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November 13th, 2010, 01:27 PM   #1
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Similar Triangles

Use one of the conditions for similarity of triangles (AA, SSS, or SAS) to prove that a midsegment of a triangle is parallel to a side of the triangle and is half the length of that side.
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November 14th, 2010, 08:40 AM   #2
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The midsegment joins the midpoints of two sides, thus "cutting off" a triangle that shares an angle with the original triangle and has sides enclosing that angle that are half as long as the corresponding sides of the original triangle, so the triangles are similar (SAS). Can you finish from there?

The theorem is known as the triangle mid-segment theorem (or, in older books, the mid-point theorem). There's a generalization where the proportionality constant is any value between 0 and 1.
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