11.2: Two angles of a triangle
In this section we will prove a weaker form of Theorem 7.4.1 which holds in any neutral plane.
Let \(\triangle ABC\) be a nondegenerate triangle in the neutral plane. Then
\(|\measuredangle CAB| + |\measuredangle ABC| < \pi.\)
Note that according to Theorem 3.3.1 , the angles \(ABC, BCA\), and \(CAB\) have the same sign. Therefore, in the Euclidean plane the theorem follows immediately from Theorem 7.4.1 .
- Proof
-
Let \(X\) be the reflection of \(C\) across the midpoint \(M\) of \([AB]\). By Proposition 7.2.1 \(\measuredangle BAX = \measuredangle ABC\) and therefore
\[\measuredangle CAX \equiv \measuredangle CAB + \measuredangle ABC.\]
Since \([BM]\) and \([MX]\) do not intersect \((CA)\), the points \(B\), \(M\), and \(X\) lie on the same side of \((CA)\). Therefore the angles \(CAB\) and \(CAX\) have the same sign. By Theorem 3.3.1 , the angles \(CAB, ABC\) have the same sign; that is all angles in 11.2.1 have the same sign.
Note that \(\measuredangle CAX \not\equiv \pi\), otherwise \(X\) would lie on \((AC)\). Therefore the identity 11.2.1 implies that
\(|\measuredangle CAB|+|\measuredangle ABC|=|\measuredangle CAX| < \pi.\)
Assume \(A,B,C\), and \(D\) are points in a neutral plane such that
\(2 \cdot \measuredangle ABC + 2 \cdot \measuredangle BCD \equiv 0.\)
Show that \((AB)\parallel (CD)\).
Note that one cannot apply the transversal property ( Theorem 7.3.1 )
- Hint
-
Arguing by contradiction, assume \(2 \cdot (\measuredangle ABC + \measuredangle BCD) \equiv 0\), but \((AB) \nparallel (CD)\). Let \(Z\) be the point of intersection of \((AB)\) and \((CD)\).
Note that \(2 \cdot \measuredangle ABC \equiv 2 \cdot \measuredangle ZBC\), and \(2 \cdot \measuredangle BCD \equiv 2 \cdot \measuredangle BCZ\).
Apply Proposition \(\PageIndex{1}\) to \(\triangle ZBC\) and try to arrive at a contradiction.
Prove the side-angle-angle congruence condition in the neutral geometry.
In other words, let \(ABC\) and \(A'B'C'\) be two triangles in a neutral plane; suppose that \(\triangle A'B'C'\) is nondegenerate. Show that \(\triangle ABC \cong \triangle A'B'C'\) if
\(AB = A'B'\), \(\measuredangle ABC = \pm \measuredangle A'B'C'\) and \(\measuredangle BCA = \pm \measuredangle B'C'A'.\)
- Hint
-
Let \(C'' \in [B'C')\) be the point such that \(B'C'' = BC\).
Note that by SAS, \(\triangle ABC \cong \triangle A'B'C''\). Conclude that \(\measuredangle B'C'A' = \measuredangle B'C''A'\).
Therefore, it is sufficient to show that \(C'' = C'\). If \(C' \ne C''\) apply Proposition \(\PageIndex{1}\) to \(\triangle A'C'C''\) and try to arrive at a contradiction.
Note that in the Euclidean plane, the above exercise follows from ASA and the theorem on the sum of angles of a triangle ( Theorem 7.4.1 ). However, Theorem 7.4.1 cannot be used here, since its proof uses Axiom V. Later ( Theorem 13.1.1 ) we will show that Theorem 7.4.1 does not hold in a neutral plane.
Assume that the point \(D\) lies between the vertices \(A\) and \(B\) of \(\triangle ABC\) in a neutral plane. Show that
\(CD < CA\) or \(CD < CB\).
- Hint
-
Use Exercise 5.2.2 and Proposition \(\PageIndex{1}\).
Alternatively, use the same argument as in the solution of Exercise 5.6.1 .