Question

use the given information to prove that $\angle 4 \cong \angle 3$.
you are not allowed to use the transitive property for parallel lines.
(you may use the transitive property for \=\ or \$\cong$\, though.)

given: $\overline{hj} \parallel \overline{fg}$
$\angle 8 \cong \angle 1$
$\overline{de} \parallel \overline{fg}$
prove: $\angle 4 \cong \angle 3$

step statement reason
2 $\angle 8 \cong \angle 1$ given
3 $\overline{de} \parallel \overline{fg}$ given
4 $\square$ reason?

Explanation:

Step1: Identify Alternate Interior Angles

Since \( \overline{DE} \parallel \overline{FG} \) (given), and the transversal is \( \overline{EF} \) (the line connecting \( E \) to \( F \)), \( \angle 2 \) and \( \angle 3 \) are alternate interior angles. By the Alternate Interior Angles Theorem, \( \angle 2 \cong \angle 3 \). Also, since \( \overline{HJ} \parallel \overline{FG} \) (given), and transversal \( \overline{FB} \), \( \angle 8 \) and \( \angle 4 \) are alternate interior angles, so \( \angle 8 \cong \angle 4 \). But we know \( \angle 8 \cong \angle 1 \) (given), so by Transitive Property of Congruence, \( \angle 1 \cong \angle 4 \). Now, from \( \overline{DE} \parallel \overline{FG} \), \( \angle 1 \) and \( \angle 4 \) - wait, no, let's correct. Wait, \( \overline{DE} \parallel \overline{FG} \), so \( \angle 1 \) and \( \angle 3 \) - no, earlier step. Wait, let's re - establish:

We know \( \overline{DE} \parallel \overline{FG} \), so when transversal \( \overline{EF} \) cuts them, \( \angle 2 \) and \( \angle 3 \) are alternate interior angles (\( \angle 2\cong\angle 3 \)). Also, \( \overline{HJ} \parallel \overline{FG} \), transversal \( \overline{FB} \), so \( \angle 8\cong\angle 4 \) (alternate interior angles). Given \( \angle 8\cong\angle 1 \), so \( \angle 1\cong\angle 4 \) (transitive). Now, since \( \overline{DE} \parallel \overline{FG} \), and transversal \( \overline{EF} \), \( \angle 1 \) and \( \angle 3 \) - no, wait \( \angle 1 \) and \( \angle 2 \) are adjacent angles? Wait, no, the line \( EF \) creates \( \angle 1 \) and \( \angle 2 \) as a linear pair? No, \( D - E \) is a straight line, \( F \) is a point, \( EF \) is a segment. Wait, maybe a better approach:

Since \( DE\parallel FG \), \( \angle 1=\angle 3 \) (alternate interior angles, transversal \( EF \)). And since \( \angle 8 = \angle 1 \) (given) and \( \angle 8=\angle 4 \) (alternate interior angles, \( HJ\parallel FG \), transversal \( FB \)), then \( \angle 1=\angle 4 \) (transitive). Then, since \( \angle 1=\angle 3 \) (from \( DE\parallel FG \)), by transitive property \( \angle 4=\angle 3 \).

But let's structure the steps properly for the proof table (step 4):

We have \( \overline{DE} \parallel \overline{FG} \) (given). So, by the Alternate Interior Angles Theorem, when a transversal (here \( \overline{EF} \)) intersects two parallel lines (\( \overline{DE} \) and \( \overline{FG} \)), alternate interior angles are congruent. Also, we know \( \angle 8\cong\angle 1 \) (given) and \( \overline{HJ} \parallel \overline{FG} \) (given) which gives \( \angle 8\cong\angle 4 \) (Alternate Interior Angles Theorem). Then by Transitive Property of Congruence, \( \angle 1\cong\angle 4 \). Then, from \( \overline{DE} \parallel \overline{FG} \), \( \angle 1\cong\angle 3 \) (Alternate Interior Angles Theorem). Then by Transitive Property of Congruence, \( \angle 4\cong\angle 3 \).

But for step 4, let's assume we are building the proof:

Statement for step 4: \( \angle 1\cong\angle 3 \) (because \( \overline{DE}\parallel\overline{FG} \), Alternate Interior Angles Theorem) and \( \angle 1\cong\angle 4 \) (from \( \angle 8\cong\angle 1 \) and \( \angle 8\cong\angle 4 \)), then \( \angle 4\cong\angle 3 \). But let's focus on the step after the given steps.

