \(B(3,3)\), \(m_{CB}=\frac{3 - 2}{3 - 0}=\frac{1}{3}\)

So, slope of \(\overline{EC}=\) slope of \(\overline{CB}\)

• For the slope of \(\overline{DE}\): Using the formula \(m=\frac{y_2 - y_1}{x_2 - x_1}\), with \(D(0,1)\) and \(E(-3,1)\), \(m_{DE}=\frac{1 - 1}{-3 - 0}=0\)
• For the slope of \(\overline{AC}\): Using the formula \(m=\frac{y_2 - y_1}{x_2 - x_1}\), with \(A(0,3)\) and \(C(0,2)\), \(m_{AC}=\frac{2 - 3}{0 - 0}\), which is undefined (vertical line)

• Since \(\triangle ABC\sim\triangle DEC\), by the property of similar triangles \(\frac{CD}{CA}=\frac{DE}{AB}\) (corresponding sides of similar triangles are in proportion)

\(CD = 1\), \(CA=1\), \(DE = 3\), \(AB = 3\), \(\frac{CD}{CA}=\frac{1}{1}\) and \(\frac{DE}{AB}=\frac{3}{3}\)

Answer:

A. slope of \(\overline{EC}=\) slope of \(\overline{CB}\); D. \(\frac{CD}{DE}=\frac{CA}{AB}\)