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Mathematics LibreTexts

11.3: Three angles of triangle

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Proposition 11.3.1

Let ABC and ABC be two triangles in the neutral plane such that AC=AC and BC=BC. Then

AB<AB if and only if |ACB|<|ACB|.

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 ACB, ACB0. In this case we need to show that

AB<ABACB<ACB.

Choose a point X so that

ACX=12(ACB+ACB).

Note that

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

ACB<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 11.3.1

Let ABC be a triangle in the neutral plane. Then

|ABC|+|BCA|+|CAB|π.

The following proof is due to Legendre [12], earler proofs were due to Saccheri [16] and Lambert [11].

Proof

Set

a=BC,     b=CA,     c=AB,α=CAB,     β=ABC,     γ=BCA.

Without loss of generality, we may assume that α,β,γ0.

截屏2021-02-23 上午10.29.47.png

Fix a positive integer n. Consider the points A0,A1,...,An on the half-line [BA), such that BAi=ic for each i. (In particular, A0=B and A1=A.) Let us construct the points C1,C2,...,Cn, so that AiAi1Ci=β and Ai1Ci=a for each i.

By SAS, we have constructed n congruent triangles

ABC=A1A0C1A2A1C2...AnAn1Cn.

Set d=C1C2 and δ=C2A1C1. Note that

α+β+δ=π.

By Proposition 11.2.1, we get that Δ0.

By construction

A1C1C2A2C2C3...An1Cn1Cn.

In particular, CiCi+1=d for each i.

By repeated application of the triangle inequality, we get that

nc=A0AnA0C1+C1C2++Cn1Cn+CnAn==a+(n1)d+b.

In particular,

cd+1n(a+bd).

Since n is arbitrary positive integer, the latter implies cd. By Proposition 11.3.1, it is equivalent to

γδ.

From 11.3.1, the theorem follows.

Exercise 11.3.1

Let ABCD be a quadrangle in the neutral plane. Suppose that the angles DAB and ABC are right. Show that ABCD.

Hint

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

Mimic the proof of Theorem 11.3.1 for the shown fence made from copies of quadrangle ABCD.


This page titled 11.3: Three angles of triangle is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Anton Petrunin via source content that was edited to the style and standards of the LibreTexts platform.

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