= Solution
A <complex representation> of a finite group $G$ is a homomorphism
$$
\rho:G\longrightarrow\operatorname{GL}(V)
$$
for a finite-dimensional complex vector space $V$. It is an <irreducible representation> when its only invariant subspaces are $0$ and $V$, and its <degree of a representation>[degree] is $\dim V$. Two representations are <isomorphic representations>[isomorphic] when there is an invertible linear map intertwining their $G$-actions.
Write the <dihedral group> as
$$
D_{2n}=\langle r,s:r^n=s^2=1,\ srs=r^{-1}\rangle.
$$
In an irreducible representation choose an <eigenvector> $0\ne v\in V$ of $\rho(r)$, which is possible because $r^n=1$. If its <eigenvalue> is $\lambda$, then
$$
\rho(r)\rho(s)v
=\rho(s)\rho(r)^{-1}v
=\lambda^{-1}\rho(s)v.
$$
Thus $W=\operatorname{span}\{v,\rho(s)v\}$ is invariant under both $r$ and $s$. Irreducibility forces $W=V$, proving $\dim V\leq2$ as in <irreducible complex representations of a finite dihedral group>.
Let $\omega=e^{2\pi i/n}$. Besides the one-dimensional representations described below, define for each indicated $j$
$$
\rho_j(r)=
\begin{pmatrix}\omega^j&0\\0&\omega^{-j}\end{pmatrix},
\qquad
\rho_j(s)=
\begin{pmatrix}0&1\\1&0\end{pmatrix}.
$$
These matrices satisfy the defining relations. Since $\omega^j\ne\omega^{-j}$, the only one-dimensional $r$-invariant subspaces are the two coordinate axes, and $s$ interchanges them. Hence $\rho_j$ is irreducible. Their <character of a representation>[characters] satisfy
$$
\chi_j(r)=2\cos\frac{2\pi j}{n},
$$
whose values are distinct in the ranges below, so the $\rho_j$ are pairwise nonisomorphic. This is the family of <two-dimensional representations of a finite dihedral group>.
If $n$ is odd, a <one-dimensional character> must send $r$ to a scalar $\varepsilon$ satisfying $\varepsilon^n=1$ and $\varepsilon=\varepsilon^{-1}$. Thus $\varepsilon=1$, while $s$ may independently map to $1$ or $-1$. These give two one-dimensional irreducibles. Taking
$$
j=1,\ldots,\frac{n-1}{2}
$$
gives $(n-1)/2$ two-dimensional irreducibles, for a total of
$$
2+\frac{n-1}{2}=\frac{n+3}{2}.
$$
If $n$ is even, $r$ may map to either $1$ or $-1$, and again $s$ may map independently to either sign. This gives four pairwise nonisomorphic one-dimensional irreducibles. Taking
$$
j=1,\ldots,\frac n2-1
$$
gives $n/2-1$ two-dimensional irreducibles, for a total of
$$
4+\left(\frac n2-1\right)=\frac{n+6}{2}.
$$
Different dimensions distinguish the one- and two-dimensional families, completing the required construction and justification.
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