BE is a transversal for AC and DF (point Be on AC, point Ee on DF), 2CBE = 75°. Insert the appropriate relational signs

Таинственный_Лепрекон_1533

34

< or =) between the following pairs of angles:

1) ∠2CBE and ∠1CBE
To solve this, we need to understand the properties of transversals and corresponding angles. When a transversal intersects two parallel lines, the corresponding angles formed are congruent.

In this case, BE is a transversal for AC and DF, so we can say that ∠2CBE and ∠1CBE are corresponding angles. Therefore, they must be congruent. We can represent this by using the equals sign (=):

∠2CBE = ∠1CBE

2) ∠2CBE and ∠3CBE
In this case, ∠2CBE and ∠3CBE are alternate interior angles. When a transversal intersects two parallel lines, alternate interior angles are congruent.

Therefore, we can represent this by using the equals sign (=):

∠2CBE = ∠3CBE

3) ∠2CBE and ∠5EED
In this case, ∠2CBE and ∠5EED are corresponding angles. Since ∠2CBE = ∠1CBE (as mentioned in the first question), and ∠1CBE = ∠5EED (since they are corresponding angles), we can conclude that ∠2CBE = ∠5EED.

Therefore, we can represent this by using the equals sign (=):

∠2CBE = ∠5EED

To summarize:

1) ∠2CBE = ∠1CBE
2) ∠2CBE = ∠3CBE
3) ∠2CBE = ∠5EED