1. **Problem Statement:** Show that any two lines $\ell_1$ and $\ell_2$ in a projective plane have the same number of points. 2. **Key Idea:** We first prove there exists a point $Q$ not on either $\ell_1$ or $\ell_2$. Then, using $Q$, we establish a one-to-one correspondence (bijection) between the points on $\ell_1$ and $\ell_2$. 3. **Step 1: Existence of $Q$** - By the axioms of projective planes, any two distinct lines intersect in exactly one point. - Since $\ell_1$ and $\ell_2$ are distinct, they intersect at exactly one point, call it $P$. - Because the projective plane is not degenerate, there must be points not on $\ell_1$ or $\ell_2$. - Hence, there exists a point $Q$ such that $Q \notin \ell_1$ and $Q \notin \ell_2$. 4. **Step 2: Constructing the bijection** - For each point $A$ on $\ell_1$, consider the unique line $m$ through $Q$ and $A$. - Since $m$ is a line distinct from $\ell_2$ and $Q \notin \ell_2$, $m$ intersects $\ell_2$ in exactly one point, call it $f(A)$. - Define the function $f: \ell_1 \to \ell_2$ by $f(A) = m \cap \ell_2$. 5. **Step 3: Show $f$ is one-to-one and onto** - **Injective:** Suppose $f(A) = f(B)$ for $A,B \in \ell_1$. Then lines $QA$ and $QB$ both pass through $f(A)$, so $QA = QB$. Since $Q$ is fixed, this implies $A = B$. - **Surjective:** For any point $C$ on $\ell_2$, the line $QC$ intersects $\ell_1$ at some point $D$. Then $f(D) = C$. 6. **Conclusion:** The function $f$ is a bijection between points on $\ell_1$ and $\ell_2$, so both lines have the same number of points. **Final answer:** Any two lines in a projective plane have the same number of points.