15.04.2013 Views

Geo_Book_Answers

Geo_Book_Answers

Geo_Book_Answers

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

<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

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!