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Answer Key
Chapter 4
Lesson
4.4
Challenge: Skills and Applications
1. a. ABD CDB (or equivalent)
b. Sample answer:
Statements
Reasons
1. ABC and CDA are
1. Given
right angles.
2. AB BC, CD DA
3. 1 and 2 are
complements; 3 and
4 are complements.
2. Def of lines
Given
2 ≅ 1, 5 ≅ 6
Def. of angle bisector
JL ≅ JL
∆JKL ≅ ∆JML
ASA congruence
postulate
Reflexive property
of congruence
KM bisects both
JKL and JML.
Given
2 ≅ 4
(2)
PT QS (1)
RQ ≅ QR
(3)
1 ≅ 3
(2)
5 ≅ 7
(2)
∆PQR ≅ ∆SRQ (4)
QR ST (1)
RS ≅ SR
(3)
R
6
Q≅Q
4
R
)(2)
3(
∆SRQ ≅ ∆RST (4)
∆PQR ≅ ∆RST (5)
3. If 2 sides of 2 adj.
acute are , then
the are complementary.
4. 1 4
4. Given
5. 2 3
5. Congruent
Complements Thm.
6. BD BD
6. Reflexive Prop. of
Congruence
7. ABD CDB
7. ASA Postulate
2. Sample answer: Since 7 12 (given)
and 8 11 (Vertical Angles Theorem), we
know that 1 6. (Third Angles Theorem)
Furthermore, since EF EI and HG HI, we
know that 1 and 2 are complements, as are
5 and 6. (If two sides of two adjacent acute
angles are perpendicular, then the angles are
complementary.) Therefore, 2 5 by the
Congruent Complements Theorem. Since IJ
bisects EIH, 3 4. Also, IJ IJ by
the Reflexive Property of Congruence. So,
EIJ HIJ by the AAS Congruence Theorem.
3. a. Sample answer: KJM KJM,
JKL JML
b. Sample answer:
JL bisects both
KJM and KLM.
4. Sample answer:
PQ RS (1)
3 ≅ 4, 7 ≅ 8
Def. of angle bisector
KM ≅ KM
Reflexive property
of congruence
∆KJM ≅ ∆KLM
ASA congruence
postulate
Reasons:
(1) Given
(2) Alternate Interior Angles Theorem
(3) Reflexive Property of Congruence
(4) ASA Congruence Postulate
(5) Transitive Property of Congruent Triangles
5. no; Sample answer: In a photographic enlargement, angles are preserved, but lengths are altered.
Therefore, two noncongruent triangles may have
congruent angles.
6. Consider any two triangles formed by two
braces, the flagpole, and the ground. The vertical
sides are the same segment, so they are congruent
by the Reflexive Property of Congruence. The base
sides are congruent, because they are radii of the
same circle. The included angles are congruent,
because they are both right angles. Therefore, the
triangles are congruent by the SAS Congruence
Postulate. Therefore, the hypotenuses are congruent,
and so the two braces are the same length. By the
Transitive Property of Congruence, all the braces
are the same length.