TOÁN TIẾNG ANH

Pythagoras'
Theorern
and Its
Applications
Theorern
L
(Pythagoras'
Theorem)
For a right-angled
triangle
with tw,o
legs
a,h
antl lrypotenuse
c, the
sum of squares of legs
is
equal
to the
s{lLtare
of its /ty-
prstennse,
i.e.
a2
,*b2
-
c2.
Theorem
ll.
(Irwerse
Theorem) If

the lengths
a,b,,c
of
three
sides of a trinngle
hqve
the
relation
a2
+
b2
:
c2, then the tiangle
must
be a right-angled
triangle
with tyvo
legs a,b
and
hypoterutse
c.
When
investigating
a right-angled
triangle
(or
shortly,
right
triangle),
the

fol-
lowing
conclusions
are
often used:
Theorenr
III"
A
triangle
is
a
right
triangle,
if
and
onty
if the median
on
one
side
is lzalf
of the
side.
Theorenr
IY.
If
a right
triangle
has an
interior

angle
of
size
30",
then its
opposite
Ieg
is half
of the
htpotenuse.
Example
Example
L.
Given
that
the
perimeter
of a
right
angled
triangle
ls (z
+
rA)
the
meclian
on the
hypotenuse
is 1
cm,

find the
area
of the
triangle.
soiution
The
TheoremIII
implies
that
AD
-
BD
:
cD
-
1,
so
y'B
[-et
AC
:
b, BC
-
a, then
c111;
_,
a2
+b2
:22
:4

and
a+b
-\/6.
Therefore
6
-
(a
+b)'
:
a2
+b2
+2ab,so
ab_T
_t,
the area
of
the triangle
ABC
it
;.
Example
2. As shown
in
the
figure,
lC
:
BD
:2.5
cm.

Find
AC.
therefore,
by
Pythagoras'
Theorem,
"
PCz
:
9
:
CQ,
+
PQz,
ICQP: 9OO.
Hence IAPB
-
ICQB
-
90o
+
45o
-
135o
\
i
90o,
lL\,,:
72, CD
-

1.5
cm,
Solution
From
D introdu
ce
D E
-L
AB,intersecti
ng
AB at
E .
When
we fold
up the
plane that
ACAD
lies
along
the line
AD,thenC
coincides
with
E, so
AC
-
AE,
DE
:
CD

-
1.5
(cm)
By applying
Pythagoras'
Theorem
to
ABED,
BE

JED,
-
DE,
-
\Ezb
-'L25-
z
(.*).
A
Letting AC
:
AE
-
n
cmand
applying
Sthagoras'
Theorem
to
LABC

leads
the equation
(r+2)':12+42',
4r-12,
.'.r-3.
Thus
AC
:3
cm.
Example
3. As shown
in
the
figure,
ABC
D is a square,
P is an
inner
point
sr:ch
thatPA
PB: PC:1:2:3.
Find
IAPB
indegrees.
Solution
Without loss
of
generality, we assume
that PA

:
1,,
PB
:
2, PC
-
3.
Rotate
the AAPB
around
B by 90o
in clock-
wise direction, such
that
P
+
Q,A
+
C,
then
AB PQ
is an isosceles
right
triangle,
therefore
D
Pe'
-
2Pr,2
-

8,cQz
-
PA2
:1,
P,/.'
rQ
a;
I
I
I
B
Exanrple 4.
(SSSMO(I)/2003)
The diagram
shows a hexagon
ABCDEf, rnade
Lip of five right-angled
isosceles
triangles ABO,
BCO,CDO,
DEO,EFO, and
a triangle AOF,
where
O
is the
point
of
intersection
of the lines BF and
AE.

