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Saint Petersburg Mathematical Olympiad
2011 Saint Petersburg Mathematical Olympiad
7
Convex quadrilateral
Convex quadrilateral
Source: St Petersburg Olympiad 2011, Grade 10, P7
September 15, 2017
geometry
Problem Statement
A
B
C
D
ABCD
A
BC
D
- convex quadrilateral.
P
P
P
is such point on
A
C
AC
A
C
and inside
△
A
B
D
\triangle ABD
△
A
B
D
, that
∠
A
C
D
+
∠
B
D
P
=
∠
A
C
B
+
∠
D
B
P
=
90
−
∠
B
A
D
\angle ACD+\angle BDP = \angle ACB+ \angle DBP = 90-\angle BAD
∠
A
C
D
+
∠
B
D
P
=
∠
A
CB
+
∠
D
BP
=
90
−
∠
B
A
D
. Prove that
∠
B
A
D
+
∠
B
C
D
=
90
\angle BAD+ \angle BCD =90
∠
B
A
D
+
∠
BC
D
=
90
or
∠
B
D
A
+
∠
C
A
B
=
90
\angle BDA + \angle CAB = 90
∠
B
D
A
+
∠
C
A
B
=
90
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