I need help with two column proofs ASAP

We start with the given statement. Then we use the segment addition postulate like so:

QT = SU

QR+RS+ST = ST+TU

QR+RS = TU

QS = TU

in the second step, I broke up QT into QR+RS+ST. Basically segment QT is composed of the smaller pieces QR, RS and ST. Similarly, SU is made of ST and TU.

The third step has us subtract ST from both sides.

Lastly, we combine QR+RS back into QS to conclude the proof. You could argue that QT = QS+ST and avoid this step if you wanted. It's up to you how you want to approach the proof.

∠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.

may i see the picture? it would a lot, xox