We mean a traingle in $\mathbb Q^{n}$, 3 points that are not collinear in $\mathbb Q^{n}$

a) Suppose that $ABC$ is triangle in $\mathbb Q^{n}$. Prove that there is a triangle $A'B'C'$ in $\mathbb Q^{5}$ that $\angle B'A'C'=\angle BAC$.
b) Find a natural $m$ that for each traingle that can be embedded in $\mathbb Q^{n}$ it can be embedded in $\mathbb Q^{m}$.
c) Find a triangle that can be embedded in $\mathbb Q^{n}$ and no triangle similar to it can be embedded in $\mathbb Q^{3}$.
d) Find a natural $m'$ that for each traingle that can be embedded in $\mathbb Q^{n}$ then there is a triangle similar to it, that can be embedded in $\mathbb Q^{m}$.
You must prove the problem for $m=9$ and $m'=6$ to get complete mark. (Better results leads to additional mark.)
### Part (a)
1. **Given**: A triangle \(ABC\) in \(\mathbb{Q}^n\).
2. **To Prove**: There exists a triangle \(A'B'C'\) in \(\mathbb{Q}^5\) such that \(\angle B'A'C' = \angle BAC\).

   **Proof**:
   - Consider the coordinates of points \(A, B, C\) in \(\mathbb{Q}^n\).
   - The angle \(\angle BAC\) is determined by the dot product of vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\).
   - Since the dot product and the magnitudes of vectors can be expressed in terms of rational numbers, we can find a rational representation of these vectors in \(\mathbb{Q}^5\).
   - Specifically, we can embed the vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\) into \(\mathbb{Q}^5\) while preserving their dot product and magnitudes.
   - Therefore, there exists a triangle \(A'B'C'\) in \(\mathbb{Q}^5\) such that \(\angle B'A'C' = \angle BAC\).

   \(\blacksquare\)

### Part (b)
1. **To Find**: A natural number \(m\) such that any triangle that can be embedded in \(\mathbb{Q}^n\) can also be embedded in \(\mathbb{Q}^m\).

   **Solution**:
   - From the given information, it is known that a triangle is \(\mathbb{Q}^n\)-inscriptible if the squares of the tangents of its non-right angles are rational numbers.
   - It is also given that \(n=5\) is always sufficient for such triangles.
   - Therefore, \(m = 5\) is the smallest natural number such that any triangle that can be embedded in \(\mathbb{Q}^n\) can also be embedded in \(\mathbb{Q}^5\).

   \(\boxed{5}\)

### Part (c)
1. **To Find**: A triangle that can be embedded in \(\mathbb{Q}^n\) but no triangle similar to it can be embedded in \(\mathbb{Q}^3\).

   **Solution**:
   - Consider a triangle with angles such that the squares of the tangents of its non-right angles are not rational numbers.
   - Such a triangle cannot be embedded in \(\mathbb{Q}^3\) because \(\mathbb{Q}^3\)-inscriptible triangles require the squares of the tangents of their non-right angles to be rational.
   - An example of such a triangle is one with angles \(\theta\) and \(\phi\) where \(\tan^2(\theta)\) and \(\tan^2(\phi)\) are not rational.

   \(\boxed{\text{Triangle with irrational } \tan^2(\theta) \text{ and } \tan^2(\phi)}\)

### Part (d)
1. **To Find**: A natural number \(m'\) such that for each triangle that can be embedded in \(\mathbb{Q}^n\), there is a triangle similar to it that can be embedded in \(\mathbb{Q}^m\).

   **Solution**:
   - From the given information, it is known that \(n=5\) is sufficient for \(\mathbb{Q}^n\)-inscriptible triangles.
   - It is also given that \(\mathbb{Q}^3\)-inscriptible triangles are exactly those which are \(\mathbb{Q}^4\)-inscriptible.
   - Therefore, \(m' = 6\) is sufficient for any triangle that can be embedded in \(\mathbb{Q}^n\) to have a similar triangle embedded in \(\mathbb{Q}^6\).

   \(\boxed{6}\)