Hw
i
品
ㄥ ㄧˇ
i
2
vW
⼤
ˇ
7
u tW
o
_ˇ
ˇ
2
2 11
3
B
it 1 1
的 BT
c蕊
c
ˇ
A
4
n AE
D
ā
3it4j
31j
乃
i
不⾔
ātb DB.DE
x
āt b
AT ā 5
万 AT ā
万 AÉ⼆ DE
万
ylǎ 5
5
yātyn
xlā 5
xātxs
ㄨ1
ā
l ty_ㄨ
ㄨ
5L1 y
1
AE
0
x
y
yl
ylǎ
5
zy
⼆
I
ㄣ⼆⽓
之
x
Thus AT and DB both have E
A
5
prove
DĚ ⼆ 之BT
E
D
E
A AE
DA 1BĀ AĒ⼆ 主AT
L
B
AT
DATE
主BĀ 专
DE
⽴
Āt AT
DÉ 之 Bi
Di 1 Bi it BE is
6
11
their midpoint
as
1 101
U
2
Tuli
Tui
1nF
1v1
1.4 0.4
uU
4
multiple of
11
4，4
U
a
F
⼆
6
1 1
21
2
6
wofo
21 h
⼆
⾔
0
45
1 111，2
v
m1
1U1
L21 1，11
Ti
1 1 4
T
e.jo
4 1 1
DE
1
1
2
21 1
111 1
1.2
U.V
L
3
7
6
1 Orthogonal
2
c
2
if ā 万
itcT.si
9
3
3 be
⾔
⼆
u.v
⼀
2
i
a
2
⾔ ⽚
⼀
⼸
us0
180
are in
the
B
opposite direction as
Dis
EA
DBT
D
1 1 31
1111
llull llull
ūmni
A
11
á
并a
l l u ll l l u l l m O
0
10
0
u.U
1
60
4
21 1
211
⾔
0
T
1
let
之
if E 570
acute
1 1
⼆
24
c
8
⾔
⽭
us0
L
the angle
不
EB 5A
B
⼆
EB BĀR
1
LCĀJ
LCT BĀR
between there is 180
B2 2CB BE BĀ
EB2 2EBBA t BA
4CBĀ
0
ciJ.BE
0
EB BĀ
so
ABCD is
A
11
AC Ant Bi
7
D
15 Bit ED
B
⼆
DT An
ˋ
ˊ
AC.BG
Bit AT JIBE An
1
BE 的
AD BE ÁB
C
⼀
AS BC
O
12
PLt l ll t b
all
兴
P t
P
装
x t2
x
Ows
0 90
0
0
c1L
tb
ts
t
c
t
号
yt
ˋ_
2
0
y 良
2
我 器
⼀
2
0
我⼗2710
z
4ㄨ 7
411b11 ll all
11all'll
0 So
我
y2
4La.b
Lab
2
AE.BG
a rectangle
11all1lb ll
when b 0
la.bl
la.bl
aiotar.otas.c
0
11all 0
11h11111.11
0
0
0
So
I a b1
11all llbll.la.bl
1141111bll