<PRE><B>Question 748157</B>
The straight line connecting midpoints of two sides of a triangle is {{{parallel}}} to the {{{third}}} side of the triangle. 

The straight line connecting midpoints of two sides of a triangle is parallel to the third side of the triangle. 

&lt;TABLE&gt;
  &lt;TR&gt;
  &lt;TD&gt;
&lt;B&gt;Proof&lt;/B&gt;

&lt;B&gt;Figure 1&lt;/B&gt; shows the triangle &lt;B&gt;ABC&lt;/B&gt; with the midpoints &lt;B&gt;D&lt;/B&gt; and &lt;B&gt;E&lt;/B&gt; that are  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp; 
located in its sides &lt;B&gt;BC&lt;/B&gt; and &lt;B&gt;AC&lt;/B&gt; respectively. The theorem states that 
the straight line &lt;B&gt;ED&lt;/B&gt;, which connects the midpoints &lt;B&gt;D&lt;/B&gt; and &lt;B&gt;E&lt;/B&gt; (green 
line in the &lt;B&gt;Figure 1&lt;/B&gt;), is parallel to the triangle side &lt;B&gt;AB&lt;/B&gt;. 


Continue the straight line segment &lt;B&gt;ED&lt;/B&gt; to its own length to the point &lt;B&gt;F&lt;/B&gt; 
(&lt;B&gt;Figure 2&lt;/B&gt;) and connect the points &lt;B&gt;B&lt;/B&gt; and &lt;B&gt;F&lt;/B&gt; by the straight line segment &lt;B&gt;BF&lt;/B&gt;.
The triangles &lt;B&gt;EDC&lt;/B&gt; and &lt;B&gt;FDB&lt;/B&gt; have the congruent vertical angles &lt;B&gt;EDC&lt;/B&gt; and 
&lt;B&gt;FDB&lt;/B&gt;, congruent sides &lt;B&gt;DC&lt;/B&gt; and &lt;B&gt;DB&lt;/B&gt; as halves of the side &lt;B&gt;BC&lt;/B&gt;, and congruent 
sides &lt;B&gt;ED&lt;/B&gt; and &lt;B&gt;FD&lt;/B&gt; by the construction. Therefore, these triangles are 
congruent in accordance to the &lt;B&gt;postulate P1 (SAS)&lt;/B&gt; of the lesson 
&lt;A HREF=http://www.algebra.com/algebra/homework/Triangles/Congruence-tests-for-triangles.lesson&gt; Congruence tests for triangles&lt;/A&gt; (which is under the topic &lt;B&gt;Triangles&lt;/B&gt; 
in the section &lt;B&gt;Geometry&lt;/B&gt; in this site).
 &lt;/TD&gt;
  &lt;TD&gt;
{{{drawing( 200, 250,  0, 4, 0, 5, 
            line( 0.3, 0.5, 3.7, 0.5), 
            line( 0.3, 0.5, 3.0, 4.5),
            line( 3.0, 4.5, 3.7, 0.5),

            locate(0.3, 0.5, A),
            locate(3.7, 0.5, B),
            locate(3.0, 4.9, C),

      green(line (1.65, 2.5, 3.35, 2.5)),

            locate(3.5, 2.7, D),
            locate(1.3, 2.7, E)
)}}}
&lt;B&gt;Figure 1&lt;/B&gt;. To the &lt;B&gt;Theorem 1&lt;/B&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;
 &lt;/TD&gt;
  &lt;TD&gt;
{{{drawing( 270, 250,  0, 5.4, 0, 5, 
            line( 0.3, 0.5, 3.7, 0.5), 
            line( 0.3, 0.5, 3.0, 4.5),
            line( 3.0, 4.5, 3.7, 0.5),

            locate(0.3, 0.5, A),
            locate(3.7, 0.5, B),
            locate(3.0, 4.9, C),

      green(line (1.65, 2.5, 3.35, 2.5)),

            locate(3.4, 2.85, D),
            locate(1.4, 2.85, E),

      green(line(3.35, 2.5, 5.15, 2.5)),
      green(line(3.7, 0.5, 5.15, 2.5)),
            locate(5.1, 2.85, F),

            line (2.5, 2.6, 2.5, 2.4),

            line (4.2, 2.6, 4.2, 2.4),

            line (3.05, 3.4, 3.25, 3.4),
            line (3.05, 3.5, 3.25, 3.5),

            line (3.4, 1.5, 3.6, 1.5),
            line (3.4, 1.6, 3.6, 1.6),

            arc(3.35, 2.5, 0.6, 0.6, 180, 255),

            arc(3.35, 2.5, 0.6, 0.6,   0,  75),


            arc(3.0, 4.5, 0.8, 0.8,  80, 120),
            arc(3.0, 4.5, 0.9, 0.9,  80, 120),

            arc(3.7, 0.5, 0.8, 0.8, 260, 300),
            arc(3.7, 0.5, 0.9, 0.9, 260, 300)
)}}}
&lt;B&gt;Figure 2&lt;/B&gt;. To the proof of the &lt;B&gt;Theorem 1&lt;/B&gt;
 &lt;/TD&gt;
 &lt;/TR&gt;
&lt;/TABLE&gt;

This means that the angles &lt;B&gt;ECD&lt;/B&gt; and &lt;B&gt;DBF&lt;/B&gt; are congruent as the corresponding angles of these triangles.
Hence, the straight lines &lt;B&gt;AC&lt;/B&gt; and &lt;B&gt;BF&lt;/B&gt; are parallel, because these angles are the alternate interior angles formed by the transverse line &lt;B&gt;BC&lt;/B&gt; 
(see the lesson &lt;A HREF=http://www.algebra.com/algebra/homework/Angles/Parallel-lines.lesson&gt; Parallel lines&lt;/A&gt; under the topic &lt;B&gt;Angles, complementary, supplementary angles&lt;/B&gt; in the section &lt;B&gt;Geometry&lt;/B&gt; in this site).
This means also that the segments &lt;B&gt;CE&lt;/B&gt; and &lt;B&gt;BF&lt;/B&gt; are of equal length as the corresponding sides of triangles &lt;B&gt;EDC&lt;/B&gt; and &lt;B&gt;FDB&lt;/B&gt;.
Since the point &lt;B&gt;E&lt;/B&gt; is the midpoint of the side &lt;B&gt;AB&lt;/B&gt; and the segments &lt;B&gt;AE&lt;/B&gt; and &lt;B&gt;CE&lt;/B&gt; are of equal length, this implies that the segments &lt;B&gt;BF&lt;/B&gt; and &lt;B&gt;AE&lt;/B&gt; are of equal length.

Thus, we have proved that in the quadrilateral &lt;B&gt;ADFE&lt;/B&gt; the two opposite sides &lt;B&gt;BF&lt;/B&gt; and &lt;B&gt;AE&lt;/B&gt; are parallel and have equal length.


At this point we can refer to the geometry fact proven in the lesson &lt;A HREF=http://www.algebra.com/algebra/homework/Triangles/Properties-of-the-sides-of-parallelograms.lesson&gt; Properties of the sides of parallelograms&lt;/A&gt; (see the &lt;B&gt;Theorem 1&lt;/B&gt; of that lesson):
&lt;B&gt;if a quadrilateral has two opposite sides parallel and of equal length, then two other opposite sides of the quadrilateral are parallel and of equal length too&lt;/B&gt;.

It implies that the straight lines &lt;B&gt;AB&lt;/B&gt; and &lt;B&gt;EF&lt;/B&gt; are parallel. 
This is exactly what we were going to prove.



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