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Geometry Difficulty 4.3 AIME Prove it Ireland

The points EE and FF are on the sides ACAC and ABAB, respectively, of triangle ABCABC such that FEFE is parallel to BCBC. The lines BEBE and CFCF intersect at GG. Prove that the line AGAG passes through the midpoint of BCBC.

Solution

Let AGAG meet BCBC at DD. We need to show that DD is the midpoint of BCBC.

Figure 1

As EFEF is parallel to BCBC, we have CEEA=BFFA\frac{|CE|}{|EA|} = \frac{|BF|}{|FA|}. Ceva's Theorem tells us that
BDDCCEEAFABF=1. \frac{|BD|}{|DC|} \cdot \frac{|CE|}{|EA|} \cdot \frac{|FA|}{|BF|} = 1.
Together these imply BD=DC|BD| = |DC|, hence DD is the midpoint of BCBC.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.