euclidean-geometrygeometry
Let A, B, C be distinct non-collinear points on the Euclidean plane.
Let D be a point between B and C.
How, only by strictly working with the axioms for Geometry, can you show that AD goes 'inside' the triangle ABC?.
Do I have to treat the triangle as a set of points and to show parametrically that every point on the segment of AD is a member of the set defined by ABC?
More generally, how is it common to define the term – 'inside'?.
Here is a more geometric proof:
$1).$ Half-planes are convex: Let $H$ be a half-plane, bounded by the line $l.$ Consider three distinct points $P,R,Q\in H$ such that $R$ is between $P$ and $Q$. If $R\notin H$ then $R$ lies on the other side of $l$ by the same sides lemma, so that $\overline {PR}$ intersects $l$ in a point $T$ such that $P,T,R$ are colinear, and $T$ is between $P$ and $R$. I will denote this situation by writing $PTR$. Now, if $PTR$ and $PRQ$ then by the Betweeness Principle, $PTQ$ which implies, on appealing to the same sides lemma once more, that $P$ and $Q$ are on opposite sides of $l,$ which is a contradiction.
$2).$ The intersection of a finite number of convex sets is convex: consider $A\cap B$, where $A$ and $B$ are any two convex sets. Let $P,Q$ be points in $A\cap B.$ Then, the segment $\overline {PQ}\subseteq A$ and $\overline {PQ}\subseteq B\Rightarrow \overline {PQ}\subseteq A\cap B.$ Now use induction to prove the general case.
$3).$ By definition, the interior of an angle is the intersection of two half-planes, and the interior of a triangle $\Delta ABC$ is the intersection of the interiors of $\angle A,\angle B$ and $\angle C.$ Now, by $2).$, we know that interiors of angles are convex. We will use this fact to prove that triangles are convex: consider the triangle $\Delta ABC$ and let $P,Q$ be distinct points in the interior of $\Delta ABC$ such that $PRQ$ for any point $R$. We need to show that $R$ is in the interior of $\Delta ABC$. But this is trivial, because if $P$ and $Q$ are interior to $\Delta ABC$ then they are by definition, interior to each of $\angle A,\angle B$ and $\angle C$.
Here is a possible proof "without inversion". (Note: Generally i am against settings of problems, which discriminate some structural part of mathematics, and ask for a solution "without" some ingredient, which would make the solution straightforward, simple, and easy to remember. But in this case i will accept the challenge. There will be some similitude argument instead... well, same two circles as in the solution using inversion are the main actors.)
Note that even the main wiki page for the inversion uses this problem to illustrate the use and usefulness of the inversion:
https://en.wikipedia.org/wiki/Inversive_geometry#Application
I will slightly change the notation to fit the order in the alphabet. So $M$ is not on $AB$, but on the side opposite to the vertex $A$, and similarly for $N,P$. (Else i would commit errors.) So let us state explicitly...
Let $\Delta ABC$ be a triangle. Let $I$ be the incenter, the intersection of the bisectors of the angles of the triangle. The incircle $(I)$ touches the sides of $\Delta ABC$ in the points $M\in BC$, $N\in CA$, and $P\in AB$.
Let $S,T,U$ be the intersections of the the angle bisectors $AI$, $BI$, $CI$ with the segments respectively perpendicular on them, $NP$, $PM$, $MN$.
Let $O$ be the circumcenter of $\Delta ABC$, the center of the circle $(ABC)$.
Let $o$ be the circumcenter of $\Delta MNP$, the center of the circle $(MNP)$. Of course, $o=I$, but it may be useful to use the notation $o$ when addressing this point for its quality of being the center of $(MNP)$. We use in the notation a small letter as possible, and with the same convention let $h$, $g$, $9$ be respectively for $\Delta MNP$ the orthocenter, the centroid, and the center of the nine point (Euler) circle $(STU)$. Here, $g=MS\cap NT\cap PU$, and $9$ is the mid point of $oh$.
We denote by $A'$ the intersection of the bisector $AI$ with the circle $(O)=(ABC)$. Construct $B',C'$ similarly.
Let $M'$ be the mid point of the segment $Mh$, so $M'\in(STU)$.
The we have:
(1) The Euler line (e) of $\Delta MNP$ is passing through $o=I$, $9$, $g$, $h$, in particular these points are colinear.
