By Sharipov R.A.

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3) admits the existence of exactly two points L on the line KM satisfying this condition. The first of them is the above point L. The ˜ second one is the point L that lies on the ray coming out from the point M in the direction opposite to the ray [M K . 3). L means that both points L and L Under this assumption we apply the axiom A13 to the ray [KL ˜ on this ray such that [K M ˜] ∼ and mark a point M = [AC]. Now ∼ ∼ ˜ ˜ from [K L] = [AB] and [K M ] = [AC], applying the item (1) of ˜M ˜] ∼ the axiom A15, we derive [L = [BC].

Then both points B and C lie between the points A and D. Proof. From (A ◮ B ◭ C) it follows that the point A lies on the line BC, while from (B ◮ C ◭ D) it follows that D also lies on the line BC. 1 all of the four points A, B, C, and D lie on one straight line. 2 we find a point E not lying on the line AD (see Fig. 1). Then we apply the axiom A10 to the points C and E. As a result on the line CE we find a point F such that the point E lies in the interior of the segment [CF ]. Let’s draw the lines AE and F B, then consider the triangle F BC.

1 does not change the signs of (k − i) and (q − j). Renumbering the points A1 , . . 6) changes these signs to opposite ones: sign(k − i) → − sign(k − i), sign(q − j) → − sign(q − j). Thus, we see that the equality sign(k −i) = sign(q −j) being valid or not does not depend on a particular choice of the monotonic sequence of points that includes the starting and ending points −−→ −−→ of the vectors AB and CD. 2. The codirectedness is a binary relation in the set of vectors lying on one line. This relation possesses the following properties: −−→ −−→ −−→ (1) AB ⇈ AB for any vector AB; −−→ −−→ −−→ −−→ (2) AB ⇈ CD implies CD ⇈ AB; −−→ −−→ −−→ −−→ −−→ −−→ (3) AB ⇈ CD and CD ⇈ EF imply AB ⇈ EF ; −−→ −−→ −−→ (4) if a vector AB is not codirected with CD, while CD is −−→ −−→ −−→ not codirected with EF , then AB ⇈ EF .

### Foundations of geometry for university students and high-school students by Sharipov R.A.

by Jeff

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