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Mathematics LibreTexts

11.3: Three angles of triangle

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    23649
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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

    截屏2021-02-23 上午10.14.09.png

    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 [12], earler proofs were due to Saccheri [16] and Lambert [11].

    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 \(\alpha, \beta, \gamma \ge 0\).

    截屏2021-02-23 上午10.29.47.png

    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 \(ABCD\).

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