SAS Theorem

11 Key excerpts on "SAS Theorem"

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  eBook - ePub

  Geometry and Its Applications

  • Walter A. Meyer(Author)
  • 2006(Publication Date)
  • Academic Press

    (Publisher)

  C ′ — that is, the triangles are congruent.

  Figure 2.2 Side-angle-side axiom (the given parts are in solid lines).

  The argument just given seems a lot like a proof of the SAS axiom, but we have not labeled the SAS principle a theorem, which we would do if we had a proof of it. The reason our argument is not a proof is that it relies on the assumption that figures can be moved without changing the sizes of their sides and angles. There is no reason that such an axiom could not be added to our set of axioms — it is, after all, quite believable. However, it is common to exclude ideas of motion from the axioms of Euclidean geometry (but not from our intuitions about geometry), and we follow this approach.

  Here is our first deduction from the SAS axiom: the well-known theorem that base angles of isosceles triangles are congruent. Recall that a triangle is an isosceles triangle if it has two congruent sides (Figure 2.3 ).

  Figure 2.3 Base angles of an isosceles triangle are congruent.

  * THEOREM 2.1

  If two sides of a triangle are congruent, say, AB = BC in Figure 2.3 , then the angles opposite these sides are congruent, m ∠A = m ∠C .

  PROOF

  The idea is to show that A → C , B →B , C →A is a congruence of the triangle to itself. Under this 1:1 correspondence, ∠B corresponds to itself and is clearly congruent to itself, so we have the “angle” part of an SAS proof. For , the corresponding side is , and these are congruent by hypothesis. For , the corresponding side is , and these are congruent by the same hypothesis. By SAS, the correspondence is a congruence. Consequently m ∠A = m ∠C since these are corresponding angles.

  Our proof of Theorem 2.1

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  College Geometry

  A Unified Approach

  • (Author)
  • 2014(Publication Date)
  • Orange Apple

    (Publisher)

  If triangle ABC is congruent to triangle DEF, the relationship can be written mathematically as: In many cases it is sufficient to establish the equality of three corresponding parts and use one of the following results to deduce the congruence of the two triangles. The shape of a triangle is determined up to congruence by specifying two sides and the angle between them (SAS), two angles and the side between them (ASA) or two angles and a corresponding adjacent side (AAS). Specifying two sides and an adjacent angle (SSA), however, can yield two distinct possible triangles. ________________________ WORLD TECHNOLOGIES ________________________ Determining congruence Sufficient evidence for congruence between two triangles in Euclidean space can be shown through the following comparisons: • SAS (Side-Angle-Side): If two pairs of sides of two triangles are equal in length, and the included angles are equal in measurement, then the triangles are congruent. • SSS (Side-Side-Side): If three pairs of sides of two triangles are equal in length, then the triangles are congruent. • ASA (Angle-Side-Angle): If two pairs of angles of two triangles are equal in measurement, and the included sides are equal in length, then the triangles are congruent. The ASA Postulate was contributed by Thales of Miletus (Greek). In most systems of axioms, the three criteria— SAS , SSS and ASA —are established as theorems. In the School Mathematics Study Group system SAS is taken as one (#15) of 22 postulates. • AAS (Angle-Angle-Side): If two pairs of angles of two triangles are equal in measurement, and a pair of corresponding non-included sides are equal in length, then the triangles are congruent. • RHS (Right-angle-Hypotenuse-Side): If two right-angled triangles have their hypotenuses equal in length, and a pair of shorter sides are equal in length, then the triangles are congruent.

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  Dr. Math Presents More Geometry

  Learning Geometry is Easy! Just Ask Dr. Math

  • (Author)
  • 2005(Publication Date)
  • Jossey-Bass

    (Publisher)

