The sum of the two interior angles of the eight angles formed at the intersection

The sum of the two interior angles of the eight angles formed at the intersection of two parallel lines of the third line is 80 °. Find each of the eight corners.

When two parallel straight lines of the third (secant) intersect, 8 angles are formed: ∠1, ∠2, ∠3, ∠4, ∠5, ∠6, ∠7 and ∠8.
1. Pairs of interior angles (∠4 and ∠6) and (∠3 and ∠5) are cross lying, their degree measures are equal.
Let ∠4 = ∠6 = x, then:
∠4 + ∠6 = 80 °;
x + x = 80 °;
2 * x = 80 °;
x = 80 ° / 2;
x = 40 °.
Then:
∠4 = ∠6 = x = 40 °.
2. Pairs of internal angles (∠4 and ∠5) and (∠3 and ∠6) are one-sided, their sum is 180 °.
Thus:
∠4 + ∠5 = 180 °;
40 ° + ∠5 = 180 °;
∠5 = 180 ° – 40 °;
∠5 = 140 °.
Then:
∠3 = ∠5 = 140 °.
3.∠1 and ∠4 are adjacent:
∠1 + ∠4 = 180 °;
∠1 + 40 ° = 180 °;
∠1 = 180 ° – 40 °;
∠1 = 140 °.
4.∠1 and ∠2 are adjacent:
∠1 + ∠2 = 180 °;
140 ° + ∠2 = 180 °;
∠2 = 180 ° – 140 °;
∠2 = 40 °.
5.∠5 and ∠8 are adjacent:
∠5 + ∠8 = 180 °;
140 ° + ∠8 = 180 °;
∠8 = 180 ° – 140 °;
∠8 = 40 °.
6.∠8 and ∠7 are adjacent:
∠8 + ∠7 = 180 °;
40 ° + ∠7 = 180 °;
∠7 = 180 ° – 40 °;
∠7 = 140 °.
Answer: ∠1 = 140 °, ∠2 = 40 °, ∠3 = 140 °, ∠4 = 40 °, ∠5 = 140 °, ∠6 = 40 °, ∠7 = 140 °, ∠8 = 40 °.