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2010 Sharygin Geometry Olympiad
21
Prove that S_{ABD}+S_{ACD} > S_{BAC}+S_{BDC} (21)
Prove that S_{ABD}+S_{ACD} > S_{BAC}+S_{BDC} (21)
Source:
October 29, 2010
geometry
geometric transformation
reflection
trigonometry
geometry proposed
Problem Statement
A given convex quadrilateral
A
B
C
D
ABCD
A
BC
D
is such that
∠
A
B
D
+
∠
A
C
D
>
∠
B
A
C
+
∠
B
D
C
.
\angle ABD + \angle ACD > \angle BAC + \angle BDC.
∠
A
B
D
+
∠
A
C
D
>
∠
B
A
C
+
∠
B
D
C
.
Prove that
S
A
B
D
+
S
A
C
D
>
S
B
A
C
+
S
B
D
C
.
S_{ABD}+S_{ACD} > S_{BAC}+S_{BDC}.
S
A
B
D
+
S
A
C
D
>
S
B
A
C
+
S
B
D
C
.
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