  But we have several theorems (or postulates) that say we don’t have to check every part; it’s enough to check just three parts, if we choose the right three. Specifically, the following: SSS tells us that if each side of one matches the correspon- ding side of the other, then the triangles are congruent. ASA tells us that if two angles of one match the correspon- ding angles of the other and the sides between those angles match, then the trian- gles are congruent. SAS tells us that if two sides of one match the correspon- ding sides of the other and the angles between them match, then the triangles are congruent. 70 Dr. Math Presents More Geometry Now, it is not enough to say that a pair of angles look the same; you have to know they are. Typically in geometry proof, you are told (or know for other reasons) that certain parts are congruent. For example, you might be told in this figure that PQ is the same length as RS and ∠PQS has the same measure as RSQ. You know without having to be told that any segment is congruent to itself, so QS ≅ SQ. So if we think about triangles PQS and RSQ (with the letters given in the order we want to compare them), we can turn the latter around in our minds so that they line up with the letters in corre- sponding spots like this: Now we have three pairs of matching parts: PQ ≅ RS ∠PQS ≅ ∠RSQ QS ≅ SQ And these are in the order SAS, with the angle between the sides. So the SAS Theorem tells us that the two triangles are congruent. We don’t have to know anything about the other side and angles; it’s like saying that if the left ears match, the right ears will match, too, so we don’t need to check them. —Dr. Math, The Math Forum Triangles: Properties, Congruence, and Similarity 71 P S Q R Q P S S R Q Hi, Qian, For many combinations of side-side-angle, two triangles will be con- gruent. But there are instances when this isn’t enough information to determine a unique triangle.

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  Exploring Geometry

  • Michael Hvidsten(Author)
  • 2016(Publication Date)
  • Chapman and Hall/CRC

    (Publisher)

  ASA: Angle-Side-Angle ,

  Prop. 26 of Book I) If in two triangles there is a correspondence in which 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

  .

  Theorem 2.12 . (AAS: Angle-Angle-Side ,

  Prop. 26 of Book I) If in two triangles there is a correspondence in which two angles and the side subtending one of the angles are congruent to two angles and the side subtending the corresponding angle of another triangle, then the triangles are congruent

  .

  Theorem 2.13 . (SSS: Side-Side-Side , Prop. 8 of Book I) If in two triangles there is a correspondence in which the three sides of one triangle are congruent to the three sides of the other triangle, then the triangles are congruent .

  We note here for future reference that the four fundamental triangle congruence results are independent of the parallel postulate; that is, their proofs do not make reference to any result based on the parallel postulate.

  Let’s see how triangle congruence can be used to analyze isosceles triangles.

  Definition 2.15 . An isosceles triangle is a triangle that has two sides congruent. The two congruent sides are called the legs of the triangle and the third side is called the base. The base angles of the triangle are those angles sharing the base as a side .

  Isosceles triangles were a critical tool for many of Euclid’s proofs, and he introduced the next result very early in the Elements . It followed immediately after SAS congruence (Proposition 4).

  Theorem 2.14 . (Prop. 5 of Book I) In an isosceles triangle, the two base angles are congruent .

  Proof: Let triangle ΔABC have sides

  A B

  ¯

  and

  B C

  ¯

  congruent, as shown in Figure 2.9 . Let

  B D

  →

  be the bisector of ∠ABC , with D inside the triangle. Let

  B D

  →

  intersect

  A C

  ¯

  at E . Then, by SAS we have triangles ΔABE and ΔCBE congruent and thus ∠EAB ≅ ∠ECB

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  Elementary Geometry for College Students

  • Daniel C. Alexander, Geralyn M. Koeberlein(Authors)
  • 2019(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  If two triangles are drawn so that two angles measure 33° and 47° while their included side measures 5 centimeters, then these triangles must be congruent. See Figure 3.8. (a) 5 cm 338 478 (b) 5 cm 338 478 Figure 3.8 POSTULATE 14 If two angles and the included side of one triangle are congruent to two angles and the included side of a second triangle, then the triangles are congruent (ASA). Although the method in Postulate 14 is written compactly as ASA, you must use cau- tion as you write these abbreviations that verify that triangles are congruent! For example, ASA refers to two angles and the included side, whereas SAS refers to two sides and the included angle. To apply any postulate, the specific conditions described in it must be satisfied. SSG EXS. 3–6 Figure 3.7 1 2 N Q M P 3.1 ■ Congruent Triangles 141 SSS, SAS, and ASA are all valid methods of proving triangles congruent. However, SSA is not a method and cannot be used; in Figure 3.9, the two triangles are marked to demonstrate the SSA relationship, yet the two triangles are not congruent. (a) 20° 5 cm 2 cm (b) 20° 5 cm 2 cm Figure 3.9 Another combination that cannot be used to prove triangles congruent is AAA. See Figure 3.10. Three pairs of congruent angles in two triangles do not guarantee three pairs of congruent sides! In Example 6, the triangles to be proved congruent overlap (see Figure 3.11). To clar- ify relationships, the triangles have been redrawn separately in Figure 3.12. In Figure 3.12, the parts indicated as congruent are established as congruent in the proof. For statement 3, Identity (or Reflexive) is used to justify that an angle is congruent to itself. EXAMPLE 6 GIVEN: AC _ DC uni22201 _ uni22202 (See Figure 3.11.) PROVE: nACE _ nDCB PROOF Statements Reasons 1. AC _ DC (See Figure 3.12.) 2. uni22201 _ uni22202 3. uni2220C _ uni2220C 4. nACE _ nDCB 1. Given 2. Given 3. Identity 4. ASA Next we consider the AAS theorem; this theorem can be proved by applying the ASA postulate.

