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Moscow Mathematical Olympiad
1952 Moscow Mathematical Olympiad
229
229
Part of
1952 Moscow Mathematical Olympiad
Problems
(1)
MMO 229 Moscow MO 1952 angle chasing in a triangle 80-80-20
Source:
8/8/2019
In an isosceles triangle
△
A
B
C
,
∠
A
B
C
=
2
0
o
\vartriangle ABC, \angle ABC = 20^o
△
A
BC
,
∠
A
BC
=
2
0
o
and
B
C
=
A
B
BC = AB
BC
=
A
B
. Points
P
P
P
and
Q
Q
Q
are chosen on sides
B
C
BC
BC
and
A
B
AB
A
B
, respectively, so that
∠
P
A
C
=
5
0
o
\angle PAC = 50^o
∠
P
A
C
=
5
0
o
and
∠
Q
C
A
=
6
0
o
\angle QCA = 60^o
∠
QC
A
=
6
0
o
. Prove that
∠
P
Q
C
=
3
0
o
\angle PQC = 30^o
∠
PQC
=
3
0
o
.
geometry
isosceles
angle
Angle Chasing