Question

figure 1
figure 2
figure 3
figure 4
in which figure is \\(\overline{de} \parallel \overline{bc}\\)?
a. figure 1
b. figure 2
c. figure 3
d. figure 4

Explanation:

Step1: Recall the Basic Proportionality Theorem (Thales' theorem)

The theorem states that if a line is drawn parallel to one side of a triangle, intersecting the other two sides, then it divides those sides proportionally. So, for \(\overline{DE} \parallel \overline{BC}\), we should have \(\frac{AD}{DB}=\frac{AE}{EC}\) (or \(\frac{AD}{AB}=\frac{AE}{AC}\)).

Step2: Check Figure 1

• \(AD = 6\), \(DB = 15\), so \(\frac{AD}{DB}=\frac{6}{15}=\frac{2}{5}\)
• \(AE = 6\) (Wait, no, looking at the figure, \(AE\) and \(EC\): Wait, the length from \(A\) to \(E\) is 6? Wait, no, the right side: \(A\) to \(E\) is 6? Wait, no, the vertical side: \(A\) to \(E\) is 6, \(E\) to \(C\) is 17.5? Wait, no, maybe I misread. Wait, \(AD = 6\), \(DB = 15\), so \(AB=AD + DB=6 + 15 = 21\). \(AE\): Wait, the top segment \(AE\) is 6? Wait, no, the vertical side: \(A\) to \(E\) is 6, \(E\) to \(C\) is 17.5? Wait, no, maybe the labels: \(A\) to \(D\) is 6, \(D\) to \(B\) is 15. \(A\) to \(E\) is 6, \(E\) to \(C\) is 17.5? Wait, then \(AE = 6\), \(EC = 17.5\). Then \(\frac{AD}{DB}=\frac{6}{15}=\frac{2}{5}\), \(\frac{AE}{EC}=\frac{6}{17.5}=\frac{60}{175}=\frac{12}{35}\). \(\frac{2}{5}=0.4\), \(\frac{12}{35}\approx0.342\). Not equal. So Figure 1: Not parallel.

Step3: Check Figure 2

• \(AD = 5\), \(DB = 12\), so \(\frac{AD}{DB}=\frac{5}{12}\approx0.4167\)
• \(AE = 6\), \(EC = 15\), so \(\frac{AE}{EC}=\frac{6}{15}=\frac{2}{5}=0.4\). Not equal. So Figure 2: Not parallel.

Step4: Check Figure 3

• \(AD = 4\), \(DB = 12\), so \(\frac{AD}{DB}=\frac{4}{12}=\frac{1}{3}\)
• \(AE = 5\), \(EC = 16\), so \(\frac{AE}{EC}=\frac{5}{16}\approx0.3125\). \(\frac{1}{3}\approx0.333\), not equal. So Figure 3: Not parallel.

Step5: Check Figure 4

• \(AD = 5.5\) (Wait, the label: \(AD = 5.5\)? Wait, the left side: \(AD = 5.5\), \(DB = 15\), so \(AB = AD + DB=5.5 + 15 = 20.5\)? Wait, no, wait the labels: \(AD = 5.5\), \(DB = 15\), so \(\frac{AD}{DB}=\frac{5.5}{15}=\frac{11}{30}\approx0.3667\)
• \(AE = 6.6\), \(EC = 18\), so \(\frac{AE}{EC}=\frac{6.6}{18}=\frac{11}{30}\approx0.3667\). Ah, here \(\frac{AD}{DB}=\frac{AE}{EC}\). So by the Basic Proportionality Theorem, \(\overline{DE} \parallel \overline{BC}\) in Figure 4.

Answer:

D. figure 4