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(a) Let be m a line and the only two semiplans determined by m. (i) Show that: If are points that do not belong to such , so and are in opposite sides of m. (ii) In the same conditions of the last item, show: and . (iii) Determine the union result , carefully justifying your answer. (b) Let be and 4 distincts points in a line such and . Show and . (c) Let be distincts points in a line m such . Under these conditions, show 2 distinct segments such the Union of both segments be equal to , carefully justifying your answer. Thanks for the help ^^

Samuel Gomes said: (a) Let be m a line and the only two semiplans determined by m. (i) Show that: If are points that do not belong to such , so and are in opposite sides of m. (ii) In the same conditions of the last item, show: and . (iii) Determine the union result , carefully justifying your answer. (b) Let be and 4 distincts points in a line such and . Show and . (c) Let be distincts points in a line m such . Under these conditions, show 2 distinct segments such the Union of both segments be equal to , carefully justifying your answer. Click to expand... @ Gomes, You are using(perhaps miss-translating) standard terms in axiomatic geometry. In any plane if $$\displaystyle m$$ is a line then there are two half-planes $$\displaystyle \mathcal{H_1}~\&~\mathcal{H_2}$$ determined by $$\displaystyle m$$ that are convex disjoint sets. If $$\displaystyle A\in \mathcal{H_1}~\&~B\in\mathcal{H_2}$$ then $$\displaystyle \exists C\in m$$ such that $$\displaystyle A-C-B$$ So the part a is simply part of the separation axioms. Having taught axiomatic geometry several times, I have never seen the notation $$\displaystyle \mathcal{P}_mA$$. You need to define. c) Given four points collinear points $$\displaystyle A,\;B,\;C,~\&~D$$ such that $$\displaystyle A-B-C~\&~A-D-C$$ then $$\displaystyle A-B-C-D$$. We cannot have $$\displaystyle A-C-B-D$$ Why? We cannot have $$\displaystyle A-D-C-B$$ Why?