3.3: Same sign lemmas

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Lemma $\PageIndex{1}$

Assume $Q' \in [PQ)$ and $Q' \ne P$. Then for any $X \in (PQ)$ the angles $PQX$ and $PQ'X$ have the same sign.

$截屏2021-02-02 下午1.37.17.png$

Proof

By Proposition 2.2.2, for any $t \in [0, 1]$ there is a unique point $Q_t \in [PQ)$ such that

\[PQ_t = (1 - t) \cdot PQ + t \cdot PQ'.\]

Note that the map $t \mapsto Q_t$ is continuous,

$Q_0 = Q$, $Q_1 = Q'$

and for any $t \in [0, 1]$, we have that $P \ne Q_t$.

Applying Corollary 3.2.1, for $P_t = P$, $Q_t$, and $X_t = X$, we get that $\angle PQX$ has the same sign as $\angle PQ'X$.

Theorem $\PageIndex{1}$ Signs of angles of a triangle

In arbitrary nondegenerate triangle $ABC$, the angles $ABC, BCA,$ and $CAB$ have the same sign.

$截屏2021-02-02 下午1.47.20.png$

Proof

Choose a point $Z \in (AB)$ so that $A$ lies between $B$ and $Z$.

According to Lemma $\PageIndex{1}$, the angles $ZBC$ and $ZAC$ have the same sign.

Note that $\measuredangle ABC = \measuredangle ZBC$ and

\[\measuredangle ZAC + \measuredangle CAB \equiv \pi.\]

Therefore, $\angle CAB$ has the same sign as $\angle ZAC$ which in turn has the same sign as $\measuredangle ABC = \measuredangle ZBC$.

Repeating the same argument for $\angle BCA$ and $\angle CAB$, we get the result.

Lemma $\PageIndex{2}$

Assume $[XY]$ does not intersect $(PQ)$, then the angles $PQX$ and $PQY$ have the same sign.

The proof is nearly identical to the one above.

$截屏2021-02-02 下午1.52.14.png$

Proof

According to Proposition 2.2.2, for any $t \in [0, 1]$ there is a point $X_t \in [XY]$, such that

$XX_t = t \cdot XY.$

Note that the map $t \mapsto X_t$ is continuous. Moreover, $X_0 = X$, $X_1 = Y$, and $X_t \not\in (QP)$ for any $t \in [0, 1]$.

Applying Corollary 3.2.1, for $P_t = P$, $Q_t = Q$, and $X_t$, we get that $\angle PQX$ has the same sign as $\angle PQY$.

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Lemma \(\PageIndex{1}\)

Theorem \(\PageIndex{1}\) Signs of angles of a triangle

Lemma \(\PageIndex{2}\)