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  The Mathematics That Every Secondary School Math Teacher Needs to Know

  • Alan Sultan, Alice F. Artzt(Authors)
  • 2017(Publication Date)
  • Routledge

    (Publisher)

  The first result we talk about is something we are all familiar with: If three sides of one triangle have the same lengths as three sides of another triangle, then the triangles are congruent. That is, all their corresponding parts match! This is quite remarkable since we have said nothing about the angles of these triangles. Yet, this follows immediately from the Law of Cosines.

  Theorem 5.2 (SSS = SSS) If the three sides of triangle ABC are congruent to the three sides of triangle DEF, then the triangles ABC and DEF are congruent .

  Proof . Let us assume that the sides that match are a and d, b and e , and c and f . (Refer to Figure 5.2 .) So a = d, b = e , and c = f . From equation (5.8) we have that

  cos C = a 2 + b 2 - c 2 2 a b

  (5.11)

  and using the same law in triangle DEF with the corresponding sides, we have

  cos F = d 2 + e 2 - f 2 2 d e .

  (5.12)

  Since a = d, b = e , and c = f we can substitute them in equation (5.11) to get

  cos C = d 2 + e 2 - f 2 2 d e

  (5.13)

  and we see from equations (5.12) and (5.13) that

  cos C = cos F .

  (5.14)

  It follows that ∡C = ∡F .

  In a similar manner using the other versions of the Law of Cosines, equations (5.9) and (5.10) , we can show that ∡A = ∡D and ∡B = ∡E .

  Thus, if three sides of one triangle are equal to three sides of another triangle, then the corresponding angles match, and so the triangles are congruent.

  Note: In the proof of Theorem 5.2 (refer to equation (5.14) ), we used the fact that, if cos C = cos F , then ∡C = ∡F . While you may have accepted this, much more is involved in this statement than meets the eye. For now, we will use this fact continually and ask you to accept it. But, we will examine the reason behind it in a later section on technical issues. We now turn to a corollary of our SSS congruence theorem.

  Corollary 5.3 (HL = HL) Two right triangles are congruent if the hypotenuse and leg of one triangle are congruent to the hypotenuse and leg of the other triangle .

  Proof . In Figure 5.3 we see two right triangles where the hypotenuse and leg of one have the same lengths as the hypotenuse and leg of the other.

  Figure 5.3

  We will show the third sides of the triangles have the same length. By the Pythagorean Theorem,

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  Course in Triangle Geometry, A

  • (Author)
  • 2014(Publication Date)
  • Learning Press

    (Publisher)

  (The included angle for any two sides of a polygon is the internal angle between those two sides.) • If three corresponding sides of two triangles are in proportion, then the triangles are similar. Two triangles that are congruent have exactly the same size and shape: all pairs of corresponding interior angles are equal in measure, and all pairs of corresponding sides have the same length. (This is a total of six equalities, but three are often sufficient to prove congruence.) Some sufficient conditions for a pair of triangles to be congruent are: • SAS Postulate: Two sides in a triangle have the same length as two sides in the other triangle, and the included angles have the same measure. • ASA: Two interior angles and the included side in a triangle have the same measure and length, respectively, as those in the other triangle. (The included side for a pair of angles is the side that is common to them.) • SSS: Each side of a triangle has the same length as a corresponding side of the other triangle. • AAS: Two angles and a corresponding (non-included) side in a triangle have the same measure and length, respectively, as those in the other triangle. • Hypotenuse-Leg (HL) Theorem: The hypotenuse and a leg in a right triangle have the same length as those in another right triangle. This is also called RHS (right-angle, hypotenuse, side). • Hypotenuse-Angle Theorem: The hypotenuse and an acute angle in one right triangle have the same length and measure, respectively, as those in the other right triangle. This is just a particular case of the AAS theorem.

