Question

Given: g ∥ h and ∠2 ≅ ∠3

Answer 1

Answer:

D

Step-by-step explanation:

Ed2020

Answer 2

Converse alternate interior angles theorem.

Answer:

In the image, you can observe a diagram representing this problem.

We know by given that $g \parallel h$ and $\angle 2 \cong \angle 3$.

From the parallelism between line g and line h, we deduct several congruence between angles.

$\angle 2 \cong \angle 1$, by corresponding angles (same side of the transversal, one interior, the other exterior to parallels).

Now, to demonstrate $e \parallel f$, 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

$\angle 2 + \angle 3+ 180-\angle 1 + x =360$

Where $x$ is the angle at the intersection line g and line f.

But, we know $\angle 2 \cong \angle 3$ and $\angle 2 \cong \angle 1$, so

$\angle 2 + \angle 2 +180 - \angle 2 +x=360\\\angle 2 + x=180$

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.

$\therefore e \parallel f$

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.