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Proportional Sequence Theorem

Lecture



Theorem.

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

  Proportional Sequence Theorem

Evidence.

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

  Proportional Sequence Theorem

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

  Proportional Sequence Theorem

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

  Proportional Sequence Theorem



  Proportional Sequence Theorem

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

  Proportional Sequence Theorem

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

  Proportional Sequence Theorem

From here

  Proportional Sequence Theorem

That contradicts the construction of the segment AD. The theorem is proved.
created: 2014-10-05
updated: 2021-03-13
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Planometry

Terms: Planometry