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  College Geometry

  A Unified Development

  • David C. Kay(Author)
  • 2011(Publication Date)
  • CRC Press

    (Publisher)

  But also, BA ′ = BA = BC by hypothesis and by (7) , Section 1.8 , A ′ = C . Thus A → C . This fact will be used in the SAS Theorem below. We are now ready for the main theorem (Euclid’s proposition I.4), which is implied by Axiom M , and a rigorous argument can be given for it. SAS Congruence Criterion from Axiom M If two sides and the included angle of one triangle are congruent to the corresponding two sides and included angle of another triangle, then the triangles are congruent. Proof Suppose that in triangles ABC and DEF we have AB = DE , AC = DF , and m ∠ BAC = m ∠ EDF (Figure 3.18). Our goal is to find a motion that maps Δ ABC to Δ DEF . Let M be the midpoint of segment AD (or if AD = α take M as any point midway between A and D ), and let l 1 be the line perpendicular to line AM at M . Then, under the reflection P → P ′ in line l 1 (by Axiom M ), A → A ′ ≡ D , B → B ′ , and C → C ′ . (The shaded triangles in the figure show the images of the given triangle under the reflections to be considered.) Now take the line l 2 containing the bisector of ∠ B ′ DE (unless B ′ = E ) and consider the reflection in l 2 . By previous observations, B ′ → B ″ ≡ E (and D ′ → D ″ ≡ D ). Now, so far, we have the product P → P ′ → P ″ of the two reflections in l 1 and l 2 mapping A to D , B to E , and C to C ″ . Next, consider line l DE 3 = arrowleftnosp arrowrightnosp combarrowextendercombarrowextender (if C ″ ≠ F ). This line contains the bisector of ∠ FEC ″ since by hypothesis m ∠ EDF = m ∠ BAC = m ∠ B ″ A ″ C ″ = m ∠ EDC ″ . (Or, if rays EF and EC are opposite rays, then l FD 3 ⊥ arrowleftnosp arrowrightnosp combarrowextendercombarrowextender and D is midway between F and C ″ .) Thus, if P ″ → P ″′ denotes the reflection in line l 3 , then C ″ maps to F , and D and E are left fixed. The product of the three reflections then maps Δ ABC to Δ DEF . Since motions preserve distance and angle measure, it follows that Δ ABC ≅ Δ DEF .

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  Elementary Geometry for College Students

  • Daniel C. Alexander, Geralyn M. Koeberlein, , , Daniel C. Alexander, Geralyn M. Koeberlein(Authors)
  • 2014(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  Identity 5. SAS ∠ s ∠ Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 142 CHAPTER 3 ■ TRIANGLES General Rule: An early statement of the proof establishes the “helping line,” such as the altitude or the angle bisector. Illustration: See the second line in the proof of Example 2. The chosen angle bisector leads to congruent triangles, which enable us to complete the proof. STRATEGY FOR PROOF ■ Using an Auxillary Line EXAMPLE 2 Give a formal proof of Theorem 3.3.3. GIVEN: Isosceles with [See Figure 3.31(a).] PROVE: NOTE: Figure 3.31(b) shows the auxiliary segment, the bisector of . PROOF Statements Reasons For the proof of Theorem 3.3.3, a different angle bisector (such as the bisector of ) would not lead to congruent triangles; that is, the choice of auxiliary line must lead to the desired outcome! Theorem 3.3.3 is sometimes stated, “The base angles of an isosceles triangle are congruent.” We apply this theorem in Example 3. EXAMPLE 3 Find the size of each angle of the isosceles triangle shown in Figure 3.32 on page 143 if: a) b) The measure of each base angle is less than twice the measure of the vertex angle SOLUTION a) . Since and and are we have Thus, and . m ∠ 2 m ∠ 3 72 m ∠ 1 36 m ∠ 2 72 2(m ∠ 2) 144 36 2(m ∠ 2) 180 , ∠ 3 ∠ 2 m ∠ 1 36 m ∠ 1 m ∠ 2 m ∠ 3 180 5 m ∠ 1 36 ∠ M ∠ MPN ∠ M ∠ N MP NP MNP P M N (a) P M N (b) Q THEOREM 3.3.3 If two sides of a triangle are congruent, then the angles opposite these sides are also congruent. Figure 3.31 1. Isosceles with 2. Draw bisector from P to 3.

