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

Answer:

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

Step-by-step 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.

Answer 2

Answer:

Option B.

Step-by-step 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.

$\overline{FH}=\frac{1}{2}\times \overline{FE}$

Multiply both sides by 2.

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

Similarly,

$\overline{FG}=\frac{1}{2}\times \overline{FI}$

Multiply both sides by 2.

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

And,

$\overline{HG}=\frac{1}{2}\times \overline{EI}$

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.