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IMAR Test
2009 IMAR Test
3
3
Part of
2009 IMAR Test
Problems
(1)
convex ABCD, AB=CB, <ABC +2<CDA = \pi, AE=EC, prove <CDE=<BDA
Source: IMAR 2009 p3
9/26/2018
Consider a convex quadrilateral
A
B
C
D
ABCD
A
BC
D
with
A
B
=
C
B
AB=CB
A
B
=
CB
and
∠
A
B
C
+
2
∠
C
D
A
=
π
\angle ABC +2 \angle CDA = \pi
∠
A
BC
+
2∠
C
D
A
=
π
and let
E
E
E
be the midpoint of
A
C
AC
A
C
. Show that
∠
C
D
E
=
∠
B
D
A
\angle CDE =\angle BDA
∠
C
D
E
=
∠
B
D
A
.Paolo Leonetti
geometry
Angle Chasing
convex quadrilateral
angles
equal angles