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Romania National Olympiad
1999 Romania National Olympiad
4
EF/BC+EG / AD=1 1999 Romania NMO VII p4
EF/BC+EG / AD=1 1999 Romania NMO VII p4
Source:
August 14, 2024
geometry
ratio
Problem Statement
In the triangle
A
B
C
ABC
A
BC
, let
D
∈
(
B
C
)
D \in (BC)
D
∈
(
BC
)
,
E
∈
(
A
B
)
E \in (AB)
E
∈
(
A
B
)
,
E
F
∥
B
C
EF \parallel BC
EF
∥
BC
,
F
∈
(
A
C
)
F \in (AC)
F
∈
(
A
C
)
,
E
G
∥
A
D
EG\parallel AD
EG
∥
A
D
,
G
∈
(
B
C
)
G\in (BC)
G
∈
(
BC
)
and
M
,
N
M,N
M
,
N
be the midpoints of
(
A
D
)
(AD)
(
A
D
)
and
(
B
C
)
(BC)
(
BC
)
, respectively. Prove that:a)
E
F
B
C
+
E
G
A
D
=
1
\frac{EF}{BC}+\frac{EG}{AD}=1
BC
EF
+
A
D
EG
=
1
b) the midpoint of
[
F
G
]
[FG]
[
FG
]
lies on the line
M
N
MN
MN
.
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