To prove that △DFE ~ △GFH by SAS similartiy theorem, then option C. ∠DFE is congruent to ∠GFH is appropriate. So that: [tex]\frac{DF}{GF}[/tex] = [tex]\frac{EF}{HF}[/tex] and ∠DFE is congruent to ∠GFH.
Given ΔDEF as shown in the diagram attached to this answer, the following can be observed:
By comparing ΔDEF and ΔGFH
DF = DG + GF
= 12 + 4
DF = 16
Also,
EF = EH + HF
= 9 + 3
EF = 12
Comparing the sides of ΔDEF and ΔGFH, we have;
[tex]\frac{DF}{GF}[/tex] = [tex]\frac{EF}{HF}[/tex]
[tex]\frac{16}{4}[/tex] = [tex]\frac{12}{3}[/tex]
4 = 4
Thus, the two triangles have similar sides.
Comparing the included angle <DFE and <GFH, then;
∠DFE is congruent to ∠GFH
So that the appropriate answer to the given question is option C. ∠DFE is congruent to ∠GFH
Therefore, to prove that △DFE ~ △GFH by the SAS similarity theorem;
[tex]\frac{DF}{GF}[/tex] = [tex]\frac{EF}{HF}[/tex] and ∠DFE is congruent to ∠GFH.
Visit: https://brainly.com/question/21782708
