QUESTION IMAGE
Question
in the diagram, dg = 12, gf = 4, eh = 9, and hf = 3. to prove that △dfe ~ △gfh by the sas similarity theorem, it can be stated that \\( \frac{df}{gf}=\frac{ef}{hf} \\) and
∠dfe is 4 times greater than ∠gfh.
∠fhg is \\( \frac{1}{4} \\) the measure of ∠fed.
∠dfe is congruent to ∠gfh.
∠fhg is congruent to ∠efd.
Step1: Calculate DF and EF
DF = DG + GF = 12 + 4 = 16, EF = EH + HF = 9 + 3 = 12.
Step2: Verify the ratio
$\frac{DF}{GF}=\frac{16}{4}=4$, $\frac{EF}{HF}=\frac{12}{3}=4$, so $\frac{DF}{GF}=\frac{EF}{HF}$.
Step3: Check the included angle
For SAS similarity theorem, the included angle between the two sides must be congruent. In $\triangle DFE$ and $\triangle GFH$, the included angle for the sides $\frac{DF}{GF}$ and $\frac{EF}{HF}$ is ∠DFE and ∠GFH respectively. So ∠DFE must be congruent to ∠GFH.
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C. ∠DFE is congruent to ∠GFH.