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Saint Petersburg Mathematical Olympiad
2015 Saint Petersburg Mathematical Olympiad
5
Pentagon angles
Pentagon angles
Source: St Petersburg Olympiad 2015, Grade 10, P5
October 17, 2017
geometry
Problem Statement
A
B
C
D
E
ABCDE
A
BC
D
E
is convex pentagon.
∠
B
C
A
=
∠
B
E
A
=
∠
B
D
A
2
,
∠
B
D
C
=
∠
E
D
A
\angle BCA=\angle BEA = \frac{\angle BDA}{2}, \angle BDC =\angle EDA
∠
BC
A
=
∠
BE
A
=
2
∠
B
D
A
,
∠
B
D
C
=
∠
E
D
A
. Prove, that
∠
D
E
B
=
∠
D
A
C
\angle DEB=\angle DAC
∠
D
EB
=
∠
D
A
C
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