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National and Regional Contests
Russia Contests
All-Russian Olympiad
1989 All Soviet Union Mathematical Olympiad
504
504
Part of
1989 All Soviet Union Mathematical Olympiad
Problems
(1)
ASU 504 All Soviet Union MO 1989 circumcenter of AEF lies on bisector of EDF
Source:
8/14/2019
A
B
C
ABC
A
BC
is a triangle. Points
D
,
E
,
F
D, E, F
D
,
E
,
F
are chosen on
B
C
,
C
A
,
A
B
BC, CA, AB
BC
,
C
A
,
A
B
such that
B
B
B
is equidistant from
D
D
D
and
F
F
F
, and
C
C
C
is equidistant from
D
D
D
and
E
E
E
. Show that the circumcenter of
A
E
F
AEF
A
EF
lies on the bisector of
E
D
F
EDF
E
D
F
.
geometry
Circumcenter
equidistant
angle bisector