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| Double Amount (Posted on 2009-11-25) |
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Let ABC be a triangle with |AB| > |AC|. Let D be a point on line AB such that |AD| + |DC| = |AB|
and point A lies between points B and D. Let E be the midpoint of line segment BC and F the point
on side AB such that CF is perpendicular to AE.
Prove that |BF| = 2|AD|.
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Submitted by Bractals
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Solution:
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Let G be the point on line AB such that |DG| = |DC| and A lies between B and G.
A is the midpoint of line segment BG since |AG| = |AD| + |DG| = |AD| + |DC| = |AB|.
Looking at triangle BCG and line segment AE, it follows that GC is parallel to AE.
Let H be the point on line GC such that DH is perpendicular to GC. Point H is the midpoint of
GC since triangle GCD is isosceles.
DH and CF are parallel since DH is perpendicular to GC, GC is parallel to AE, and AE is perpendicular to CF.
Looking at triangle GCF and line segment DH, it follows that point D is the midpoint of FG. Thus,
|FD| = |DG| = |DC|.
Therefore, |FA| + |AD| = |FD| = |DC| = |AB| - |AD|.
Thus, 2|AD| = |AB| - |FA| = |BF|.
See post by Harry for an alternate solution.
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Subject |
Author |
Date |
 | Solution | Harry | 2009-11-26 12:34:50 |
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