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Triangle D E F is reflected across D F to form triangle E G F. The lengths of sides E F and F G are congruent. To prove that ΔDEF ≅ ΔDGF by SAS, what additional information is needed? ∠DEF ≅ ∠ DGF ∠DFE ≅ ∠ DFG DE ≅ DG DG ≅ GF

Sagot :

MrRoyalidn

Answer:

[tex]\angle DFE = \angle DFG[/tex]

Step-by-step explanation:

Given

See attachment for complete question

Required

What makes DEF and DGF congruent

We have:

[tex]EF = GF[/tex] --- this is indicated by the single line on both sides

Also:

[tex]DF = DF[/tex] --- both triangle share same side

For SAS to be true;

2 sides and 1 angle must be equal in either triangles

So far, we have:

[tex]EF = GF[/tex] ---- S

[tex]DF = DF[/tex] ---- S

The additional to complete the proof is:

[tex]\angle DFE = \angle DFG[/tex] ---- angle between the above sides

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Olayemiolakunle65idn

Reflection is a type of rigid transformation which requires the turning of an object, shape or figure about a reference point or line. Therefore, the needed additional information is ∠DFE ≅ ∠ DFG. Option B.

Reflection implies turning the given triangle DEF about its side DF, so as to produce an image with the same dimensions but different orientation.

The required proof by Side-Angle-Side (SAS) implies that the relations will be in respect of two of its sides and their included angle.

So that,

GF ≅ FE (given)

DF is the common side to triangles DEF and DFG.

DFG is the included side.

Thus;

∠DFE ≅ ∠ DFG (Side-Angle-Side postulate, SAS)

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