Question

in the following diagram, $overline{fg}paralleloverline{bc}paralleloverline{de}$, $b$ is on $overline{ad}$; $c$ is on $overline{ae}$; $a$ is on both $overline{ef}$ and $overline{dg}$. consider the following statements: 1. $af = 10$ 2. $ce = 15$ 3. $de = 18$ which must be true? a. only statement i b. only statement ii c only statements i and ii d. only statements ii and iii e statements i, ii, and iii

Explanation:

Step1: Use similar - triangle property

Since $\overline{FG}\parallel\overline{BC}\parallel\overline{DE}$, we have similar triangles. $\triangle AFG\sim\triangle ABC\sim\triangle ADE$. For $\triangle AFG$ and $\triangle ABC$, the ratio of their sides is given by $\frac{AF}{AB}=\frac{AG}{AC}$. We know that $\frac{AF}{4}=\frac{8}{5}$, so $AF=\frac{4\times8}{5}=6.4
eq10$.

Step2: For $\triangle ABC$ and $\triangle ADE$

The ratio of corresponding sides of similar triangles $\triangle ABC$ and $\triangle ADE$ is $\frac{AB}{AD}=\frac{AC}{AE}=\frac{BC}{DE}$. We know that $\frac{AB}{AB + 12}=\frac{AC}{AC+CE}=\frac{6}{DE}$. Also, from $\triangle ABC$ and $\triangle AFG$, we can find the ratio of sides. Since $\frac{AC}{AG}=\frac{AB}{AF}$, and using the fact that for $\triangle ABC$ and $\triangle ADE$, $\frac{AC}{AE}=\frac{5}{5 + CE}=\frac{4}{4 + 12}$. Cross - multiply: $4(5 + CE)=5\times16$. Expand: $20+4CE = 80$. Solve for $CE$: $4CE=60$, so $CE = 15$.

Step3: For the length of $DE$

Using the ratio of sides of $\triangle ABC$ and $\triangle ADE$, $\frac{BC}{DE}=\frac{AB}{AD}$. We have $\frac{6}{DE}=\frac{4}{4 + 12}$, cross - multiply: $4DE=6\times16$, so $DE = 24
eq18$.

Answer:

B. Only statement II