Question

ΔEFI is dilated by a scale factor of one half with the center of dilation at point F. Then, it is reflected over line a to create ΔHFG. Based on these transformations, which statement is true?

Line segments EG and HI intersect at point F, forming triangles EFI and HFG. Line a intersects with both triangles at point F.

segment FH = two segment FE, segment FG = two segment FI, and segment HG = two segment EI; ΔEFI ~ ΔHFG
two segment FH = segment FE, two segment FG = segment FI, and two segment HG = segment EI; ΔEFI ~ ΔHFG
segment FH = two segment FI, segment FG = two segment FE, and segment HG = two segment EI; ΔEFI ~ ΔGFH
two segment FH = segment FI, two segment FG = segment FE, and two segment HG = segment EI; ΔEFI ~ ΔGFH

Answer 1

Option B.

Explanation:

It is given that ΔEFI is dilated by a scale factor of one half with the center of dilation at point F. Then, it is reflected over line a to create ΔHFG.

`\overline{FH}=(1)/(2)* \overline{FE}`

Multiply both sides by 2.

`2\overline {FH}=\overline{FE}`

Similarly,

`\overline{FG}=(1)/(2)* \overline{FI}`

Multiply both sides by 2.

`2\overline {FG}=\overline{FI}`

And,

`\overline{HG}=(1)/(2)* \overline{EI}`

Multiply both sides by 2.

`2\overline {HG}=\overline{EI}`

By SSS property of similarity,

ΔEFI ~ ΔGFH

Since,
`2\overline {FH}=\overline{FE}`,
`2\overline {FG}=\overline{FI}`,
`2\overline {HG}=\overline{EI}`, therefore ΔEFI ~ ΔGFH.

Therefore, the correct option is B.

Answer 2

two segment FH = segment FE, two segment FG = segment FI, and two segment HG = segment EI; ΔEFI ~ ΔHFG

Explanation:

Dilation by a factor of 1/2 means twice the reduced segment is equal to the original.

The reflection maps E'F to HF, so ΔEFI ~ ΔHFG.