Will number of edges in convex polygon of n extreme points be n?
Here convex polygon means convex hull of finitely many points in $\mathbb{R}^2$.
An extreme point is point which can't be written as strict linear combinations of 2 points from polygon. And an edge of polygon is line segment such that if line joining two points from polygon $x$ and $y$ intersect that segment, then line segment contains line segment joining $x$ and $y$
Let's have a closer look at the conditions mentioned by gerw.
Each edge is incident to two vertices.
This is the simple part: If there were three (or more) vertices, then one of them would be between the other two (Hilbert's axioms of order). This point would be a strict convex combination of the other two and as such it would not be a vertex.
Each vertex is incident to two edges.
Let $x_0$ be a vertex with three (or more) edges. Wolog $x_0=0.$ One edge connects $x_0$ with vertex $x_1,$ another edge connects $x_0$ with vertex $x_2,$ and yet another edge connects $x_0$ with vertex $x_3.$ As $x_0,$ $x_i$ and $x_j$ cannot be on the same line for $(i,j)\in\{(1,2),(2,3),(1,3)\},$ we can find $\lambda_1\neq 0,$ $\lambda_2\neq 0$ and $\lambda_3\neq 0$ such that $$ \lambda_1x_1+\lambda_2x_2+\lambda_3x_3=0 $$ and $\lambda_i>0$ for at least two indices $i\in\{1,2,3\}.$
If $x_i=(u_i,v_i),$ we can use $$ \lambda_1=\varepsilon \begin{vmatrix} u_2 & u_3 \\ v_2 & v_3\end{vmatrix} \;\;,\;\; \lambda_2=\varepsilon \begin{vmatrix} u_3 & u_1 \\ v_3 & v_1\end{vmatrix} \;\;,\;\; \lambda_3=\varepsilon \begin{vmatrix} u_1 & u_2 \\ v_1 & v_2\end{vmatrix} $$ and choose $\varepsilon\in\{-1,1\}$ such that at least two elements in $\{\lambda_1,\lambda_2, \lambda_3\}$ are positive.
If $\lambda_1,$ $\lambda_2$ and $\lambda_3$ are all positive then $$ x_0 = 0 = \frac{\lambda_1}{\lambda_1+\lambda_2+\lambda_3}x_1 +\frac{\lambda_2}{\lambda_1+\lambda_2+\lambda_3}x_2 +\frac{\lambda_3}{\lambda_1+\lambda_2+\lambda_3}x_3 $$ which makes $x_0$ a strict convex combination of $x_1,$ $x_2$ and $x_3.$ This contradicts the fact that $x_0$ is a vertex.
If one of $\lambda_1,$ $\lambda_2$ and $\lambda_3$ is negative (wolog $\lambda_3<0$), then we can easily find a point $y_1$ on the edge $x_0x_1$ and a point $y_2$ on the edge $x_0x_2$ such that $y_1y_2$ intersects $x_0x_3$ which contradicts the fact that $x_0x_3$ is an edge:
Let $$ m=|\lambda_1|+|\lambda_2|+|\lambda_3| \;,\;\; y_1 = \frac{\lambda_1}{m} \,x_1 \;,\;\; y_2 = \frac{\lambda_2}{m} \,x_2 $$ then $$ \frac{1}{2}(y_1+y_2) = \frac{1}{2m}(\lambda_1 x_1+\lambda_2 x_2) =\frac{-\lambda_3}{2m}\,x_3 $$ which means that the center of the line segment $y_1y_2$ lies on $x_0x_3.$