A surface is a cone iff all tangent planes intersects at a common point.
the $\rightarrow$ part is easy. But is there proof of the $\leftarrow$ part? that is, if a regular surface $S$ whose all tangent plane intersects at a common point $q$, then it is a cone.
For any $p$ of the surface, consider the curve $r(t)$ in $S$ which is the intersection of $S$ and the plane $\Pi$, where $\Pi$ is the plane passeses though $q$, $p$ and the normal of $S$ at $p$. Can we show that the tangeent line of $r(t)$ passes though $p$? If it is OK, then $r(t)$ should be a line, and proves that $S$ is a cone.
All tangent planes contains a common point the origin $o$. If $\gamma$ is a geodesic in the surface $\Sigma$ s.t. $|\gamma'|=1$, then tangent planes along $\gamma$ are generated by $\{\gamma'(t),\gamma (t)\}$.
Hence since $\gamma$ is a geodesic, then $$ \gamma''(t)\cdot \gamma (t)=0 $$
Since $\frac{d}{dt}\ \gamma'(t)\cdot \gamma (t) =\gamma '(t)\cdot \gamma ' (t) =1 $, then $$ \gamma'(T)\cdot \gamma (T) - \gamma'(0)\cdot \gamma (0) = \int_0^T\ \frac{d}{dt} \gamma'(t)\cdot \gamma (t) \ dt =T $$
Here if $f(T)=\frac{1}{2}|\gamma (T)|^2$, then $f'(T)- f'(0)=T$ Hence $$f(T)=\frac{1}{2}T^2 + f'(0)T +\frac{1}{2}|\gamma (0)|^2$$ so that $$ |\gamma (T)|= \sqrt{ T^2 +2f'(0)T +|\gamma (0)|^2 }$$
Consider a unit vector $\frac{\gamma (0)}{ |\gamma (0)|} \in T_{\gamma (0)}\Sigma $ and we have a geodesic $\gamma$ s.t. $\gamma'(0)=\frac{\gamma (0)}{ |\gamma (0)|}$.
Hence $|\gamma (T)| = T+|\gamma (0)|$ so that $\gamma ([0,T])$ is a line segment.