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  Mathematics for Elementary Teachers

  A Contemporary Approach

  • Gary L. Musser, Blake E. Peterson, William F. Burger(Authors)
  • 2013(Publication Date)
  • Wiley

    (Publisher)

  Is Roberto correct? Discuss. GEOMETRIC PROBLEM SOLVING USING TRIANGLE CONGRUENCE AND SIMILARITY The two parallelograms below have congruent corresponding sides, but the parallelograms themselves are not congruent. The two tri- angles below have congruent corresponding sides and, by the SSS triangle congruence, the triangles are congruent. These two ideas are related to some principles of fence building. When building a gate for a fence, a crosspiece (A), like the one shown, is added to ensure a stron- ger more stable gate that won’t sag. Why does the diagonal (A) add so much strength that the two parallel pieces (B and C ) couldn’t provide? How is this strengthening crosspiece related to the SSS triangle congruence? 766 Chapter 14 Geometry Using Triangle Congruence and Similarity Applications of Triangle Congruence In this section we apply triangle congruence and similarity properties to prove properties of geometric shapes. Many of these results were observed informally in Chapter 12. Our first result is an application of the SAS congruence property that establishes a property of the diagonals of a rectangle. Problem-Solving Strategy Draw a Picture Show that the diagonals of a rectangle are congruent. SOLUTION Suppose that ABCD is a rectangle [Figure 14.50(a)]. Figure 14.50 A B D C AC = BD (a) A B D C (b) Then ABCD is a parallelogram (from Chapter 12). By a theorem in Section 14.1, the opposite sides of ABCD are congruent [Figure 14.50(b)]. In particular, AB DC ≅ . Consider nABC and n n DCB; ABC C ≅ ∠D B, since they are both right angles. Also, CB BC ≅ . Hence, n n ABC DCB ≅ by the SAS congruence property. Consequently, AC BD ≅ , as desired. ■ The next result is a form of a converse of the theorem in Example 14.11. In parallelogram ABCD, if the diagonals are congruent, it is a rectangle. SOLUTION Suppose that ABCD is a parallelogram with AC BD ≅ [Figure 14.51(a)]. Figure 14.51 Since ABCD is a parallelogram, AD BC ≅ .

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  Methods of Geometry

  • James T. Smith(Author)
  • 2011(Publication Date)
  • Wiley-Interscience

    (Publisher)

  Theorem 9. The area of a triangle is half the product of a base b by the corresponding altitude a. Proof. Consult figure 3.8.4. There are two cases. • Corollary 10. The area of a trapezoid is half the product of an altitude by the sum of the corresponding bases. Proof. Draw a diagonal. • Corollary 11. The area of a parallelogram is the product of a base by the corresponding altitude. Pythagoras' theorem that the sum of the squares of the legs of a right triangle equals the square of its hypotenuse is often called the most important in mathematics. While that may be mere grandiloquence, the result does prove fundamental for many different subjects. Set in modern algebraic language, it may seem to be about lengths, hence inappropriate for inclusion in this section. But it was originally viewed as a theorem about the areas of squares erected on the triangle's edges. Therefore, it's appropriate to present a proof based on area. Section 3.9 will give another proof, based on similarity theory. Theorem 12 (Pythagoras' theorem). The square of the hypotenuse of a right triangle is the sum of the squares of its legs. 86 ELEMENTARY EUCLIDEAN GEOMETRY Figure 3.8.5 Proof of Pythagoras' theorem Proof. Consult figure 3.8.5. Start with AOPQ where LO is a right angle. Construct R, S, T, U, and V as shown. Verify that V, Q, and Ο are collinearand VQ = b. Note that AOPQ=ARSP=ATUS=AVQU and that ORTV and PSUQ are squares. Then (PQ) 2 = aPSUQ = &ORTV - A&AOPQ = (a + b) 2 - 4( ι Λα6) = 2 + 6 2 . • The following result is a converse of Pythagoras' theorem: A triangle is right if its edges satisfy Pythagoras' equation. For example, a triangle with edges of length 3, 4, and 5 is right because 3 2 + 4 2 = 5 2 . This provided ancient surveyors a practical method for constructing right angles—form a loop of rope with 3 + 4 + 5 = 12 equally spaced knots, and have assistants hold appropriate knots, stretching the rope taut into a right triangle.

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