MathDB
Problems
Contests
International Contests
Baltic Way
1998 Baltic Way
12
The point D on the side BC for the right triangle ABC
The point D on the side BC for the right triangle ABC
Source: Baltic Way 1998
January 11, 2011
geometry
circumcircle
geometry proposed
Problem Statement
In a triangle
A
B
C
ABC
A
BC
,
∠
B
A
C
=
9
0
∘
\angle BAC =90^{\circ}
∠
B
A
C
=
9
0
∘
. Point
D
D
D
lies on the side
B
C
BC
BC
and satisfies
∠
B
D
A
=
2
∠
B
A
D
\angle BDA=2\angle BAD
∠
B
D
A
=
2∠
B
A
D
. Prove that
2
A
D
=
1
B
D
+
1
C
D
\frac{2}{AD}=\frac{1}{BD}+\frac{1}{CD}
A
D
2
=
B
D
1
+
C
D
1
Back to Problems
View on AoPS