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# 11.3: Three angles of triangle

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Proposition $$\PageIndex{1}$$

Let $$\triangle ABC$$ and $$\triangle A'B'C'$$ be two triangles in the neutral plane such that $$AC = A'C'$$ and $$BC = B'C'$$. Then

$$AB < A'B'$$ if and only if $$|\measuredangle ACB| < |\measuredangle A'C'B'|$$.

Proof Without loss of generality, we may assume that $$A = A'$$, $$C = C'$$, and $$\measuredangle ACB$$, $$\measuredangle ACB' \ge 0$$. In this case we need to show that

$$AB < AB' \Leftrightarrow \measuredangle ACB < \measuredangle ACB'.$$

Choose a point $$X$$ so that

$$\measuredangle ACX = \dfrac{1}{2} \cdot (\measuredangle ACB + \measuredangle ACB').$$

Note that

• $$(CX)$$ bisects $$\angle BCB'$$.
• $$(CX)$$ is the perpendicular bisector of $$[BB']$$.
• $$A$$ and $$B$$ lie on the same side of $$(CX)$$ if and only if

$$\measuredangle ACB < \measuredangle ACB'$$.

From Exercise 5.2.1, $$A$$ and $$B$$ lie on the same side of $$(CX)$$ if and only if $$AB < AB'$$. Hence the result.

Theorem $$\PageIndex{1}$$

Let $$\triangle ABC$$ be a triangle in the neutral plane. Then

$$|\measuredangle ABC| + |\measuredangle BCA| + |\measuredangle CAB| \le \pi.$$

The following proof is due to Legendre , earler proofs were due to Saccheri  and Lambert .

Proof

Set

$$\begin{array} {rclcrclcrcl} {a} & = & {BC,} & \ \ \ \ \ & {b} & = & {CA,} & \ \ \ \ \ & {c} & = & {AB,} \\ {\alpha} & = & {\measuredangle CAB,} & \ \ \ \ \ & {\beta} & = & {\measuredangle ABC,} & \ \ \ \ \ & {\gamma} & = & {\measuredangle BCA.} \end{array}$$

Without loss of generality, we may assume that $$\alpha, \beta, \gamma \ge 0$$. Fix a positive integer $$n$$. Consider the points $$A_0, A_1, ..., A_n$$ on the half-line $$[BA)$$, such that $$BA_i = i \cdot c$$ for each $$i$$. (In particular, $$A_0 = B$$ and $$A_1 = A$$.) Let us construct the points $$C_1, C_2, ..., C_n$$, so that $$\measuredangle A_iA_{i-1}C_i = \beta$$ and $$A_{i-1} C_i = a$$ for each $$i$$.

By SAS, we have constructed n congruent triangles

$$\triangle ABC = \triangle A_1A_0C_1 \cong \triangle A_2A_1C_2 \cong ... \cong \triangle A_nA_{n-1} C_n.$$

Set $$d = C_1C_2$$ and $$\delta = \measuredangle C_2A_1C_1$$. Note that

$\alpha + \beta + \delta = \pi.$

By Proposition 11.2.1, we get that $$\Delta \ge 0$$.

By construction

$$\triangle A_1C_1C_2 \cong \triangle A_2C_2C_3 \cong ... \cong \triangle A_{n - 1} C_{n - 1} C_n.$$

In particular, $$C_i C_{i + 1} = d$$ for each $$i$$.

By repeated application of the triangle inequality, we get that

$$\begin{array} {rcl} {n \cdot c} & = & {A_0A_n \le} \\ {} & \le & {A_0 C_1 + C_1 C_2 + \cdots + C_{n - 1} C_n + C_n A_n =} \\ {} & = & {a + (n - 1) \cdot d + b.} \end{array}$$

In particular,

$$c \le d + \dfrac{1}{n} \cdot (a + b - d).$$

Since $$n$$ is arbitrary positive integer, the latter implies $$c \le d$$. By Proposition $$\PageIndex{1}$$, it is equivalent to

$$\gamma \le \delta.$$

From 11.3.1, the theorem follows.

Exercise $$\PageIndex{1}$$

Let $$ABCD$$ be a quadrangle in the neutral plane. Suppose that the angles $$DAB$$ and $$ABC$$ are right. Show that $$AB \le CD$$.

Hint

Set $$a = AB, b = BC, c = CD$$, and $$d = DA$$; we nned to show that $$c \ge a$$.

Mimic the proof of Theorem $$\PageIndex{1}$$ for the shown fence made from copies of quadrangle $$ABCD$$.