Geo_Book_Answers
Geo_Book_Answers
Geo_Book_Answers
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<strong>Answers</strong> to Exercises<br />
5. See flowchart below.<br />
6. Given: ABC with BA BC, CD AD <br />
Show: BD is the angle bisector of ABC.<br />
B<br />
1<br />
2<br />
A<br />
C<br />
BA BC CD AD BD BD<br />
Given Given<br />
Same segment<br />
ABD CBD<br />
SSS<br />
1 2<br />
CPCTC<br />
→<br />
BD bisects ABC<br />
Definition of angle<br />
bisector<br />
7. The angle bisector does not go to the midpoint<br />
of the opposite side in every triangle, only in an<br />
isosceles triangle.<br />
8. NE, because it is across from the smallest angle<br />
in NAE. It is shorter than AE, which is across<br />
from the smallest angle in LAE.<br />
9. The triangles are congruent by SSS, so the two<br />
central angles cannot have different measures.<br />
10. PRN by ASA; SRE by ASA<br />
11. Cannot be determined. Parts do not<br />
correspond.<br />
56 ANSWERS TO EXERCISES<br />
D<br />
5. (Lesson 4.7)<br />
1<br />
2<br />
SA NE<br />
?<br />
?<br />
Given<br />
SE NA<br />
Given<br />
3<br />
4<br />
5<br />
3 4<br />
AIA Conjecture<br />
SN SN<br />
12. ACK by SSS<br />
13. a 72°, b 36°, c 144°, d 36°, e 144°,<br />
f 18°, g 162°, h 144°, j 36°, k 54°,<br />
m 126°<br />
14. The circumcenter is equidistant from all three<br />
vertices because it is on the perpendicular bisector<br />
of each side. Every point on the perpendicular<br />
bisector of a segment is equidistant from the<br />
endpoints. Similarly, the incenter is equidistant<br />
from all three sides because it is on the angle<br />
bisector of each angle, and every point on an angle<br />
bisector is equidistant from the sides of the angle.<br />
15. ASA. The fishing pole forms the side.<br />
“Perpendicular to the ground” forms one angle.<br />
“Same angle on her line of sight” forms the other<br />
angle.<br />
16. 2<br />
7 <br />
17.<br />
18. y<br />
4<br />
Y<br />
X<br />
B O<br />
B' Y'4<br />
?<br />
? 6<br />
<br />
? <br />
? 7<br />
?<br />
?<br />
?<br />
1 2<br />
ESN ANS<br />
SA NE <br />
AIA Conjecture ASA<br />
CPCTC<br />
Same segment<br />
O'<br />
This proof shows that in a parallelogram,<br />
opposite sides are congruent.<br />
X'<br />
x