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 $\triangle AFG\sim\triangle ABC\sim\triangle ADE$. For similar triangles, the ratios of corresponding sides are equal. Consider $\triangle AFG$ and $\triangle ABC$. The ratio of the sides is $\frac{AF}{AB}=\frac{AG}{AC}$. Given $AB = 4$ and $AG=8$, $AC = 5$. Let $AF=x$. Then $\frac{x}{4}=\frac{8}{5}$, cross - multiply gives $5x = 32$, $x=\frac{32}{5}
eq10$. So statement 1 is false.

Step2: Use similar - triangle property for $CE$

Consider $\triangle ABC$ and $\triangle ADE$. Let $CE = y$. We know that $\frac{AB}{AD}=\frac{AC}{AE}$. Given $AB = 4$, $AD=4 + 12=16$, $AC = 5$, and $AE=5 + y$. Then $\frac{4}{16}=\frac{5}{5 + y}$. Cross - multiply: $4(5 + y)=16\times5$, $20+4y = 80$, $4y=60$, $y = 15$. So $CE = 15$ and statement 2 is true.

Step3: Use similar - triangle property for $DE$

Since $\triangle ABC\sim\triangle ADE$, $\frac{BC}{DE}=\frac{AB}{AD}$. Given $BC = 6$, $AB = 4$, $AD=16$. Then $\frac{6}{DE}=\frac{4}{16}$, cross - multiply: $4DE=6\times16$, $DE = 24
eq18$. So statement 3 is false.

Answer:

B. Only statement II