In the diagram, DG = 12, GF = 4, EH = 9, and HF = 3. Triangle D E F is shown. Line G H is drawn parallel to side D E within the triangle to form triangle G F H. The length of D G is 12, the length of G F is 4, the length of E H is 9, and the length of H F is 3. To prove that △DFE ~ △GFH by the SAS similarity theorem, it can be stated that StartFraction D F Over G F EndFraction = StartFraction E F Over H F EndFraction and ∠DFE is 4 times greater than ∠GFH. ∠FHG is One-fourth the measure of ∠FED. ∠DFE is congruent to ∠GFH. ∠FHG is congruent to ∠EFD.

To prove that △DFE ~ △GFH by SAS similartiy theorem, then option C. ∠DFE is congruent to ∠GFH is appropriate. So that: = 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;

=

=

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;

= and ∠DFE is congruent to ∠GFH.

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