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ITAMO
2000 ITAMO
2
(DB+BC)^2=AD^2+AC^2
(DB+BC)^2=AD^2+AC^2
Source: Lithuanian TST 2005 p2 - Italy 200
December 10, 2022
geometry
angles
Problem Statement
Let
A
B
C
D
ABCD
A
BC
D
be a convex quadrilateral, and write
α
=
∠
D
A
B
\alpha=\angle DAB
α
=
∠
D
A
B
,
β
=
∠
A
D
B
\beta=\angle ADB
β
=
∠
A
D
B
,
γ
=
∠
A
C
B
\gamma=\angle ACB
γ
=
∠
A
CB
,
δ
=
∠
D
B
C
\delta= \angle DBC
δ
=
∠
D
BC
and
ϵ
=
∠
D
B
A
\epsilon=\angle DBA
ϵ
=
∠
D
B
A
. Assuming that
α
<
π
/
2
\alpha<\pi/2
α
<
π
/2
,
β
+
γ
=
π
/
2
\beta+\gamma=\pi /2
β
+
γ
=
π
/2
, and
δ
+
2
ϵ
=
π
\delta+2\epsilon=\pi
δ
+
2
ϵ
=
π
, prove that
(
D
B
+
B
C
)
2
=
A
D
2
+
A
C
2
(DB+BC)^2=AD^2+AC^2
(
D
B
+
BC
)
2
=
A
D
2
+
A
C
2
.
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