SAS Theorem

3 Key excerpts on "SAS Theorem"

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  eBook - ePub

  Geometry For Dummies

  • Mark Ryan(Author)
  • 2016(Publication Date)
  • For Dummies

    (Publisher)

  (If you fill in numbers, you can see that if and were both, and would both be.) Here’s the formal proof: © John Wiley & Sons, Inc. ASA: Taking the angle-side-angle tack ASA (Angle-Side-Angle): The ASA postulate says that if two angles and the included side of one triangle are congruent to two angles and the included side of another triangle, then the triangles are congruent. (The included side is the side between the vertices of the two angles.) See Figure 9-5. © John Wiley & Sons, Inc. FIGURE 9-5: The congruence of two pairs of angles and the side between them make these triangles congruent. Here’s a congruent-triangle proof with the ASA postulate: © John Wiley & Sons, Inc. Here’s my game plan: Note any congruent sides and angles in the diagram. First and foremost, notice the congruent vertical angles (I introduce vertical angles in Chapter 2). Vertical angles are important in many proofs, so you can’t afford to miss them. Next, midpoint E gives you. So now you have a pair of congruent angles and a pair of congruent sides. Determine which triangle postulate you need to use. To finish with SAS, you’d need to show ; and to finish with ASA, you’d need. A quick glance at the bisected angles in the givens (or the title of this section, but that’s cheating!) makes the second alternative much more likely. Sure enough, you can get because a given says that half of () is congruent to half of (). That’s a wrap. Here’s how the formal proof plays out: © John Wiley & Sons, Inc. STICKING TO THE BASICS: THE WORKINGS BEHIND SSS, SAS, AND ASA The idea behind the SSS postulate is pretty simple. Say you have three sticks of given lengths (how about 5, 7, and 9 inches?) and then you make a triangle out of them. Now you take three more sticks of the same lengths and make a second triangle

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  eBook - ePub

  Spherical Geometry and Its Applications

  • Marshall A. Whittlesey(Author)
  • 2019(Publication Date)
  • Chapman and Hall/CRC

    (Publisher)

  π, and this sum is not generally the same from one triangle to another.

  However, the angle sum theorem in the plane depends on the parallel postulate, and in fact the SAA congruence theorem in the plane does not depend on the parallel postulate. The reader will find a proof of SAA congruence in the plane in [Mo1963] (or [MD1982]) which makes use of the exterior angle theorem in the plane (which states that an exterior angle of a triangle has measure greater than the measures of either of the opposite interior angles). The spherical analogue of this theorem also turns out to be false.

  In plane geometry the SSA correspondence does not guarantee congruence of triangles, although we might say that it almost does. Knowledge of the measures of two sides and an angle that is not included leads to the so-called “ambiguous case” in determining the other sides of the triangle. The other sides are uniquely determined if the angle is a right angle, but otherwise there are either two possibilities for the other sides and angles, or the triangle cannot be constructed at all.

  We can make an immediate observation for spherical geometry: either SSA and SAA both guarantee congruence of triangles or neither does. The reason is that if one guarantees congruence, then we could prove that the other does as well by employing the same argument with polar triangles used in the proofs of ASA and AAA congruence. For example, suppose there were an SAA congruence theorem. To prove that an SSA correspondence guarantees congruence, we would consider two triangles which have an SSA correspondence. Their polar triangles would have an SAA correspondence by Theorem 11.19, and hence would be congruent. The congruence of the pair of polar triangles would in turn guarantee (by Theorem 11.19 again) that the original triangles be congruent. A similar line of reasoning would allow us to use an SSA congruence theorem to prove an SAA congruence theorem.

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  Lapses in Mathematical Reasoning

  • V. M. Bradis, L. Minkovskii, A. K. Kharcheva(Authors)
  • 2016(Publication Date)
  • Dover Publications

    (Publisher)

  III. Stories, with Explanations, of Causes of Erroneous Reasoning

  63. Similar triangles with equal sides

  Take two similar scalene triangles and denote the sides of the first in the order of increasing size by the letters a, b, c (a < b < c), and the corresponding sides of the second by the letters a1 , b1 , c1 . By virtue of proportionality of the corresponding sides of similar polygons we have: a1 = aq, b1 = bq, c1 = cq, where q is the coefficient of proportionality, and therefore a1 < b1 < c1 .

  If q = 1, then all the sides of the two triangles are respectively equal, and the triangles are congruent. The congruence of the triangles is thus a special case of similarity.

  It may seem, that if the triangles are similar, but not congruent, then they have no equal sides. The erroneousness of such a conclusion is shown by a simple consideration of triangles with the sides 8, 12, 18 cm, and 12, 18, 27 cm. The sides of the second are one-and-one-half times as large as the corresponding sides of the first, and therefore the triangles are similar (but not congruent). As we see, these two triangles have two pairs of respectively equal sides.

  We shall establish the conditions which have to be satisfied by two similar, but not congruent, triangles having two pairs of respectively equal sides.

  Suppose that q > 1. The smallest side a of the first triangle (with the sides a, b, c) is less than the smallest side a1 of the second triangle (with sides a1 = aq, b1 = bq, c1 = cq), and cannot be equal to any of the sides of the latter.

  The middle side b of the first triangle may be equal only to the smallest side a1 of the second triangle (since b is less than b1 = bq and much smaller than c1 = cq), and the greatest side c of the first triangle is equal either to the smallest side a1 or to the middle side b1 of the second triangle. If two sides of the first triangle are equal to two sides of the second triangle, then b = a1 , c = b1 . Hence b = aq, c = bq = aq × q = aq2 . Thus the sides of the first triangle form a geometric progression a, aq, aq2 . The sides of the second triangle are equal to aq, aq2 , aq3

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