=53
01
D
A
PC
B
1
1
3
H
ABCD AB⊥AD AD=DC=1 AB=3 P
. BD l AP=xAB+yAD(x,y∈R) x+ylD (1,
D’
B’
E
Fx+y= AFAE
AE= 1·310= 310
EF=
12·1·1=
12
= ·BD·r12
r= 110
2
10
AF=AE+EF= 5
10
x+y= AFAE
53 )
S△DCB=
1
02 ABCD AB=1 AD=2 P C . BD l AP=λ
AB+μAD λ+μl ( )
3 22 5 2
B
A D
C
1
2
P
B’
D’
H
H’
△A △ABDB’D’
02
△ABC D BCl E,F AD l + BA·CA=4 BF·CF=-1
BE·CElA
CB
E
F
D
78
AB·AC FB·FC
EB·ECx
x
x
14(3x)2- lBCl2=4
14x
2- lBCl2=-1
x2= 58
lBCl2=528144x2- ·lBCl2=
l A
a·b=14 [(a+b)2-(a-b)2]
A
O B
a
b
14= [4lOCl
2-lABl2]
C
14=lOCl2- lABl2
1
_2
03 B C
△ABC 2l + P ABC PA·(PB+PC)l
32-
43--2 -1
A
B C
P
D
2PA·PD
△PAD
lADl= 3
E
=2·(lPEl2- ·lADl2)14
=2·(lPEl2- ·3)14
=2·lPEl2- 32
04 B C
ABCD AB⊥BC AD⊥CD ∠BAD=120° AB=AD=1
CD l AE·BEl ( )32
32116
2516
A
B
C
E
D
11
EA·EB
F
EA·EB=lEFl2- lABl214
=lEFl2- 14
D
E
A
F
112
14
1�( )2- =A54
14
05 B C
∠ABC=90° ∠BAC=90° AB=AC AC=BC
△ABC P0 AB P0B= AB . AB P
PB·PC�P0B·P0C ( )
14
B
A
C
PP0
D
lPDl2- lBCl214
�lP0Dl2- lBCl214
lPDl�lP0DlE
_ A
+ l
PG= (PA+PB+PC)13
G △ABCl PA+PB+PC=0 P + ABCl
PA·PB=PB·PC=PC·PA P △ABCl
λ(ABlAB
AClAC )(λ≠0) △ABCl ( ∠BACl )
1
2
3
3G △ABCl PA+PB+PC=0 P + ABCl
PA·PB=PB·PC=PC·PA P △ABCl2
PA·PB-PB·PC=0
PB·(PA-PC)=0
PB·CA=0
A
CB
P
+ l
PG= (PA+PB+PC)13
G △ABCl PA+PB+PC=0 P + ABCl
PA·PB=PB·PC=PC·PA P △ABCl
λ(ABlAB
AClAC )(λ≠0) △ABCl ( ∠BACl )
1
2
3
B
C
A
lABlAB
lAClACλ( )
03
O,N,P △ABC .lOAl=lOBl=lOCl,NA+NB+NC=0,
.PA·PB=PB·PC·PA O,N,P + ABCl( )
06 B C
O △ABCl .OA+OB+CO=0 △ABCl C
2OD
2OD=OC
A
C
BD
O
OA=OB=OC
120°
+CO=0
07 B C
P ABC l .3PA+5PB+2PC=0 △ABCl S
△PACl
P
B
A
C
E
PC=- PA- PB32
52
l- +(- )l=32
52
μλ
4H
lPHllPCl =4
Hx
4x
S△ABCS△APB 1
5=
PB=- PA- PC35
25
B
A
C
P
H
lPHllPBl =1
x
x
S△ABCS△PAC 1
2=
12 S
l- - l=3525 1