I tried to do this from a purely coordinate geometric stand point. WLOG (like you said) assume that one of the vertices of our triangle is the orgin $\displaystyle O$. Thus we are really examining $\displaystyle \triangle OBC$, where $\displaystyle B,C$ are the other verticies. We may assume that $\displaystyle B=\left(2a,2b\right)$ for some $\displaystyle a,b\in\mathbb{Q}$. From here we break this into two cases, namely: $\displaystyle a,b\ne0$, or $\displaystyle a\text{ or }b=0$. Clearly both cannot be zero since the origin and $\displaystyle B$ are distinct points.

**Case 1**, $\displaystyle a\text{ or }b=0$: It is clear that we assume WLOG that $\displaystyle a=0$ and $\displaystyle b>0$ for all other cases are analgous. Thus this triangle will have $\displaystyle B=(0,2b)=(0,c)$ and $\displaystyle C=(x,y)$ for some $\displaystyle x,y$. Clearly seeing though that the length of $\displaystyle \overline{OB}$ is $\displaystyle |c|=c$ (since they are both on the y-axis) we may conclude that the length of $\displaystyle \overline{OC},\overline{BC}$ is as well. Realizing though that the lengths of $\displaystyle \overline{OC},\overline{BC}$ are given independently by the distance equations $\displaystyle d\left(O,C\right)=\sqrt{x^2+y^2},d\left(B,C\right) =\sqrt{x^2+(y-c)^2}$ and appealing once again to the equilaterallness of the triangle we may conclude that, in fact, $\displaystyle d\left(O,C\right)=\sqrt{x^2+y^2}=\sqrt{x^2+(y-c)^2}=d\left(B,C\right)$. Solving this equation yields $\displaystyle y=\frac{c}{2}$. As stated earlier though $\displaystyle d\left(O,B\right)=d\left(O,C\right)=d\left(B,C\rig ht)=c$ and focusing mainly on $\displaystyle d\left(O,C\right)$ we see that $\displaystyle \sqrt{x^2+y^2}=\sqrt{x^2+\frac{c^2}{4}}=c\implies x=\pm\frac{\sqrt{3}c^2}{4}$ which is clearly irrational for any rational $\displaystyle c$.

**Case 2**, $\displaystyle a,b\ne0$: Now we have that $\displaystyle B=(2a,2b)$ where neither $\displaystyle a$ or $\displaystyle b$ is zero. Constructing the midpoint $\displaystyle M$ of $\displaystyle \overline{OB}$ gives us that $\displaystyle M=(a,b)$ and that the length of the segment $\displaystyle \overline{OM}$ is $\displaystyle d\left(O,M\right)=\sqrt{a^2+b^2}$. Now, form $\displaystyle \overline{MC}$ to be the median connecting $\displaystyle M$ and $\displaystyle C$. If we consider the two segments $\displaystyle \overline{OB},\overline{MC}$ to be the restrictions of the lines $\displaystyle \ell_1,\ell_2$ respectively we can see that $\displaystyle \text{slope of }\overline{OB}=\frac{b}{a}$ and since $\displaystyle \overline{OB}\perp\overline{MC}$ we may conclue that $\displaystyle \text{slope of }\overline{MC}=\frac{-a}{b}$ (now you see why we needed $\displaystyle a,b\ne0$). Letting $\displaystyle C=(x,y)$ for some $\displaystyle x,y$ again and realizing that $\displaystyle M,C$ both lie on $\displaystyle \ell_2$ we may conclude that $\displaystyle C=\left(a+\Delta x,b+\Delta y\right)=\left(a+ \Delta x,b+\frac{-a}{b}\Delta x\right)\quad\color{red}\star$ where $\displaystyle \Delta x$ represents the change in $\displaystyle x$ between the points $\displaystyle M$ and $\displaystyle C$. Appealing to this we see that $\displaystyle \text{length of }\overline{MC}=\sqrt{\left(a+\Delta x-a\right)^2+\left(b+\frac{-a}{b}\Delta x-b\right)^2}=\left|\Delta x\right|\sqrt{1+\frac{a^2}{b^2}}$. Remembering though that forming the median $\displaystyle \overline{MC}$ bisected $\displaystyle \angle OBC$ we see that the new triangle formed, $\displaystyle \triangle OMC$ is a $\displaystyle 30-60-90$ triangle. In particular, this tells us that $\displaystyle \text{length of }\overline{MC}=\sqrt{3}\cdot\text{length of }\overline{OB}=\sqrt{3}\cdot\sqrt{a^2+b^2}$. Equating our two lengths of $\displaystyle \overline{MC}$ gives us $\displaystyle \left|\Delta x\right|\sqrt{1+\frac{a^2}{b^2}}=\sqrt{3}\sqrt{a^2 +b^2}\implies \left|\Delta x\right|\sqrt{a^2+b^2}=\sqrt{3}\cdot b\cdot\sqrt{a^2+b^2}$ and since $\displaystyle a,b\ne0\implies\sqrt{a^2+b^2}\ne0$ we may conclude that $\displaystyle \left|\Delta x\right|=\sqrt{3}b\implies \Delta x=\pm\sqrt{3}b$. Thus, substitution into $\displaystyle \color{red}\star$ gives us that $\displaystyle C=\left(a\pm\sqrt{3}b,b\mp\sqrt{3}a\right)$ which is clearly irrational since $\displaystyle a,b\in\mathbb{Q}$.

The conclusion follows. $\displaystyle \square$