11.2: Two angles of a triangle

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In this section we will prove a weaker form of Theorem 7.4.1 which holds in any neutral plane.

Theorem $\PageIndex{1}$

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

$截屏2021-02-23 上午9.30.47.png$

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

$|\measuredangle CAB|+|\measuredangle ABC|=|\measuredangle CAX| < \pi.$

Exercise $\PageIndex{1}$

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.

Exercise $\PageIndex{2}$

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

$截屏2021-02-23 上午10.01.46.png$

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.

Exercise $\PageIndex{3}$

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.

Theorem \(\PageIndex{1}\)

Exercise \(\PageIndex{1}\)

Exercise \(\PageIndex{2}\)

Exercise \(\PageIndex{3}\)