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Russia Contests
Sharygin Geometry Olympiad
2010 Sharygin Geometry Olympiad
23
23
Part of
2010 Sharygin Geometry Olympiad
Problems
(1)
Prove that the lines AD, BE and CF are concurrent (23)
Source:
10/29/2010
A cyclic hexagon
A
B
C
D
E
F
ABCDEF
A
BC
D
EF
is such that
A
B
⋅
C
F
=
2
B
C
⋅
F
A
,
C
D
⋅
E
B
=
2
D
E
⋅
B
C
AB \cdot CF= 2BC \cdot FA, CD \cdot EB = 2 DE \cdot BC
A
B
⋅
CF
=
2
BC
⋅
F
A
,
C
D
⋅
EB
=
2
D
E
⋅
BC
and
E
F
⋅
A
D
=
2
F
A
⋅
D
E
.
EF \cdot AD = 2FA \cdot DE.
EF
⋅
A
D
=
2
F
A
⋅
D
E
.
Prove that the lines
A
D
,
B
E
AD, BE
A
D
,
BE
and
C
F
CF
CF
are concurrent.
geometry proposed
geometry