Let $D_{4n}$ be the dihedral group of the order $4n$.
Prove that $D_{4n}/\langle T^n \rangle$ is isomorphic to $D_{2n}$.
We tried to configure an action on the diagonals of the $n$-gon and prove that there is a homomorphism from $D_{4n}$ to $D_{2n}$ and the kernel of the homomorphism is $\langle T^{2n} \rangle$ but we're stuck.
Let $D_{4n}=\langle a,b\rangle$ with $a^{2n}=1$ and $b^2=1$ and let $D_{2n}=\langle c,d\rangle$ with $c^{n}=1$ and $b^2=1$.
Define $\phi : D_{4n} \to D_{2n}$ by $\phi(a)=c^2$ and $\phi(b)=d$.
This is a homomorphism with kernel $\{a^ib^j \in D_{4n} : \phi(a^ib^j)=1\}$, and $$a^ib^j \in \ker\phi \iff 1=\phi(a^ib^j)=\phi(a)^i\phi(b)^j=c^{2i}d^j$$
If $j$ is odd, you have a rotation, $1$, equal to a reflection, $c^{2i}d^j$, so $j$ must be even.
Thus, $1=c^{2i}$, which implies that $n \mid 2i$, or $2i=nk$ for some integer $k$.
This means that any element of the kernel is an integer power of $a^n$, so $\ker\phi = \langle a^n\rangle$.
EDIT: to show the homomorphism is well-defined, we need to check that $\phi(bab)=\phi(a^{-1})$, as Jyrki Lahtonen pointed out.
$$\phi(bab)=\phi(b)\phi(a)\phi(b)=dc^2d=d(dc^{-2})=(c^2)^{-1}=(\phi(a))^{-1}=\phi(a^{-1})$$
ANOTHER EDIT: Note that $\phi$ is also surjective; now the first isomorphism theorem gives you the isomorphism.