The point P lies inside the triangle ABC. A line is drawn through P parallel to each side of the triangle. The lines divide AB into three parts length c, c′, c′′ (in that order), and BC into three parts length a, a′, a′′ (in that order), and CA into three parts length b, b′, b′′ (in that order). Show that abc=a′b′c′=a′′b′′c′′.
Solution
Solution:
The three small triangles are similar, so a/a′′=c′/c=b′′/b′ and a/a′=c′/c′′=b′′/b. Hence (a/a′′)(b/b′′)=(c′/c)(c′′/c′)=c′′/c, so abc=a′′b′′c′′. Similarly, (a/a′)(c/c′)=(b′′/b)(b′′/b′′)=b/b, so abc=a′b′c′.
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