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International Contests
Tournament Of Towns
1985 Tournament Of Towns
(106) 6
(106) 6
Part of
1985 Tournament Of Towns
Problems
(1)
TOT 106 1985 Autumn S6 <BEA=45^o prove <EHC=45^o
Source:
8/26/2019
In triangle
A
B
C
,
A
H
ABC, AH
A
BC
,
A
H
is an altitude (
H
H
H
is on
B
C
BC
BC
) and
B
E
BE
BE
is a bisector (
E
E
E
is on
A
C
AC
A
C
) . We are given that angle
B
E
A
BEA
BE
A
equals
4
5
o
45^o
4
5
o
.Prove that angle
E
H
C
EHC
E
H
C
equals
4
5
o
45^o
4
5
o
. (I. Sharygin , Moscow)
geometry
angles
altitude
angle bisector