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The Defect of a Triangle

Since the angle sum of any triangle in neutral geometry is not more than tex2html_wrap_inline11150, we can compute the difference between the number 180 and the angle sum of a given triangle.

Definition: The defect  of a triangle tex2html_wrap_inline11270 is the number
displaymath15084

In euclidean geometry we are accustomed to having triangles whose defect is zero. Is this always the case? The Saccheri-Legrendre Theorem indicates that it may not be so. However, what we wish to see is that the defectiveness of triangles is preserved. That is, if we have one defective triangle, then all of the triangles are defective. By defective, we mean that the triangles have positive defect.

Theorem 10.4:[Additivity of Defect]  Let tex2html_wrap_inline11270 be any triangle and let D be a point between A and B. Then tex2html_wrap_inline15108.

 figure2501
Figure 11.3:  

Proof: Since tex2html_wrap_inline14770 lies in tex2html_wrap_inline11432, we know that
displaymath15085
and since tex2html_wrap_inline15122 and tex2html_wrap_inline15124 are supplementary angles tex2html_wrap_inline15126. Therefore,
eqnarray2521

Corollary 1: tex2html_wrap_inline15128 if and only if tex2html_wrap_inline15130.

A rectangle  is a quadrilateral all of whose angles are right angles. We cannot prove the existence or non-existence of rectangles in Neutral Geometry. Nonetheless, the following result is extremely useful.

Theorem 10.4:  If there exists a triangle of defect 0, then a rectangle exists. If a rectangle exists, then every triangle has defect 0.

Let me first outline the proof in five steps.

  1. Construct a right triangle having defect 0.
  2. From a right triangle of defect 0, construct a rectangle.
  3. From one rectangle, construct arbitrarily large rectangles.
  4. Prove that all right triangles have defect 0.
  5. If every right triangle has defect 0, then every triangle has defect 0.
Having outlined the proof, each of the steps is relatively straightforward.

  1. Construct a right triangle having defect 0.
    Let us assume that we have a triangle tex2html_wrap_inline11270 so that tex2html_wrap_inline15128. We may assume that tex2html_wrap_inline11270 is not a right triangle, or we are done. Now, at least two angles are acute since the angle sum of any two angles is always less than tex2html_wrap_inline11150. Let us assume that tex2html_wrap_inline11392 and tex2html_wrap_inline11394 are acute. Also, let D be the foot of C on tex2html_wrap_inline12754. We need to know that tex2html_wrap_inline15150.

     figure2552
    Figure 11.4: Right triangle with defect 0 

    Assume not; i.e., assume that tex2html_wrap_inline15160. See Figure 11.4. This means that tex2html_wrap_inline13320 is exterior to tex2html_wrap_inline15164 and, therefore, tex2html_wrap_inline15166. This makes tex2html_wrap_inline11392 obtuse, a contradiction. Similarly, if tex2html_wrap_inline12520 we can show that tex2html_wrap_inline11394 is obtuse. Thus, we must have that tex2html_wrap_inline15150.

    This makes tex2html_wrap_inline15176 and tex2html_wrap_inline15178 right triangles. By Corollary 1 above, since tex2html_wrap_inline11270 has defect 0, each of them has defect 0, and we have two right triangles with defect 0.

  2. From a right triangle of defect 0, construct a rectangle.
    We now have a right triangle of defect 0. Take tex2html_wrap_inline15182 from Step 1, which has a right angle at D. There is a unique ray tex2html_wrap_inline15186 on the opposite side of tex2html_wrap_inline11308 from D so that
    displaymath15086
    from Axiom C-4. From Axiom C-1, there is a unique point E on tex2html_wrap_inline15186 such that tex2html_wrap_inline15196.

     figure2572
    Figure 11.5: Rectangles 

    Thus, tex2html_wrap_inline15208 by SAS. Then tex2html_wrap_inline15210 and tex2html_wrap_inline15212 must also have defect 0. Now, clearly, since tex2html_wrap_inline15214
    displaymath15087
    and, hence,
    displaymath15088
    Likewise, tex2html_wrap_inline15216 and tex2html_wrap_inline15218 is a rectangle.

  3. From one rectangle, construct arbitrarily large rectangles.
    Given any right triangle tex2html_wrap_inline15220, we can construct a rectangle tex2html_wrap_inline15222 so that PS>XZ and RS>YZ. By applying Archimedes Axiom, we can find a number n so that we lay off segment BD in the above rectangle on the ray tex2html_wrap_inline15232 to reach the point P so that tex2html_wrap_inline15236 and tex2html_wrap_inline15238. We make n copies of our rectangle sitting on PZ=PS. This gives us a rectangle with vertices P, Z=S, Y, and some other point. Now, using the same technique, we can find a number m and a point R on tex2html_wrap_inline15254 so that tex2html_wrap_inline15256 and tex2html_wrap_inline15258. Now, constructing m copies of the long rectangle, gives us the requisite rectangle containing tex2html_wrap_inline15220.
  4. Prove that all right triangles have defect 0.
    Let tex2html_wrap_inline15220 be an arbitrary right triangle. By Step 3 we can embed it in a rectangle tex2html_wrap_inline15222.

     figure2598
    Figure 11.6: Defect 0 for all right triangles 

    Since tex2html_wrap_inline15278, we have that tex2html_wrap_inline15280 and then, tex2html_wrap_inline15282 has defect 0. Using Corollary 1 to Theorem 11.4 we find tex2html_wrap_inline15284 and thus, tex2html_wrap_inline15286. Therefore, each triangle has defect 0.

  5. If every right triangle has defect 0, then every triangle has defect 0.
    As in the first step, use the foot of a vertex to decompose the triangle into two right triangles, each of which has defect 0, from Step 4. Thus, the original triangle has defect 0.

Corollary 1: If there is a triangle with positive defect, then all triangles have positive defect.

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