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Kyiv City MO
Kyiv City MO - geometry
Kyiv City MO Juniors 2003+ geometry
2004.7.3
2004.7.3
Part of
Kyiv City MO Juniors 2003+ geometry
Problems
(1)
<DOE=90^o if BD=AD, CB=CE, right triangle (Kyiv City Olympiad 2004 7.3)
Source:
6/27/2021
Given a right triangle
A
B
C
ABC
A
BC
(
∠
A
<
4
5
o
\angle A <45^o
∠
A
<
4
5
o
,
∠
C
=
9
0
o
\angle C = 90^o
∠
C
=
9
0
o
), on the sides
A
C
AC
A
C
and
A
B
AB
A
B
which are selected points
D
,
E
D,E
D
,
E
respectively, such that
B
D
=
A
D
BD = AD
B
D
=
A
D
and
C
B
=
C
E
CB = CE
CB
=
CE
. Let the segments
B
D
BD
B
D
and
C
E
CE
CE
intersect at the point
O
O
O
. Prove that
∠
D
O
E
=
9
0
o
\angle DOE = 90^o
∠
D
OE
=
9
0
o
.
geometry
right triangle
right angle