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All-Russian Olympiad
1989 All Soviet Union Mathematical Olympiad
492
492
Part of
1989 All Soviet Union Mathematical Olympiad
Problems
(1)
ASU 492 All Soviet Union MO 1989 simila triangles wanted
Source:
8/13/2019
A
B
C
ABC
A
BC
is a triangle.
A
′
,
B
′
,
C
′
A' , B' , C'
A
′
,
B
′
,
C
′
are points on the segments
B
C
,
C
A
,
A
B
BC, CA, AB
BC
,
C
A
,
A
B
respectively.
∠
B
′
A
′
C
′
=
∠
A
\angle B' A' C' = \angle A
∠
B
′
A
′
C
′
=
∠
A
,
A
C
′
C
′
B
=
B
A
′
A
′
C
=
C
B
′
B
′
A
\frac{AC'}{C'B} = \frac{BA' }{A' C} = \frac{CB'}{B'A}
C
′
B
A
C
′
=
A
′
C
B
A
′
=
B
′
A
C
B
′
. Show that
A
B
C
ABC
A
BC
and
A
′
B
′
C
′
A'B'C'
A
′
B
′
C
′
are similar.
geometry
similar triangles
equal ratios
equal angles