Geometry: An Interactive Journey to Mastery



Solutions

7. i. Label lengths as shown in Figure S.8.5/DEHOWKHƒDQJOHJLYHQ
   ii. a 1 = a because GT = GO.
   iii. xah yah  and EHFDXVHRIWKH3\WKDJRUHDQ
   theorem.
   iv. x = y because a 1 = a.
   v. A bisects TO because x = y.


8. i. Label angles w and z as shown in Figure S.8.6.
   ii. a = z because alternate interior angles
   for BE CD&.
   iii. z = w because vertical angles.
   iv. w = b because alternate interior angles
   for AC ED&.
   v. a = b because of algebra.


9. This one is tricky!
   i. Label angles x 1 , x, x 3 , y 1 , y as shown
   in Figure S.8.7.
   ii. x 1 = x because KG bisects ‘EBH.
   iii. y 1 = y because FL bisects ‘NCH.
   iv. xx mCBI ‘ because they are
   vertical angles.
   v. mCBI y y‘  because they are
   alternate interior angles for EI NJ&.
   vi. x 1 + x = y 1 + y because of the previous
   two steps.

G

TA Oxy

a (^1) h a 2
Figure S.8.5
A
zBw
a
b
C D

E

Figure S.8.6

A

F

G H

I

J

K

L

M

N

E

B

D

C

x 1 x^2 x

3

y 1 y^2

Figure S.8.7