Question

1. (03.02 mc)$delta efi$ is dilated by a scale factor of $\frac{1}{2}$ with the center of dilation at point f. then, it is reflected over line a to create $delta hfg$. based on these transformations, which statement is true? (1 point)$circ overline{fh}=\frac{1}{2}overline{fi}, overline{fg}=\frac{1}{2}overline{fe},$ and $overline{hg}=\frac{1}{2}overline{ei}; delta efi sim delta hfgLXB0circ overline{fh}=\frac{1}{2}overline{fi}, overline{fg}=\frac{1}{2}overline{fe},$ and $overline{hg}=overline{ei}; delta efi sim delta gfh$$circ overline{fh}=overline{fi}, overline{fg}=overline{fe},$ and $overline{hg}=overline{ei}; delta efi sim delta gfh$

Explanation:

Step1: Analyze dilation scale factor

Dilation scale factor is $\frac{1}{2}$ from center $F$, so $\overline{FH} = \frac{1}{2}\overline{FE}$, $\overline{FG} = \frac{1}{2}\overline{FI}$, $\overline{HG} = \frac{1}{2}\overline{EI}$.

Step2: Analyze reflection over line a

Reflection preserves segment lengths, so the segment relationships stay $\overline{FH} = \frac{1}{2}\overline{FE}$, $\overline{FG} = \frac{1}{2}\overline{FI}$, $\overline{HG} = \frac{1}{2}\overline{EI}$. Reflection and dilation are similarity transformations, so $\triangle EFI \sim \triangle HFG$.

Step3: Match to correct option

The relationships and similarity match the second option.

Answer:

$\boldsymbol{\overline{FH} = \frac{1}{2}\overline{FE}, \overline{FG} = \frac{1}{2}\overline{FI}, \text{and } \overline{HG} = \frac{1}{2}\overline{EI}; \triangle EFI \sim \triangle GFH}$