Me with this geometry proof asap - image attached -

1) in first place you need to Know what kind of figure is it, in order to find solve the problem, as you can see it is a parallelogram, and the condition for this geometric figure the opposites  are equal ,

2) if you have a non -rectangular parallelogram and draw a diagonal  from A to C, you obtain two equal triangles, for this reason the line AC IS CALLED DIAGONAL,  and divide the parallelogram

3) in first place to determine that the angles are congruents for two triangles the Diagonal is comun for both, and the vertex  ∡CAB IS THE SAME ∡ACD

4) If the diagonal AB AND CD are equal , AD AND BC are equal too, then the triangles are congruent

5)  segment AB Parallel to CD THEY HAVE THE SAME SLOPE AND MAGNITUDE

(-1,-0.5)

step-by-step explanation:

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.