=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