Given
that
O A
:
B
cm, find the
area of A.AO F
in cmz.
Solution From
oc
_
#roB-(i)roA-toA,
oE
_
hoc-IoA-2(cm).
Since RiAt.F'O
-
RIAABO,
EF:oF-!o"-J-gn.
4
4\/z
Let FG
-L
AE
at
G,
then f'G
-
hOF
-

tOA:
1cm. Thus, the area of AAOF,
SaAoF,is
givenby
steop
-t
oo.FG- 4
(cm2).
Example
5.
(Formula
for median) In
LABC,
AM is
the median on the side
BC.
Frove that
AB2
+
AC2
:
2(AM2
+
n
N127.
Solution
Suppose
that AD
-L
BC at D.

By Pythagoras'
Theorem,
A82
BD2+AD2-(BM+MD)z+AD2
BIUI2
+28X,[
.MD
+
MD2
+
AM2
_
MD2
B]VTZ
+
AM2
+2BM
.
]/ID,
Similarly,
we have
AC2
:
CIVT2
+
A]VIz
_zMC
.]VID.
B
Thus,

by adding
the
two equalitie,s up, since
B
M
-
C M
,
AB2
+
ACz
:2(AM'
+
BM'\.
Note:
When
AM
isextended
to.E such
that
ABEC
is
a
parallg.iogram,
then
the
fornrula
of median
is the
same as the

parallelogram
rule:
AB2
+
BEz
+
EC2
+
C4'
:
AE2
+
BC2.
Exarnple
6.
In the
figure
,
lC
-
90o
,
lA
-
30o, D
is the
mid-point
of AB
and
DE

I
AB,
AE
-
4
cm. Find
BC.
Solution
Connect
B,E.
Since ED
is
the
perpendicular
bisector
of
AB,
-
300
-
300,
Now
let
BC
_
r crn,
then
from
Pythagoras'
B

Theorem,
(2")':tr2+621a2-!2
+r-{r2-2tfrcm.
CEA
Thus,
BC
:
ZtR
cm.
Exarnple
7.
For
aABC,o
is
an innerpoint,
and
D,
E,F
are
on BC,CA,AB
respectively,
such
that
OD L BC,
OE L
CA,
and
OF L AB.
prove
that

AF2
+
BD2
+
CE2
:
BF2
+
DC2
+
AE2.
Solution
By
applying
the Pythagoras'
Theorem
to
the triangles
O AF,
O
B F
,
OBD,
OC
D,,
OC
E and
OAE,
it
follows

that
A
BE
:
AE,
sa
IEBD
-
IEBA
-
lA
-
30o
,ICBE:
60o
.'.CE
:
*na
-
DE
-
+AE
-2
cm.
AFz+BD2+CE2
-
AO2
-
OFz
+

BO2
-
OD2
+
CO2
-
OE2
:
(BQ'
-
OF,)
+
(CO2
-
ODr)
+
(AO2
-
OE
)
-
BF2
+
DC2
+
A82.
T'he
conclusioir
is
proven.

Exarnple
8. In
the
diagram given
below
P
is an
interior
point
of aABC
,
p
p1
_L
AB,
PPz
)- BC,
PP3
L
AC,
and BP1
-
BPz,
CP2
-
Cps,
prove
thar
AP7
-

APs.
Solution
For
the
quadrilatenl
AP1BP,
since
its
two
diagonals
are perpen-
dicular
to
each
other,
AP?
+
BPz
_
AF2
*
ptF2
+
BF2
+
pF2
_
AP2
+
BP?.

By
considering
AhC
P
and
PC
PzB
respec-
tively,
it
follows
similarly
that
.
AP2+Cp|-Ap|+pC2,
BP?
+
PCz
:
pBz
+
Cpi.
B
Then
adding
up
rhe
rhree
equalities
yields

Ap?
-
Ap|,

Apt
-
Ap*
Ps
Exarnple
9.
In square
ABCD,
M
is
the
midpoint
of AD
and
l/
is the
midpoint
of
l\,'I
D . Prove
that
IIY
BC
:
2lAB
M

.
Solution
LetAB
-
BC
:
CD
-
DA
-
a.LetEbethemidporntof
CD.
Let the lines
AD
and
BE
tntersect
at
F.
By symmetry,
we
have
DF
-
CB
:
a. Since
A
nght
triangles

AB
M and
C
B
E are
syfilmetric
in
the
line BD,
IABIVI
:
ICBE.
lt suffices
to show
INBE
-
IEBC,
and
for
this
we
only
need to
show
INBF
-
IBFIV
since IDFE
:
IEBC.

