Prove that a group generated by two elements of order $2$, $x$ and $y$, is isomorphic to $D_{2n}$, where $n = |xy|.$

I am completely stuck at the question

Let $G$ be a finite group and let $x$ and $y$ be distinct elements of order 2 in $G$ that generate $G$. Prove that $G \cong D_{2n}$, where $n = |xy|.$

I have proved that

Let $x$ and $y$ be elements of order 2 in any group $G$. If $t = xy$ then $tx = xt^{-1}$.

Can I get some hints?

Hint: Take the homomorphism that sends $xy$ to a rotation, and $y$ to a reflection. Show that this homomorphism is an isomorphism.

By introducing the letter $t$, you have proven that you group can be written as a quotient of the dihedral group of order $2n$, $D_{2n}=\langle x, t; x^2, t^n, x^{-1}tx=t^{-1}\rangle$ (can you see why?). It also proves that every element has the form $xt^i$ or $t^i$. If the quotient was non-trivial then there would exist some integer $i$ with $0<i<n$ such that either $xt^i=1$ or $t^i=1$. However, $t^i\neq 1$ (why?). So $xt^i=1$, and so $x=t^i$. Can you see the required contradiction which completes the proof?

Personally, I like the following presentation for $D_{2n}$:

$$\langle x,y\mid x^n=y^2=(xy)^2=1\rangle$$ Now assume that the following presentation is given: $$G=\langle x,y\mid x^2=y^2=(xy)^n=1\rangle$$ So we have: $$ \begin{align*} G=\langle x,y&\mid x^2=y^2=(xy)^n=1\rangle\\ \cong \langle x,y,a&\mid x^2=y^2=(xy)^n=1,a=xy\rangle\\ \cong \langle x,y,a&\mid x^2=y^2=(xy)^n=a^n=1,a=xy\rangle\\ \cong \langle x,y,a&\mid x^2=y^2=a^n=1,a=xy\rangle\\ \cong \langle x,y,a&\mid x^2=y^2=a^n=1,a=xy,(ay^{-1})^2=1\rangle\\ \cong \langle x,y,a&\mid y^2=a^n=1,a=xy,(ay^{-1})^2=1\rangle\\ \cong \langle y,a&\mid y^2=a^n=1,(ay^{-1})^2=1\rangle\\ \cong \langle y,a,b&\mid y^2=a^n=1,(ay^{-1})^2=1,b=y^{-1}\rangle\\ \cong \langle y,a,b&\mid y^2=a^n=1,(ay^{-1})^2=1,b=y^{-1},b^2=1\rangle\\ \cong \langle y,a,b&\mid a^n=1,(ay^{-1})^2=1,b=y^{-1},b^2=1\rangle\\ \cong \langle y,a,b&\mid a^n=b^2=(ab)^2=(ay^{-1})^2=1,b=y^{-1}\rangle\\ \cong \langle y,a,b&\mid a^n=b^2=(ab)^2=1,b=y^{-1}\rangle\\ \cong \langle a,b&\mid a^n=b^2=(ab)^2=1\rangle \end{align*}$$ Which is $D_{2n}$.

We call the two elements $s$, $t$ instead, so $G=\langle s,t\rangle$.

Note that $G=\langle st, t\rangle$ since $(st)t=s$. Since $G$ is finite, $st$ has a finite order, say $m$, so that $(st)^m=1_G$. We also have $[(st)t]^2=s^2=1$.

We claim that there are no other relations, other than $(st)^m=t^2=[(st)t]^2=1$.

Suppose to the contrary $sts=1$. Then $sstss=ss$, i.e.\ $t=1$, a contradiction. Similarly if $ststs=1$, $tsststsst=tsst$ implies $s=1$, a contradiction. Inductively, $(st)^ks\neq 1$ and $(ts)^kt\neq 1$ for any $k\geq 1$.

Thus $$G\cong D_{2m}=\langle a,b|a^m=b^2=(ab)^2=1\rangle.$$