(2) $OA'\| oM\|S9M'$, and similarly $OB'\| oN\|T9N'$, $OC'\| oP\|U9P'$.
(3) The triangles $\Delta STU$ and $\Delta A'B'C'$ are perspective / similar / homothetic, the center of the perspectivity being $o=I$, the intersection of the bisectors $ASIA'$, $BTIB'$, $CUIC'$.
(4) The circles $(STU)$ and $(A'B'C')=(ABC)$ respect the same homothety. In particular, the center of homothety $o$ is on the line their centers $9$ and $O$.
(5) The points $o=I$, $9$, $g$, $h$, $O$ are on the Euler line $(e)$ of $\Delta MNP$.
Proof: All points are clear. (2) deserves maybe a slight explanation. The lines $oM=IM$, and $OA'$ are perpendicular on $BC$. ($M$ is the tangency point of $(I)$, and $OA'$ is the side bisector of $BC$.) So $oM\|OA'$. Then in $\Delta MNP$ we have the parallelogram $oMM'S$. (For instance, $IS$, $MM'$ are perpendicular on $NP$, it remains to show they have the same length. Use now that $9$ is the mid point of $oh$, $9$ is also the mid point of the diameter $M'S$, so $ShM'o$ parallelogram, so $SI=So=hM'=M'M$.)
$\square$
Best Answer
Pasch's Axiom states that, if a line $l$ intersects $\triangle ABC$ on the side $\overline{AB}$ between its endpoints, but does not contain $A$, $B$, or $C$, then $l$ intersects precisely one of $\overline{AC}$ or $\overline{BC}$ between its endpoints. Pasch's Axiom is Hilbert's Axiom II.4.
Using Pasch's axiom, it is possible to define half-planes. Namely, each line $l$ separates the plane into two convex sets, called half-planes, such that $A$ and $B$ are in the same half-plane if and only if $\overline{AB}$ does not intersect $l$, and in opposite half-planes if and only if $\overline{AB}$ does intersect $l$.
The interior of the angle $\angle BAC$ is defined as the intersection of two half-planes: the half-plane of $\overleftrightarrow{AB}$ containing $C$, and the half-plane of $\overleftrightarrow{AC}$ containing $B$. As the intersection of two convex sets, the interior of the angle is itself convex.
The interior of the triangle may then be defined as the intersection of the interiors of its three angles.
Now consider $\triangle ABC$ and a point $D$ between $B$ and $C$. Note that the segment $\overline{CD}$ does not intersect the line $\overleftrightarrow{AB}$, because the line $\overleftrightarrow{CD}$ intersects $\overleftrightarrow{AB}$ at $B$, hence cannot intersect $\overleftrightarrow{AB}$ at a point between $C$ and $D$. Similarly, the segment $\overline{BD}$ does not intersect the line $\overleftrightarrow{AC}$. It follows that $D$ is in the interior of $\angle BAC$.
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Let $E$ be any point between $A$ and $D$. $E$ is on $\overrightarrow{AD}$, so $E$ is in the interior of $\angle BAC$. If it were not, then, since $D$ is in the interior, the segment $\overline{DE}$ would have to intersect either $\overleftrightarrow{AB}$ or $\overleftrightarrow{AC}$, but the line $\overleftrightarrow{DE}$ intersects each line at $A$, hence cannot intersect either at a point between $D$ and $E$.
Now consider $\angle ABC$. $\overline{DE}$ does not intersect $\overleftrightarrow{AB}$, as just explained, so $E$ is in half-plane of $\overleftrightarrow{AB}$ containing $D$. It follows that $E$ and $C$ are in the same half-plane of $\overleftrightarrow{AB}$ as well. On the other hand, the segment $\overline{AE}$ does not intersect $\overleftrightarrow{BC}$, because the line $\overleftrightarrow{AE}$ intersects $\overleftrightarrow{BC}$ at $D$, hence cannot intersect $\overleftrightarrow{BC}$ at any point between $A$ and $E$. It follows that $E$ is in the half-plane of $\overleftrightarrow{BC}$ containing $A$. So $E$ is in the interior of $\angle ABC$.
Similarly, $E$ is in the interior of $\angle CAB$.
Because $E$ is in the interior of each of the three angles, it must be in the interior of $\triangle ABC$.