MathDB
Problems
Contests
International Contests
Baltic Way
1993 Baltic Way
18
18
Part of
1993 Baltic Way
Problems
(1)
Angle OLK = Angle OLM
Source: Baltic Way 1993
6/15/2012
In the triangle
A
B
C
ABC
A
BC
,
∣
A
B
∣
=
15
,
∣
B
C
∣
=
12
,
∣
A
C
∣
=
13
|AB|=15,|BC|=12,|AC|=13
∣
A
B
∣
=
15
,
∣
BC
∣
=
12
,
∣
A
C
∣
=
13
. Let the median
A
M
AM
A
M
and bisector
B
K
BK
B
K
intersect at point
O
O
O
, where
M
∈
B
C
,
K
∈
A
C
M\in BC,K\in AC
M
∈
BC
,
K
∈
A
C
. Let
O
L
⊥
A
B
,
L
∈
A
B
OL\perp AB,L\in AB
O
L
⊥
A
B
,
L
∈
A
B
. Prove that
∠
O
L
K
=
∠
O
L
M
\angle OLK=\angle OLM
∠
O
L
K
=
∠
O
L
M
.
trigonometry
geometry proposed
geometry