In projective geometry, there is a nice way to see that the set of lines through a point $p$ in $\Bbb{P}^3$ is naturally a projective plane: Simply choose a plane $H$ that does not contain $p$. Then every line through $p$ intersects $H$ in a unique point, and every point in $H$ corresponds to a unique line through $p$.
Is there a similar nice way to see that the set of lines in a plane $H$ is naturally a projective plane?
EDIT: I'm aware of the fact that duality of the plane yields a bijection between the lines in $H$ and the points in $H$. However, I'm looking for a geometric construction in $\Bbb{P}^3$, if such a thing exists.
Sorry, this is not very elementary but I think this is very nice. Since the isomorphism you asked is not canonical, we will have some choice to do, we will pick two distinct lines $A,B$ in $H$.
We begin by a lemma :
For definition of a blow-up and the proof of the lemma you can see these notes.
Construction of the isomorphism $\text{Line}(H) \cong \Bbb P^2$ :
Let $p = A \cap B$. Any line $L$ is determined by $L \cap A$ and $L \cap B$ if $p \notin L$. This give a rational map $f : \text{Line}(H) \dashrightarrow \Bbb P^1 \times \Bbb P^1$. We can make the map an isomorphism the following way : first we will blow-up $(p,p)$ for obtaining a well-defined map $g : \text{Line}(H) \to Bl_{(p,p)}\Bbb P^1 \times \Bbb P^1$. Now, it is not surjective precisely for the point on the form $p \times \Bbb P^1$ or $\Bbb P^1 \times p$, but we can contract these lines by $k : Bl_{(p,p)}(\Bbb P^1 \times \Bbb P^1) \to Y$. By the lemma, $Y \cong H$ so $k \circ g : \text{Line}(H) \to H$ is an isomorphism.