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| Centroid Line (Posted on 2010-08-10) |
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Let AA' and CC' be medians of ΔABC intersecting in point G.
Let m be any line through G intersecting sides AB and AC.
Let P, Q, and R be the feet of perpendiculars to m from
A, B, and C respectively.
Prove that |AP| = |BQ| + |CR|.
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Submitted by Bractals
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Solution:
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Let D be the foot of the perpendicular from A' to m.
Look at trapezoid BCRQ. A' is the midpoint of side BC.
Therefore, 2|A'D| = |BQ| + |CR|.
Look at similar triangles APG and A'DG. |AG| = 2|A'G|.
Therefore, |AP| = 2|A'D| = |BQ| + |CR|.
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Subject |
Author |
Date |
 | Solution | Harry | 2010-08-15 17:15:15 |
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