I am trying to prove that a circle is convex in neutral geometry. i.e. If $A$ and $B$ are inside a circle $C$, than any point in $AB$ is also in $C$. But I have difficulty in proving it. The case $ABCO$ is colinear is easy, where $O$ is the centre of $O$.
I am following marvin greenberg's Euclidean Geometry 3rd Editon.
Thank you so much!
On
You can assume w.l.o.g. that AB is a chord. For if not, just take A' and B' to be the points at which the unique line containing AB intersects C. Now let P be such that A-P-B. We have to prove that OP < OB (= OA). Angle OPB is an exterior angle of triangle OAP. So, it is larger than angle OAP by the exterior angle inequality (valid in neutral geometry). But OA = OB, so that, by the isosceles triangle theorem, angle OBA = angle OAB. So, angle OPB > angle OBA = angle OBP. In triangle OPB, this means that OB > OP (larger side opposes larger angle), which is what we wanted to show.
I know this question is more than a year old, but hopefully the solution below is still helpful!
As stated in the problem, let $P$ be a point on $\overline{AB}$. The key claim is the following.
CLAIM: $OP<\max\{OA,OB\}$.
Proof. Let $D$ be the foot of the perpendicular from $O$ to $AB$. Note that there are two cases to consider depending on the location of $P$.
CASE 1: $P$ lies on $\overline{AD}$.
In this case, remark that $AD>PD$, so $$\sqrt{AD^2+DO^2}>\sqrt{PD^2+DO^2}\quad\implies\quad AO>PO.$$
CASE 2: $P$ lies on $\overline{BD}$.
This case is similar: from $PD<BD$ we have $$\sqrt{BD^2+DO^2}>\sqrt{PD^2+DO^2}\quad\implies\quad BO>PO.$$
Combining both cases yields $PO<\max\{AO,BO\}$ as desired. $\blacksquare$
Let $R$ be the radius of the circle. Note that since $AO<R$ and $BO<R$, we have $PO<R$ as well, so $P$ must lie within the circle. (If that is not a detailed enough finish, you could extend $PO$ to intersect the circle at $C$ and $D$, then use your "$ABCO$ collinear" case to resolve the matter.)