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Pearson Unit 1 Topic 4: Congruent Triangles 4-2: Triangle Congruence by SSS and SAS Pearson Texas Geometry ©2016 Holt Geometry Texas ©2007.

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Presentation on theme: "Pearson Unit 1 Topic 4: Congruent Triangles 4-2: Triangle Congruence by SSS and SAS Pearson Texas Geometry ©2016 Holt Geometry Texas ©2007."— Presentation transcript:

1 Pearson Unit 1 Topic 4: Congruent Triangles 4-2: Triangle Congruence by SSS and SAS Pearson Texas Geometry ©2016 Holt Geometry Texas ©2007

2 TEKS Focus: (6)(B) Prove two triangles are congruent by applying the Side-Angle-Side, Angle-Side-Angle, Side-Side-Side, Angle-Angle-Side, and Hypotenuse-Leg congruence conditions. (1)(C) Select tools, including real objects, manipulatives paper and pencil, and technology as appropriate, and techniques, including mental math, estimations, and number sense as appropriate, to solve problems. (1)(E) Create and use representations to organize, record, and communicate mathematical ideas. (1)(F) Analyze mathematical relationships to connect and communicate mathematical ideas. (1)(G) Display, explain, or justify mathematical ideas and arguments using precise mathematical language in written or oral communication. (5)(A) Investigate patterns to make conjectures about geometric relationships, including angles formed by parallel lines cut by a transversal, criteria required for triangle congruence, special segments of triangles, diagonals of quadrilaterals, interior and exterior angles of polygons, and special segments and angles of circles choosing from a variety of tools. (5)(C) Use the constructions of congruent segments, congruent angles, angle bisectors, and perpendicular bisectors to make conjectures about geometric relationships.

3 An included angle is an angle formed by two adjacent sides of a polygon.
B is the included angle between sides AB and BC.

4 The property of triangle rigidity gives you a shortcut for proving two triangles congruent. It states that if the side lengths of a triangle are given, the triangle can have only one shape.

5 For example, you only need to know that two triangles have three pairs of congruent corresponding sides. This can be expressed as the following postulate.

6 It can also be shown that only two pairs of congruent corresponding sides are needed to prove the congruence of two triangles if the included angles are also congruent.

7 The letters SAS are written in that order because the congruent angles must be between pairs of congruent corresponding sides. Caution Adjacent triangles share a side, so you can apply the Reflexive Property to get a pair of congruent parts. Remember!

8 EXAMPLE 1: The triangles can be proved congruent by SSS.
They cannot be proven congruent by SAS because the angles are in different positions.

9 Example 2: LB is congruent to itself by the Reflexive Property.
That gives us 2 out of the 3 parts. We must have the side that makes ELB and NBL the included angle. Therefore, we need LE  NB.

10 Example: 3 1. Given 2. Given 4. SSS
Use SSS to explain why ∆ABC  ∆DBC. Statements Reasons 1. Given 2. Given 3. Reflexive Property of Congruence 4. SSS

11 Example: 4 1. Given 2. Given 4. SSS Use SSS to explain why
∆ABC  ∆CDA. Statements Reasons 1. Given 2. Given 3. Reflexive Property of Congruence 4. SSS

12 Example: 5 The diagram shows part of the support structure for a tower. Use SAS to explain why ∆XYZ  ∆VWZ. Statements Reasons 1. Given 2. Vertical s Thm. 3. Given 4. SAS

13 Example: 6 Use SAS to explain why ∆ABC  ∆DBC. 1. Given 2. Given
Statements Reasons 1. Given 2. Given 3. Reflexive Property of Congruence 4. SAS

14 Example: 7 ∆MNO  ∆PQR by SSS. S→ S→ S→ ∆MNO  ∆PQR, when x = 5. PQ
Show that the triangles are congruent for the given value of the variable. ∆MNO  ∆PQR, when x = 5. PQ = x + 2 = = 7 QR = x = 5 PR = 3x – 9 = 3(5) – 9 = 6 S→ PQ  MN, QR  NO, PR  MO S→ S→ ∆MNO  ∆PQR by SSS.

15 Example: 8 ∆STU  ∆VWX by SAS. S→ A→ S→ ∆STU  ∆VWX, when y = 4. ST
Show that the triangles are congruent for the given value of the variable. ∆STU  ∆VWX, when y = 4. ST = 2y + 3 = 2(4) + 3 = 11 TU = y + 3 = = 7 mT = 20y + 12 = 20(4)+12 = 92° S→ ST  VW T  W TU  WX A→ S→ ∆STU  ∆VWX by SAS.

16 Example: 9 ∆ADB  ∆CDB by SAS. Show that ∆ADB  ∆CDB, t = 4. DA
= 3(4) + 1 = 13 DC = 4t – 3 = 4(4) – 3 = 13 mD = 2t2 = 2(16)= 32° DA  DC Def. of  segments. ADB  CDB Def. of . DB  DB Reflexive Prop. of . ∆ADB  ∆CDB by SAS.

17 Example: 10 Given: BC ║ AD, BC  AD Prove: ∆ABD  ∆CDB Statements
Reasons 1. BC  AD 1. Given 2. BC || AD 2. Given 3. CBD  ABD 3. Alt. Int. s Thm. 4. BD  BD 4. Reflex. Prop. of  5. ∆ABD  ∆CDB 5. SAS

18 Example: 11 Given: QP bisects RQS. QR  QS Prove: ∆RQP  ∆SQP
Statements Reasons 1. QR  QS 1. Given 2. QP bisects RQS 2. Given 3. RQP  SQP 3. Def. of angle bisector 4. QP  QP 4. Reflexive Prop. of  5. ∆RQP  ∆SQP 5. SAS


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