Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Name ________________________________________ Date __________________ Class__________________ LESSON 4-7 Reteach Triangle Congruence: CPCTC Corresponding Parts of Congruent Triangles are Congruent (CPCTC) is useful in proofs. If you prove that two triangles are congruent, then you can use CPCTC as a justification for proving corresponding parts congruent. Given: AD ≅ CD, AB ≅ CB Prove: ∠A � ∠C Proof: Complete each proof. 1. Given: ∠PNQ � ∠LNM, PN ≅ LN, N is the midpoint of QM . Prove: PQ ≅ LM Proof: 2. Given: UUXW and UUVW are right Us. UX ≅ UV Prove: ∠X � ∠V Proof: Statements Reasons 1. UUXW and UUVW are rt. Us. 1. Given 2. UX ≅ UV 2. a. __________________________ 3. UW ≅ UW 3. b. __________________________ 4. c. __________________________ 4. d. __________________________ 5. ∠X � ∠V 5. e. __________________________ Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor. 4-54 Holt McDougal Geometry Name ________________________________________ Date __________________ Class__________________ LESSON 4-7 Reteach Triangle Congruence: CPCTC continued You can also use CPCTC when triangles are on the coordinate plane. Given: C(2, 2), D(4, �2), E(0, �2), F(0, 1), G(�4, �1), H(�4, 3) Prove: ∠CED � ∠FHG Step 1 Plot the points on a coordinate plane. Step 2 Find the lengths of the sides of each triangle. Use the Distance Formula if necessary. d = ( x2 − x1 )2 + ( y 2 − y1 )2 CD = (4 − 2)2 + ( −2 − 2)2 = 4 + 16 = 2 5 FG = ( −4 − 0)2 + ( −1 − 1)2 = 16 + 4 = 2 5 DE = 4 GH = 4 EC = (2 − 0)2 + [2 − ( −2)]2 HF = [0 − ( −4)]2 + (1 − 3)2 = 4 + 16 = 2 5 = 16 + 4 = 2 5 So, CD ≅ FG, DE ≅ GH, and EC ≅ HF . Therefore UCDE � UFGH by SSS, and ∠CED � ∠FHG by CPCTC. Use the graph to prove each congruence statement. 3. ∠RSQ � ∠XYW 4. ∠CAB � ∠LJK _________________________________________ ________________________________________ _________________________________________ ________________________________________ _________________________________________ ________________________________________ _________________________________________ ________________________________________ 5. Use the given set of points to prove ∠PMN � ∠VTU. M(�2, 4), N(1, �2), P(�3, �4), T(�4, 1), U(2, 4), V(4, 0) _________________________________________________________________________________________ _________________________________________________________________________________________ Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor. 4-55 Holt McDougal Geometry 3. Statements ∠GHF, and ∠IHF are congruent, so FH bisects ∠IFG and ∠IHG. Similar reasoning shows that GI bisects ∠FGH and ∠FIH. Reasons 1. FGHI is a rectangle. 1. Given 2. FI ≅ GH , ∠FIH and ∠GHI are right angles. 2. Def. of rectangle 3. The diagonals of a rectangle bisect each other. 3. ∠FIH ≅ ∠GHI 3. Rt. ∠ ≅ Thm. 4. IH ≅ IH 4. Reflex. Prop. of ≅ 4. The diagonals of a square are congruent perpendicular bisectors that bisect the vertex angles of the square. 5. UFIH ≅ UGHI 5. SAS 5. The diagonals are congruent. 6. FH ≅ GI 6. CPCTC 7. FH = GI 7. Def. of ≅ segs. Reteach 1. Practice C 1. Possible answer: From the definition of a parallelogram, DC is congruent to AB and DC is parallel to AB . By the Alternate Interior Angles Theorem, ∠BAC is congruent to ∠DCA and ∠CDB is congruent to ∠ABD. Therefore UABE is congruent to UCDE by ASA. By CPCTC, DE is congruent to BE and AE is congruent to CE . Congruent segments have equal lengths, so the diagonals bisect each other. 2. Possible answer: From the definition of a rhombus, IH is congruent to FG , IF is congruent to GH , and IH is parallel to FG . By Alternate Interior Angles Theorem, ∠GFH is congruent to ∠IHF and ∠FGI is congruent to ∠HIG. Therefore UFGJ is congruent to UHIJ by ASA. By CPCTC, FJ is congruent to HJ and GJ is congruent to IJ . So UFJI is congruent to UGHJ by SSS. But UHIJ is also congruent to UFIJ by SSS. And so all four triangles are congruent by the Transitive Property of Congruence. By CPCTC and the Segment Addition Postulate, FH is congruent to GI . By CPCTC and the Linear Pair Theorem, ∠FJI, ∠GJF, ∠HJG, and ∠IJH are right angles. So FH and GI are perpendicular. By CPCTC, ∠GFH, ∠IFH, 2. Statements Reasons 1. UUXW and UUVW are rt. Us. 1. Given 2. UX ≅ UV 2. a. Given 3. UW ≅ UW 3. b. Reflex. Prop. of ≅ 4. c. UUXW ≅ UUVW 4. d. HL 5. ∠X ≅ ∠V 5. e. CPCTC 3. QR = WX = 13 , RS = XY = 7, SQ = YW = 34 . So UQRS ≅ UWXY by SSS, and ∠RSQ ≅ ∠XYW by CPCTC. 4. AB = JK = 5, BC = KL = 10, CA = LJ = 53. So UABC ≅ UJKL by SSS, and ∠CAB ≅ ∠LJK by CPCTC. 5. MN = TU = 3 5 , NP = UV = 2 5 , PM = VT = 65 . So UMNP ≅ UTUV by SSS, and ∠PMN ≅ ∠VTU by CPCTC. Challenge Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor. A42 Holt McDougal Geometry