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
o ∠dfe is 4 times greater than ∠gfh
o ∠fhg is \\( \frac{1}{4} \\) the measure of ∠fed
o ∠dfe is congruent to ∠gfh
o ∠fhg is congruent to ∠efd

Explanation:

Step1: Calculate the ratio of sides

For \(\triangle DFE\) and \(\triangle GFH\), we have \(DF = DC+CF=12 + 4=16\), \(EF=EH + HF=9+3 = 12\).
The ratio of \(DF\) to \(GF\) is \(\frac{DF}{GF}=\frac{16}{4} = 4\), and the ratio of \(EF\) to \(HF\) is \(\frac{EF}{HF}=\frac{12}{3}=4\).

Step2: Check the included angle

\(\angle DFE\) and \(\angle GFH\) are the same angle (common angle).
By the SAS (Side - Angle - Side) similarity theorem, if \(\frac{DF}{GF}=\frac{EF}{HF}\) and the included angles are equal, then \(\triangle DFE\sim\triangle GFH\).

Answer:

\(\angle DFE\) is \(4\) times greater than \(\angle GFH\) is incorrect. The correct statement is that \(\triangle DFE\sim\triangle GFH\) by SAS similarity as \(\frac{DF}{GF}=\frac{16}{4} = 4\), \(\frac{EF}{HF}=\frac{12}{3}=4\) and \(\angle DFE=\angle GFH\) (common angle). So the answer is \(\angle DFE\) is congruent to \(\angle GFH\) is wrong. The correct is \(\triangle DFE\sim\triangle GFH\) by SAS similarity