By
assumption
we
have
MN
F
D
l _,
I

I

12.
'ln
nx
-
Xo,
.'.
rrB
-
On
ttre other
hand,
NF:!no*o-Xo,
so
l/F
-
BIU{,hence
lI{BF
-

IBFN.
rXa'+
o'
-
Xo.
Testing
Questions
(A)
r.{
(cirnl,ry1995)
ln LABC,
lA
-
90o,
AB
-
AC, D
is
a
point on
EC.
Prove that BDz
+
C
D2
-
2AD2
.
2.
{

Given
that RTAABC
has
a
perimeter
of 30
cm
and an
area of 30
cm2.
Find
the lengths of
its three
sides.
:1.(
trn tlre RIAABC,
lC:
90o,
AD
isthe
angle bisector
of lAwhich
inter-
sects BC atD. Given
AB
-
15 cm,
AC:9cm,
BD:
DC

:5:3.
Find
the
distance
of
D from
AB.
4.
In the right triangle ABC,
lC
:
g0o
,
BC
:
L2
cm, AC
-
6
cm, the
per-
pendicular
bisector
of
AB
intersects
AB and BC at
D
and
-E

respectively.
FtndCE.
*-
/1
,f u^
5.
In the rectangle
'ABCD,
CE
L DB
a,t E, BE
:
1U,
and
CE-
5
cm.
Find the length of AC
.
-'
1
2, i
'*'
4
6.F
In
AABC,IC:
90o,
D is the
mid-point

of
AC.Prove
that
AB2
+BBC2
-
48D2.
In
the right
triangle
ABC, lC
:90o,,
E,
D
arc
points
an AC and
BC,
respectively.
Prove
that
AD2+BE2:AB2+D82.
(CHNMOL/1990)
LABC is an isosceles
triangle
with AB
-
AC
-
Z.

Thereare
100points
Pt,Pzl
,Proo
ontheside
BC.Wntemi
-
AP?
+
BPt,.PiC
(i:1,2, ,100),
findthevalue
of
m1
lm,z +
*mrc0.
In LABC,
lC
-
90o,
D
is the midpoint
of AB, E,
F are
two
points
on
AC
andBC
respectively,

and
DE L DF.kovethat
EF2
:
AE2+8F2.
(CHINA/1996)
Given that P
is an rnner point
of the equilateral
triangle
ABC
,
such
that
PA
:
2, PB
-
2tE,
PC
:4.
Find the
iengttr
of the
side
of LABC.
Testing
Questions
(B)
(SSSMO(I)/2003/Q8)

AB is
a chord in
a
circle
with center
O
and
radius
52 cm.
The
point
M
dlides the
chord
AB
such
that
AM
-
63 cm
and
NIB
-
33 cm.
Find the
length
OM incm.
(CHINA/I996)
ABC
D is

a
rectangle,
P
is an
inner point
of the
rectangle
such
that
PA:3,
PB
:
4, PC
:
5, find
PD.
Determine
whether
such a right-angled
triangle
exists:
each
side
is an
integer
and
one
leg
is a
multiple

of the
other
leg of
the
right
angle.
(AHSME
lLgg6)
In
rectangle
ABC
D, lC
is
trisecte
dby
C
F
and,
C
E,where
closest
to the
area
of the'rectangle
ABC
D?
(A)
110,
(B)
120,

(C)
130,
(D)
140,
(E)
150.
(Hungary
11912)Let
ABC D
bea convex
quadrilateral.
prove
that
AC
L B D
if
and
only
if
AB2
+
CD2
:
ADz
+
BC2.
-{
(.
8.'
I.t

10.
1.
2.
J.
4.
5.