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Old Kyiv MO Geometry
Kyiv City MO 1984-93 - geometry
1993.9.2
<C = 135^o if (ABC)=1/4 (c^2 - a^2 - b^2) 1993 Kyiv City MO 9.2
<C = 135^o if (ABC)=1/4 (c^2 - a^2 - b^2) 1993 Kyiv City MO 9.2
Source:
July 20, 2021
geometry
angles
area
Problem Statement
Let
a
,
b
,
c
a, b, c
a
,
b
,
c
be the lengths of the sides of a triangle, and let
S
S
S
be its area. We know that
S
=
1
4
(
c
2
−
a
2
−
b
2
)
S = \frac14 (c^2 - a^2 - b^2)
S
=
4
1
(
c
2
−
a
2
−
b
2
)
. Prove that
∠
C
=
13
5
o
\angle C = 135^o
∠
C
=
13
5
o
.
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