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Concept. Example 1 Congruent Segments and Angles A. Name two unmarked congruent angles. Answer:  BCA and  A  BCA is opposite BA and  A is opposite.

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Presentation on theme: "Concept. Example 1 Congruent Segments and Angles A. Name two unmarked congruent angles. Answer:  BCA and  A  BCA is opposite BA and  A is opposite."— Presentation transcript:

1 Concept

2 Example 1 Congruent Segments and Angles A. Name two unmarked congruent angles. Answer:  BCA and  A  BCA is opposite BA and  A is opposite BC, so  BCA   A. ___

3 Example 1 Congruent Segments and Angles B. Name two unmarked congruent segments. Answer: BC  BD ___ BC is opposite  D and BD is opposite  BCD, so BC  BD. ___

4 A.A B.B C.C D.D Example 1a A.  PJM   PMJ B.  JMK   JKM C.  KJP   JKP D.  PML   PLK A. Which statement correctly names two congruent angles?

5 A.A B.B C.C D.D Example 1b B. Which statement correctly names two congruent segments? A.JP  PL B.PM  PJ C.JK  MK D.PM  PK

6 Concept

7 Since QP = QR, QP  QR. By the Isosceles Triangle Theorem, base angles P and R are congruent, so m  P = m  R. Use the Triangle Sum Theorem to write and solve an equation to find m  R. Example 2 Find Missing Measures A. Find m  R. Triangle Sum Theorem m  Q = 60, m  P = m  R Simplify. Subtract 60 from each side. Divide each side by 2. Answer: m  R = 60

8 Since all three angles measure 60, the triangle is equiangular. Because an equiangular triangle is also equilateral, QP = QR = PR. Since QP = 5, PR = 5 by substitution. Example 2 Find Missing Measures B. Find PR. Answer: PR = 5 cm

9 A.A B.B C.C D.D Example 2a A.30° B.45° C.60° D.65° A. Find m  T.

10 A.A B.B C.C D.D Example 2b A.1.5 B.3.5 C.4 D.7 B. Find TS.

11 Example 3 Find Missing Values ALGEBRA Find the value of each variable. Since  E =  F, DE  FE by the Converse of the Isosceles Triangle Theorem. DF  FE, so all of the sides of the triangle are congruent. The triangle is equilateral. Each angle of an equilateral triangle measures 60°.

12 Example 3 Find Missing Values m  DFE= 60Definition of equilateral triangle 4x – 8 = 60Substitution 4x= 68Add 8 to each side. x= 17Divide each side by 4. The triangle is equilateral, so all the sides are congruent, and the lengths of all of the sides are equal. DF= FEDefinition of equilateral triangle 6y + 3= 8y – 5Substitution 3= 2y – 5Subtract 6y from each side. 8= 2yAdd 5 to each side.

13 Example 3 Find Missing Values 4= yDivide each side by 2. Answer: x = 17, y = 4

14 A.A B.B C.C D.D Example 3 A.x = 20, y = 8 B.x = 20, y = 7 C.x = 30, y = 8 D.x = 30, y = 7 Find the value of each variable.

15 Example 4 Apply Triangle Congruence NATURE Many geometric figures can be found in nature. Some honeycombs are shaped like a regular hexagon. That is, each of the six sides and interior angle measures are the same. Given: HEXAGO is a regular polygon. ΔONG is equilateral, N is the midpoint of GE, and EX || OG. Prove:ΔENX is equilateral. ___

16 Example 4 Apply Triangle Congruence Proof: ReasonsStatements 1.Given1.HEXAGO is a regular polygon. 5.Midpoint Theorem 5.NG  NE 6.Given 6.EX || OG 2.Given 2.ΔONG is equilateral. 3. Definition of a regular hexagon 3. EX  XA  AG  GO  OH  HE 4. Given 4.N is the midpoint of GE

17 Example 4 Apply Triangle Congruence Proof: ReasonsStatements 7. Alternate Exterior Angles Theorem 7.  NEX   NGO 8.ΔONG  ΔENX 8. SAS 9.OG  NO  GN 9. Definition of Equilateral Triangle 10. NO  NX, GN  EN 10. CPCTC 11. XE  NX  EN 11. Substitution 12. ΔENX is equilateral. 12. Definition of Equilateral Triangle

18 Example 4 Proof: ReasonsStatements 1.Given1.HEXAGO is a regular hexagon. 2.Given 2.  NHE   HEN   NAG   AGN ___ Given: HEXAGO is a regular hexagon.  NHE   HEN   NAG   AGN Prove: HN  EN  AN  GN ___ 3.HE  EX  XA  AG  GO  OH 3.Definition of regular hexagon 4.ΔHNE  ΔANG 4.ASA

19 A.A B.B C.C D.D Example 4 Proof: ReasonsStatements 5.HN  AN, EN  NG 6.HN  EN, AN  GN 6.Converse of Isosceles Triangle Theorem 7.HN  EN  AN  GN 7.Substitution 5.CPCTE ___ Given: HEXAGO is a regular hexagon.  NHE   HEN   NAG   AGN Prove: HN  EN  AN  GN ___


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