I found this significantly easier than A5.

Welp. Better late than never...

Trigonometry does this. Let $\angle BAC=\alpha, \angle CAD=\beta, \angle BAD=\gamma$. Now, from cosine law in triangles $ABC, ACD, ABD$ $2 \cdot AB \cdot AC \cdot \cos \alpha=AB^2+AC^2-BC^2$ $2 \cdot AB \cdot AD \cdot \cos \gamma=AB^2+AD^2-BD^2$ $2 \cdot AC \cdot AD \cdot \cos \beta=AC^2+AD^2-CD^2$. That means that $\cos^2 \alpha \in Q \implies \sin^2 \in Q$ So, $2 \cdot AB \cdot AD \cdot \cos \gamma=2 \cdot AB \cdot AD \cdot \cos \alpha \cos \beta-2 \cdot AB \cdot AD \cdot \sin \alpha \sin \beta$ Both of this terms are $\in Q$, so $2 \cdot AB \cdot AD \cdot \sin \alpha \sin \beta \in Q$ Now, $\frac{A_{ABC}}{A_{ABD}}=\frac{AB \cdot AC \cdot \sin \alpha}{AC \cdot AD \cdot \sin \beta}=\frac{2\cdot AB \cdot AD \cdot \sin \alpha \sin \beta}{2 \cdot AD^2 \cdot \sin^2\beta} \in Q$. And that is all

We prove the following stronger result. Stronger wrote: Four points are given in the plane, with no three collinear, such that the squares of the $\binom{4}{2}= 6$ pairwise distances are all rational. Show that the ratio of the areas between any two of the $\binom{4}{3}= 4$ triangles determined by these points is also rational. Let the points be $A,B,C,D$. WLOG the triangles are $\triangle ABC,\triangle ABD$ in that order. Apply a transformation to take $A$ to $(0,0)$ on the coordinate plane and a rotation then dilation to take $B$ to $(1,0)$ on the coordinate plane; note that this transformation affects neither the given information nor the desired result. Then, let $C = (x,y)$ and $D=(a,b)$. We have \[\frac{[ABC]}{[ABD]} = \frac{y}{b}.\]Note the information \begin{align*} AC^2&=x^2+y^2\in\mathbb{Q}\\ BC^2&=(x-1)^2+y^2\in\mathbb{Q}. \end{align*}Subtract the first equation from the second to see that $-2x+1\in\mathbb{Q}$, so $x\in\mathbb{Q}$ and $y^2\in\mathbb{Q}$. Similarly, $a,b^2\in\mathbb{Q}$. Now, note that \[CD^2=(a-b)^2+(b-y)^2=(a-x)^2+b^2-2by+y^2\in\mathbb{Q}.\]As $a,b^2,x,y^2\in\mathbb{Q}$, we get that $2by\in\mathbb{Q}$. Then, note that $2b^2\in\mathbb{Q}$, which implies that \[\frac{y}{b}\in\mathbb{Q}\]as desired.

Let the points have coordinates $(0,0),(a,0), (x_1,y_1),(x_2,y_2)$. Note that $a^2$ is rational. Also, \[x_1^2+y_1^2, (a-x_1)^2+y_1^2\in \mathbb{Q} \Longrightarrow a^2-2ax_1\in \mathbb{Q} \Longrightarrow ax_1 \in \mathbb{Q}\]Based on that, we also have $x_1^2 = \frac{(ax_1)^2}{a^2}\in \mathbb{Q}$. Also, $y_1^2 = (x_1^2+y_1^2)-x_1^2\in \mathbb{Q}$. Similar results give $x_2^2, y_2^2, ax_2 \in \mathbb{Q}$. Based on that, $x_1x_2 = \frac{(ax_1)(ax_2)}{a^2}\in \mathbb{Q}$, so \[(x_1-x_2)^2+(y_1-y_2)^2 \in \mathbb{Q} \Longrightarrow y_1y_2 \in \mathbb{Q}\]Thus, $\frac{[(0,0)-(a,0)-(x_1,y_1)]}{[(0,0)-(a,0)-(x_2,y_2)]}=\frac{y_1}{y_2} = \frac{y_1y_2}{y_2^2}\in \mathbb{Q}$ and we've shown that the ratio between any two triangles which share a side is rational, and since by pigeonhole all pairs of triangles share a side, we're done.

Sorry for the bump, but is there a solution with Heron's?

Bump-this got buried

Bump Bump

Amkan2022 wrote: Sorry for the bump, but is there a solution with Heron's? I'm not aware of any such solution (though of course it's hard to disprove the existence of a solution, in general...).

Walkthrough : Observe that \[\frac{\mathrm{area}(\triangle ABC)}{\mathrm{area}(\triangle ABD)}\]$$=\frac{\frac{1}{2}.AB.BC.sin(\angle{ABC})}{\frac{1}{2}.AB.AD.sin(\angle{BAD})}=\frac{\frac{1}{4}.2.AB.BC.sin(\angle{ABC})}{\frac{1}{4}.2.AB.AD.sin(\angle{BAD})}$$$$=\frac{(AB^2+BC^2-AC^2)tan(\angle{ABC})}{(AB^2+AD^2-BD^2)tan(\angle{BAD})}.$$Now its easy to get that $\frac{tan(\angle{ABC})}{tan(\angle{BAD})}$ is also rational by further trigonometric manipulation .And since the squares of the lengths of the line segments $AB$, $AC$, $AD$, $BC$, $BD$, and $CD$ are rational numbers therefore \[\frac{\mathrm{area}(\triangle ABC)}{\mathrm{area}(\triangle ABD)}\]is also rational .

Just toss it on the coordinate plane, lol. By scaling, we can assume that $A$ is at $(0, 0)$ and $B$ is at $(1, 0)$. (We can scale by the square root of any rational number as both areas then scale by rational numbers.) Let $C$ be at $(a, b)$ and $D$ be at $(a, d)$. Then $a^2+b^2$ and $(a-1)^2+b^2$ are both rational, hence $a$ is also rational. Similarly, $c$ is rational. But because $(a-c)^2+(b-d)^2$ is rational, $ac+bd$ is also rational, so $bd$ is rational. The ratio of the areas is then $\frac db = \frac{bd}{b^2}$ which is also rational.

A6?? Let $A:(-1,0), B:(1,0), C:(x,y), D:(m,n)$ $AD^2=(m+1)^2+n^2, AC^2=(x+1)^2+y^2, BC^2=(x-1)^2+y^2, BD^2=(m-1)^2+n^2, CD^2=(x-m)^2+(y-n)^2$ and they are all rational. Note $AD^2-BC^2=4x$, it is rational, so $x$ is rational; similar reason yields that $4m$ is rational, $m$ is rational. Now, we get $n^2, y^2$ are rational respectively. Note $CD^2=(x-m)^2+y^2+n^2-2yn$ is rational, so $yn$ is rational, the area ratio is equivalent to $\frac{y}{n}=\frac{yn}{n^2}$, which is rational and we are done!