TEKS: G10B, G9B, G5B, G2B The student will justify and apply triangle congruence relationships. The student will formulate and test conjectures about the properties and attributes of polygons and their component parts based on explorations. The student will use numeric and geometric patterns to make generalizations about properties of polygons. The student will make conjectures about angles, lines, and polygons
Geometric figures are congruent if they are the same size and shape. Corresponding angles and corresponding sides are in the same position in polygons with an equal number of sides. Two polygons are congruent polygons if and only if their corresponding sides are congruent. Thus triangles that are the same size and shape are congruent.
CPCTC is an abbreviation for the phrase Corresponding Parts of Congruent Triangles are Congruent. It can be used as a justification in a proof after you have proven two triangles congruent.
Helpful Hint Two vertices that are the endpoints of a side are called consecutive vertices. For example, P and Q are consecutive vertices. P R Q
To name a polygon, write the vertices in consecutive order. For example, you can name polygon PQRS as QRSP or SRQP, but not as PRQS. In a congruence statement, the order of the vertices indicates the corresponding parts. P Q S R
Helpful Hint When you write a statement such as ABC DEF, you are also stating which parts are congruent.
Given: PQR STW Identify all pairs of corresponding congruent parts. Angles: P S, Q T, R W Sides: PQ ST, QR TW, PR SW
If polygon LMNP polygon EFGH, identify all pairs of corresponding congruent parts. Angles: L E, M F, N G, P H Sides: LM EF, MN FG, NP GH, LP EH
Some hikers come to a river in the woods. They want to cross the river but decide to find out how wide it is first. So they set up congruent right triangles. The figure shows the river and the triangles. Find the width of the river, GH. Explain. 5 meters, by using CPCTC
Example: 4 Given: ABC DBC. Find the value of x. BCA and BCD are rt. s. BCA BCD m BCA = m BCD (2x 16) = 90 2x = 106 x = 53 Def. of lines. Rt. Thm. Def. of s Substitute values for m BCA and m BCD. Add 16 to both sides. Divide both sides by 2.
Example: 5 Given: ABC DBC. Find m DBC. m ABC + m BCA + m A = 180 m ABC + 90 + 49.3 = 180 m ABC + 139.3 = 180 m ABC = 40.7 Sum Thm. Substitute values for m BCA and m A. Simplify. Subtract 139.3 from both sides. DBC ABC m DBC = m ABC Corr. s of s are. Def. of s. m DBC 40.7 Trans. Prop. of =
Example: 6 Given: ABC DEF Find the value of x. AB DE AB = DE 2x 2 = 6 2x = 8 x = 4 Corr. sides of s are. Def. of parts. Substitute values for AB and DE. Add 2 to both sides. Divide both sides by 2.
Example: 7 Given: ABC DEF Find m F. m EFD + m DEF + m FDE = 180 ABC DEF m ABC = m DEF Sum Thm. Corr. s of are. Def. of s. m DEF = 53 Transitive Prop. of =. m EFD + 53 + 90 = 180 Substitute values for m DEF and m FDE. m F + 143 = 180 m F = 37 Simplify. Subtract 143 from both sides.
Example: 8 Given: YWX and YWZ are right angles. YW bisects XYZ. W is the midpoint of XZ. XY YZ. Prove: XYW ZYW
Statements Reasons 1. YWX and YWZ are rt. s. 1. Given 2. YWX YWZ 2. Rt. Thm. 3. YW bisects XYZ 3. Given 4. XYW ZYW 4. Def. of bisector 5. W is mdpt. of XZ 6. XW ZW 7. YW YW 8. X Z 9. XY YZ 10. XYW ZYW 5. Given 6. Def. of mdpt. 7. Reflex. Prop. of 8. Third s Thm. 9. Given 10. Def. of
Example: 9 The diagonal bars across a gate give it support. Since the angle measures and the lengths of the corresponding sides are the same, the triangles are congruent. Given: PR and QT bisect each other. PQS RTS, QP RT Prove: QPS TRS
Statements Reasons 1. QP RT 2. PQS RTS 3. PR and QT bisect each other. 4. QS TS, PS RS 5. QSP TSR 6. QSP TRS 7. QPS TRS 1. Given 2. Given 3. Given 4. Def. of bisector 5. Vert. s Thm. 6. Third s Thm. 7. Def. of s
Example: 10 Use the diagram to prove the following. Given: MK bisects JL. JL bisects MK. JK ML. JK ML. Prove: JKN LMN
Statements Reasons 1. JK ML 2. JK ML 3. JKN NML 4. JL and MK bisect each other. 5. JN LN, MN KN 6. KNJ MNL 7. KJN MLN 8. JKN LMN 1. Given 2. Given 3. Alt int. s are. 4. Given 5. Def. of bisector 6. Vert. s Thm. 7. Third s Thm. 8. Def. of s
Example: ON HW Given: AD bisects BE. BE bisects AD. AB DE, A D Prove: ABC DEC
1. A D Statements 1. Given Reasons 2. BCA DCE 3. ABC DEC 4. AB DE 5. AD bisects BE, 2. Vertical s are. 3. Third s Thm. 4. Given 5. Given BE bisects AD 6. BC EC, AC DC 7. ABC DEC 6. Def. of bisector 7. Def. of s