Proportional Sequence Theorem

Theorem.

Parallel straight lines intersecting the angle cut off proportional segments from the angle sides.



Evidence.

Let the sides of angle A intersect parallel lines at points B, B ₁ , C, C ₁ . The theorem states that



Divide the segment AC into n equal parts. Let δ be the length of the division segment and AC = nδ.

Two cases are possible:

1) There is n such that B is a division point. That is, there is m <n such that AB = mδ. We draw through the dividing points of the segment AC straight lines parallel to the straight line CC ₁ . By the Thales theorem, these straight lines divide the segment AC1 into equal segments of some length δ ₁ . We get AB ₁ = mδ ₁ , AC ₁ = nδ ₁ . From this



2) For no n, B 1 is not a division point. Assume that







Let us put on line AC ₁ segment AD = (AC ₁ / AC) * AB. Moreover, AD <AB ₁ . We divide AC ₁ into a sufficiently large number n equal parts. We draw through the dividing points straight lines parallel to SS ₁ . If n is sufficiently large, on the segment DB _{1 there} will be division points. Denote one of them as a point Y and draw through it a straight parallel SS ₁ that intersects the ray AC at the point X. By the proven



Replace AY with a smaller value AD, and AX with a larger value AB. Then



From here



That contradicts the construction of the segment AD. The theorem is proved.