Download Reteach

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