The converse is also true: If points
D
,
E
and
F
are chosen on
BC
,
AC
and
AB
respectively
so that
BF
AF
∙
CD
BD
∙
AE
CE
= 1, then
AD
,
BE
and
CF
are concurrent.
See Lecture “
Geometry
Menelaus and Ceva
” for details.
Example 16
:
2
2
b
a
,
2
2
4
b
a
,
2
2
4
b
a
are three sides of a triangle, where
a
,
b
,
and
c
are positive numbers. Find the area of the triangle.
Solution:
We observe that the lengths of the three sides are
2
2
b
a
,
2
2
)
2
(
b
a
, and
2
2
)
2
(
b
a
.
We can construct a rectangle
ABCD
such that
EF =
2
2
b
a
, EC =
2
2
)
2
(
b
a
,
and
CF =
2
2
)
2
(
b
a
.
Therefore, the area of the triangle will be:
CEF
S
=
ABCD
S
–
AEF
S
–
BCE
S
–
CDF
S
= 4
ab
–
2
1
ab
–
ab
–
ab
=
2
3
ab
.

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50 AMC Lectures
Chapter 10 Area And Area Method
206
Example 17
: Find the area of triangle
ABC
if three medians are known.
Solution:
We know that
AP
=
3
2
AD
.
So
APF
=
2
1
ABP
=
6
1
ABC
.
Take
M
, the midpoint of
AP
and connect
FM
:
a
m
MP
3
1
,
c
m
FP
3
1
,
b
m
FM
3
1
MPF
S
=
2
1
APF
S
=
12
1
ABC
S
By Heron’s Formula, we have
12
1
ABC
S
=
MPF
S
=
)
)(
)(
(
9
1
c
b
a
m
m
m
m
m
m
m
where
m
=
)
(
2
1
c
b
a
m
m
m
.
Therefore,
ABC
S
=
)
)(
)(
(
3
4
c
b
a
m
m
m
m
m
m
m
.
Example 18
: In triangle
ABC, AD, BE,
and
CF
meet at
O.
Let the areas of six small
triangles be
p, q, r, m, n,
and
k,
respectively, as shown in the figure. Show that
pqr = mnk
(1)
k
n
m
r
q
p
1
1
1
1
1
1
(2)
Solution:
We begin by proving (1) first:
From Example
15, we proved Ceva’s Theorem, which stated:
1
EA
CE
DC
BD
FB
AF
(3)
By Theorem 6, we have
k
p
FB
AF
(4)

50 AMC Lectures
Chapter 10 Area And Area Method
207
m
q
DC
BD
(5)
n
r
EA
CE
(6)
Substituting (4), (5), and (6) into (3) yields
1
n
r
m
q
k
p
pqr = mnk
Now we will prove (2):
By Theorem (5), we have:
r
n
k
p
CD
BD
m
q
qn + qr = mp + mk
(7)
Similarly we get
rp + rk = nq + nm
(8)
pq + pm = kn + kr
(9)
(7) + (8) + (9):
qr + rp + pq = kn + mk + nm
(10)
Dividing left hand side by
pqr
and right hand side by
mnk
:
k
n
m
r
q
p
1
1
1
1
1
1
.

50 AMC Lectures
Chapter 10 Area And Area Method
208
PROBLEMS
Problem 1
:
D
and
C
trisect the arc of the half circle as shown in the figure. Find the
shaded area if the area of the half circle is 9
.
Problem 2
: As shown in the figure, right triangle
ABC
with
BC =
20
cm. BDC
is a half
circle with the diameter
BC
. The difference between two shaded areas I and II is 23.
Find
AC
in terms of
.
Problem 3:
The larger one of the two squares in the figure below has the side length of 10
cm
. Find the shaded area.
Problem 4
:
As shown in the figure, in
ABC, D
is any point on
AB. DE//BC
and meets
BC
at
F.
The area of
ADE
is
S
1.
The area of
DBF
is
S
2
. Find the area of parallelogram
DFCE.

50 AMC Lectures
Chapter 10 Area And Area Method
209
Problem 5
: In trapezoid
ABCD,
The area of
ABC
is 253
cm
2
.
The area of
AOB
is 42
cm
2
larger than the area of
COD.
Find the area of
trapezoid
ABCD.