Answer :
Converse alternate interior angles theorem.
Answer:
In the image, you can observe a diagram representing this problem.
We know by given that [tex]g \parallel h[/tex] and [tex]\angle 2 \cong \angle 3[/tex].
From the parallelism between line g and line h, we deduct several congruence between angles.
[tex]\angle 2 \cong \angle 1[/tex], by corresponding angles (same side of the transversal, one interior, the other exterior to parallels).
Now, to demonstrate [tex]e \parallel f[/tex], we must demonstrate a congruence between angle 2 and an angle on the intersection between line g and line f.
In the parallelogram formed, we know
[tex]\angle 2 + \angle 3+ 180-\angle 1 + x =360[/tex]
Where [tex]x[/tex] is the angle at the intersection line g and line f.
But, we know [tex]\angle 2 \cong \angle 3[/tex] and [tex]\angle 2 \cong \angle 1[/tex], so
[tex]\angle 2 + \angle 2 +180 - \angle 2 +x=360\\\angle 2 + x=180[/tex]
Notice that we don't have a congruence, however there's theorem which states that the same-side interior angles of parallels are supplementary.
In this case, we use the corolary of that theorem, which states if two same-side interior angles are supplementary, then the lines are parallels.
[tex]\therefore e \parallel f[/tex]
However, according to the choices of the problem, the missin proof is "converse alternate interior angles theorem", because the problem was demonstrate using transitive property, to show that angles 1 and 3 are congruent, there by converse alternate interior angles theorem, lines e and f are parallels.
This is the same case we used, but using converse alternate interior angles theorem.
Answer:
the answer is c-
Step-by-step explanation:
converse alternate interior angles theorem
hope this is helpful
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