# Triangle midsegment theorem proof

The triangle midsegment theorem proof is easy to follow in this lesson. This lesson will give a coordinate proof of the triangle midsegment theorem. What is the triangle midsegment theorem?

If a segment joins the midpoints of the sides of a triangle, then the segment is parallel to the third side and the segment is half the length of the third side.

## Here is the triangle midsegment theorem proof

Proof of the theorem.

Consider the following triangle on the coordinate system.

 Given:S is the midpoint of  OQ R is the midpoint of  PQ Prove: SR || OP SR =   OP / 2
 Given:S is the midpoint of  OQ R is the midpoint of  PQ Prove: SR || OP SR =   OP / 2

To prove that SR || OP, we can just show that their slopes are equal.

 S: ( b + 0 / 2 , c + 0 / 2 ) = ( b / 2 , c / 2 )
 S: ( b + 0 / 2 , c + 0 / 2 ) = ( b / 2 , c / 2 )

 R: ( a + b / 2 , c + 0 / 2 ) = ( a + b / 2 , c / 2 )
 R: ( a + b / 2 , c + 0 / 2 ) = ( a + b / 2 , c / 2 )

Since the y-coordinate is the same for both points, the slope of SR is zero. The same is true for OP. Since the y-coordinate is the same, the slope is also zero. Since the slope is the same for SR and OP , SR || OP

To prove that SR is half OP, we can use the distance formula to find SR and OP.

 $$OP = \ {\sqrt{(a - 0)^2 + (0 - 0)^2 } }$$
 $$OP = \ {\sqrt{(a)^2 + (0)^2 } }$$
 $$OP = \ {\sqrt{(a)^2 } } = a$$
 $$SR = \ {\sqrt{(\frac{a+b} {2} - \frac{b} {2} )^2 + (\frac{c} {2} -\frac{c} {2} )^2 } }$$
 $$SR = \ {\sqrt{(\frac{a} {2}+ \frac{b} {2} - \frac{b} {2} )^2 + (0)^2 } }$$
 $$SR = \ {\sqrt{(\frac{a} {2} + 0)^2 + (0)^2 } }$$
 $$SR = \ {\sqrt{(\frac{a} {2})^2 } = \frac{a} {2} }$$
Therefore, SR =
OP / 2

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