Question

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.

Answer 1

"∠DFE is congruent to ∠GFH."

Reason: G lies on DF and H lies on EF, so rays FD and FG are the same and rays FE and FH are the same. Thus the two angles at F are formed by the same pair of rays and are congruent. With DF/GF = EF/HF = 4 and this included-angle congruence, △DFE ~ △GFH by SAS similarity.