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In the figure below, ∠DEC ≅ ∠DCE, ∠B ≅ ∠F, and segment DF is congruent to segment BD. Point C is the point of intersection between segment AG and segment BD while point E is the point of intersection between segment AG and segment DF.


*look at image attached*


Prove ΔABC ≅ ΔGFE.

HELPI will give brainliestIn the figure below DEC DCE B F and segment DF is congruent to segment BD Point C is the point of intersection between segment AG and class=

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nevaehspratley2003

Answer:

∠ABC ≅ ∠DEC

∠DEC = ∠GEF

thus:

∠ABC = ∠GEF

∠GFE ≅ ∠DCE

∠DCE = ∠ACB

thus:

∠ACB = ∠GFE

If two of their angles are equal, then the third angle must also be equal, because angles of a triangle always add to make 180°.

thus: ΔABC ∼ ΔGEF by AA similar triangle postulate

I hope this helps

Prove [tex]\Delta ABC\cong \Delta GFE[/tex]

The sides opposite to equal angles are equal

∴[tex]DE=DC[/tex]

Subtract [tex]DC[/tex] and [tex]DE[/tex] from [tex]DB[/tex] and [tex]DF[/tex] respectively.

[tex]DB-DC=DF-DE[/tex]

We get, [tex]BC=EF[/tex]------Eq [tex]1[/tex]

[tex]\angle DEC\cong \angle DCE[/tex] (Given)

[tex]\angle DEC= \angle DCE[/tex] (Vertically opposite angle)

[tex]\angle DEC= \angle GEF[/tex] (Vertically opposite angle)

Hence, [tex]\angle ACB= \angle GEF[/tex]------Eq [tex]2[/tex]

In [tex]\Delta ABC[/tex] and [tex]\Delta GFE[/tex]

[tex]BC=EF[/tex] (From eq [tex]1[/tex])

[tex]\angle ACB= \angle GEF[/tex] ( From eq [tex]2[/tex])

[tex]\angle B\cong \angle F[/tex] (Given)

Therefore By AAS congruency [tex]\Delta ABC\cong \Delta GFE[/tex].

Hence Proved.

Learn More about congruent triangles here:

https://brainly.com/question/25063346?referrer=searchResults

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