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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 ^^
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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?
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