20.1: Solid triangles
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We say that the point X lies inside a nondegenerate triangle ABC if the following three condition hold:
- A and X lie on the same side of the line (BC);
- B and X lie on the same side of the line (CA);
- C and X lie on the same side of the line (AB).
The set of all points inside △ABC and on its sides [AB], [BC], [CA] will be called solid triangle ABC and denoted by ▴ABC.
Show that any solid triangle is convex; that is, for any pair of points X,Y∈▴ABC, then the line segment [XY] lies in ▴ABC.
- Hint
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Assume the contrary; that is, there is a point W∈[XY] such that W∉▴ABC.
Without loss of generality, we may assume that W and A lie on the opposite sides of the line (BC).
It imples that both segments [WX] and [WY] intersect (BC). By Axiom II, W∈(BC) — a contradiction.
The notations △ABC and ▴ABC look similar, they also have close but different meanings, which better not to confuse. Recall that △ABC is an ordered triple of distinct points (see page ), while ▴ABC is an infinite set of points.
In particular, ▴ABC=▴BAC for any triangle ABC. Indeed, any point that belongs to the set ▴ABC also belongs to the set ▴BAC and the other way around. On the other hand, △ABC≠△BAC simply because the ordered triple of points (A,B,C) is distinct from the ordered triple (B,A,C).
Note that ▴ABC≅▴BAC even if △ABC≆△BAC, where congruence of the sets ▴ABC and ▴BAC is understood the following way:
Two sets S and T in the plane are called congruent (briefly S≅T) if T=f(S) for some motion f of the plane.
If △ABC is not degenerate and
\(\blacktriangle ABC\cong \blacktriangle A'B'C',\)
then after relabeling the vertices of △ABC we will have
△ABC≅△A′B′C′.
Indeed it is sufficient to show that if f is a motion that maps ▴ABC to ▴A′B′C′, then f maps each vertex of △ABC to a vertex △A′B′C′. The latter follows from the characterization of vertexes of solid triangles given in the following exercise:
Let △ABC be nondegenerate and X∈▴ABC. Show that X is a vertex of △ABC if and only if there is a line ℓ that intersects ▴ABC at the single point X.
- Hint
-
To prove the "only if" part, consider the line passing thru the vertex that is parallel to the opposite side.
To prove the "if" part, use Pasch’s theorem (Theorem 3.4.1).