Wait, the given steps in the table: step 2 is \( \angle 8\cong\angle 1 \) (given), step 3 is \( \overline{DE}\parallel\overline{FG} \) (given). So step 4: We can say \( \angle 1\cong\angle 3 \) (Alternate Interior Angles Theorem, since \( \overline{DE}\parallel\overline{FG} \) and transversal \( EF \)). Also, \( \angle 8\cong\angle 4 \) (Alternate Interior Angles Theorem, \( \overline{HJ}\parallel\overline{FG} \), transversal \( FB \)). Since \( \angle 8\cong\angle 1 \), then \( \angle 1\cong\angle 4 \) (Transitive Property of Congruence). Then, since \( \angle 1\cong\angle 3 \) and \( \angle 1\cong\angle 4 \), by Transitive Property of Congruence, \( \angle 4\cong\angle 3 \).

But to fill step 4 statement: Let's see, after step 3 (\( \overline{DE}\parallel\overline{FG} \) given), we can use Alternate Interior Angles Theorem to say \( \angle 1\cong\angle 3 \) (because \( DE\parallel FG \), transversal \( EF \), so \( \angle 1 \) and \( \angle 3 \) are alternate interior angles). Also, from \( \angle 8\cong\angle 1 \) (step 2) and \( \overline{HJ}\parallel\overline{FG} \) (given) which gives \( \angle 8\cong\angle 4 \) (Alternate Interior Angles Theorem), so \( \angle 1\cong\angle 4 \) (Transitive). Then, since \( \angle 1\cong\angle 3 \) and \( \angle 1\cong\angle 4 \), \( \angle 4\cong\angle 3 \).

So step 4 statement: \( \angle 1\cong\angle 3 \) (Reason: Alternate Interior Angles Theorem, since \( DE\parallel FG \)) and then later steps, but for the first part after step 3, the key is to use the parallel lines \( DE\parallel FG \) to get a congruence related to \( \angle 1 \) or \( \angle 4 \).

Wait, maybe a simpler path:

1. \( \overline{HJ}\parallel\overline{FG} \) (Given)
2. \( \angle 8\cong\angle 1 \) (Given)
3. \( \overline{DE}\parallel\overline{FG} \) (Given)
4. \( \angle 8\cong\angle 4 \) (Alternate Interior Angles Theorem, \( \overline{HJ}\parallel\overline{FG} \), transversal \( FB \))
5. \( \angle 1\cong\angle 4 \) (Transitive Property of Congruence, from steps 2 and 4)
6. \( \angle 1\cong\angle 3 \) (Alternate Interior Angles Theorem, \( \overline{DE}\parallel\overline{FG} \), transversal \( EF \))
7. \( \angle 4\cong\angle 3 \) (Transitive Property of Congruence, from steps 5 and 6)

So in step 4 (the one in the table, after step 3), the statement could be \( \angle 8\cong\angle 4 \) with reason Alternate Interior Angles Theorem (since \( \overline{HJ}\parallel\overline{FG} \)). But the table has step 2 as \( \angle 8\cong\angle 1 \) (given), step 3 as \( \overline{DE}\parallel\overline{FG} \) (given). So step 4:

Statement: \( \angle 8\cong\angle 4 \)

Reason: Alternate Interior Angles Theorem (because \( \overline{HJ}\parallel\overline{FG} \), and \( \overline{FB} \) is the transversal, so \( \angle 8 \) and \( \angle 4 \) are alternate interior angles)

Then step 5: \( \angle 1\cong\angle 4 \) (Reason: Transitive Property of Congruence, from \( \angle 8\cong\angle 1 \) and \( \angle 8\cong\angle 4 \))

Then step 6: \( \angle 1\cong\angle 3 \) (Reason: Alternate Interior Angles Theorem, \( \overline{DE}\parallel\overline{FG} \), transversal \( \overline{EF} \))

Then step 7: \( \angle 4\cong\angle 3 \) (Reason: Transitive Property of Congruence, from \( \angle 1\cong\angle 4 \) and \( \angle 1\cong\angle 3 \))

So for step 4 in the table (the empty one), the statement is \( \angle 8\cong\angle 4 \) and reason is Alternate Interior Angles Theorem (since \( \overline{HJ}\parallel\overline{FG} \)).

Answer:

Step 4 Statement: \( \angle 8 \cong \angle 4 \)
Step 4 Reason: Alternate Interior Angles Theorem (because \( \overline{HJ} \parallel \overline{FG} \), and \( \overline{FB} \) is the transversal, so \( \angle 8 \) and \( \angle 4 \) are alternate interior angles)

(Then subsequent steps to reach \( \angle 4 \cong \angle 3 \))