Geometry :: Notes :: Neutral Geometry Theorems
1 Neutral Geometry Theorems
The previous chapter replaced Euclid's original first four axioms, as well as all of Euclid's unstated assumptions, we have "patched up" Euclid's errors. We are now ready to start proving theorems from Book I of the Elements. Later we will add a parallel axiom, either Euclidean or Hyperbolic.
Let \(\triangle ABC\) be a triangle. The angles \(\angle CAB, \angle ABC\), and \(\angle BCA\) are called interior angles of the triangle. An angle that forms a linear pair with one of the interior angles is called an exterior angle for the triangle. If the exterior angle forms a linear pair with the interior angle at one vertex, then the other two interior angles at the other two vertices are referred to as remote interior angles (for the original vertex).
The measure of an exterior angle for a triangle is strictly greater than the measure of either remote interior angles.
You need not prove this theorem.
For every line \(l\) and for every point \(P\), there exists a unique line \(m\) such that \(P\) lies on \(m\) and \(m \perp l\).
You need not prove this theorem.
If one interior angle of a triangle is a right angle or is an obtuse angle, then both the other interior angles are acute.
If \(\triangle ABC\) and \(\triangle DEF\) are two triangles such that \(\angle CAB \cong \angle FDE, \overline{AB} \cong \overline{DE}\), and \(\angle ABC \cong \angle DEF\), then \(\triangle ABC \cong \triangle DEF\).
If \(\triangle ABC\) is a triangle such that \(\angle ABC \cong \angle ACB\), then \(\overline{AB} \cong \overline{AC}\).
If \(\triangle ABC\) and \(\triangle DEF\) are two triangles such that \(\angle ABC \cong \angle DEF, \angle BCA \cong \angle EFD\), and \(\overline{AC} \cong \overline{DF}\) then \(\triangle ABC \cong \triangle DEF\).
A triangle is a right triangle if one of its interior angles is a right angle. The side opposite the right angle is called the hypotenuse and the two sides adjacent to the right angle are called legs.
If the hypotenuse and one leg of a right triangle are congruent to the hypotenuse and a leg of a second right triangle, then the two triangles are congruent.
If \(\triangle ABC\) is a triangle and \(\overline{DE}\) is a segment congruent to \(\overline{AB}\), and \(H\) is a half-plane bounded by \(\overleftrightarrow{DE}\), then there is a unique point \(F \in H\) such that \(\triangle DEF \cong \triangle ABC\).
If \(\triangle ABC\) and \(\triangle DEF\) are two triangles such that \(\overline{AB} \cong \overline{DE}\), \(\overline{BC} \cong \overline{EF}\), and \(\overline{CA} \cong \overline{FD}\), the \(\triangle ABC \cong \triangle DEF\).
You need not prove this theorem.
A scalene triangle is a triangle where each of the three sides has a different length.
Let \(\triangle ABC\) be a triangle. Then \(AB > BC\) if and only if \(\mu(\angle ACB) > \mu(\angle BAC)\).
If \(\triangle ABC\) and \(\triangle DEF\) are two triangles such that \(AB = DE, AC = DF\), and \(\mu(\angle BAC) < \mu(\angle EDF)\), then \(BC < EF\).
You need not prove this theorem.
Let \(l\) be a line and \(P\) a point. A perpendicular from \(P\) to \(l\) is the line \(m\) from Theorem Existence and Uniqueness of Perpendiculars. The intersection of \(l\) and \(m\) is called the foot of the perpendicular.
Let \(l\) be a line, \(P\) an external point, and \(F\) the foot of the perpendicular from \(l\) to \(P\). If \(R\) is any other point on \(l\) distinct from \(F\), then \(PR > PF\).
The distance from a point \(P\) to a line \(l\), denoted \(d(P,l)\), is the distance from \(P\) to the foot of the perpendicular from \(P\) to \(l\).
Let \(A,B\), and \(C\) be three non-collinear points and let \(P\) be a point on the interior of \(\angle BAC\). Then \(P\) lies on the angle bisector of \(\angle BAC\) if and only if \(d(P, \overleftrightarrow{AB}) = d(p, \overleftrightarrow{AC})\).
Let \(A\) and \(B\) be distinct points. A point \(P\) lies on the perpendicular bisector of \(\overline{AB}\) if and only if \(PA = PB\).