It seems pretty intuitive, and I have seen many sources make this claim. From what I've seen, congruence in geometry is defined as a direct isometry between two shapes, that is, an isometry that preservers handedness.
Here is what Wikipedia says: "The direct isometries comprise a subgroup of E(n), called the special Euclidean group. They include the translations and rotations, and combinations thereof; including the identity transformation, but excluding any reflections."
However, I haven't found any proof that the set of all direct isometries is composed solely of rotations and translations, or that these transformations do preserve handedness.
This might be a well-known result, but I'm new to geometry, so if anyone could suggest some books related to this topic, that would be greatly appreciated.
I don't have a book on hand for this, but it's a pleasant exercise to verify that isometries of Euclidean space are of the form $x \mapsto Mx + b$ where $b$ is the translation vector and $M$ is an orthogonal matrix (I'm ignoring orientation for now).
It requires a bit of vector algebra (not only to prove, but even to formulate). Let $x \mapsto f(x)$ be an isometry on $\mathbb{R}^n$. Write $g(x) = f(x) - f(0)$; then $g$ is an isometry such that $g(0) = 0$. It remains to show that $g$ is represented by an orthogonal matrix. Notice
$$(g(x) - g(y)) \cdot (g(x) - g(y)) = (x - y) \cdot (x - y) \qquad (1)$$ (both sides represent squared distances); in particular, using $g(0) = 0$,
$$g(x) \cdot g(x) = x \cdot x, \qquad g(y) \cdot g(y) = y \cdot y. \qquad (2)$$ Combining (1) and (2) appropriately, and using distributivity and symmetry properties of the dot product, derive
$$g(x) \cdot g(y) = x \cdot y \qquad (3)$$ for all $x, y \in \mathbb{R}^n$. Then
$$g(x + x') \cdot g(y) = (x + x') \cdot y = x \cdot y + x' \cdot y = g(x) \cdot g(y) + g(x') \cdot g(y) = (g(x) + g(x')) \cdot g(y)$$ and so
$$[g(x) + g(x') - g(x + x')] \cdot g(y) = 0.$$ Considering the cases $y = x, x', x+x'$ and combining them, one easily derives
$$\|g(x) + g(x') - g(x+x')\|^2 = 0$$ whence $g(x) + g(x') = g(x+x')$ for all $x, x' \in \mathbb{R}^n$. Similar sorts of arguments show that $g(rx) = r g(x)$ for all scalars $r \in \mathbb{R}$. So $g$ is a linear transformation, represented by a matrix $M$ which by (3) satisfies $Mx \cdot My = x \cdot y$, and therefore $M$ is an orthogonal matrix, completing the argument.