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Hungary Contests
Kürschák Math Competition
1953 Kurschak Competition
3
3
Part of
1953 Kurschak Competition
Problems
(1)
<A=<D, <B=<E, <C=<F if ABCDEF is convex 6-gon, <A+<C+<E=<B+<D+<F, equal sides
Source: 1953 Hungary - Kürschák Competition p3
10/10/2022
A
B
C
D
E
F
ABCDEF
A
BC
D
EF
is a convex hexagon with all its sides equal. Also
∠
A
+
∠
C
+
∠
E
=
∠
B
+
∠
D
+
∠
F
\angle A + \angle C + \angle E = \angle B + \angle D + \angle F
∠
A
+
∠
C
+
∠
E
=
∠
B
+
∠
D
+
∠
F
. Show that
∠
A
=
∠
D
\angle A = \angle D
∠
A
=
∠
D
,
∠
B
=
∠
E
\angle B = \angle E
∠
B
=
∠
E
and
∠
C
=
∠
F
\angle C = \angle F
∠
C
=
∠
F
.
geometry
equal segments
equal angles
hexagon