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**Proof that the medians of any triangle intersect at a single point.**

Using synthetic geometry.

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**Proof that the three medians of a triangle meet at one point.**

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median In a triangle, the median is the line segment between a vertex and the midpoint of the opposite side.

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**In this proof, we will need to make a couple of small assumptions**

In this proof, we will need to make a couple of small assumptions. First, we will accept without proof that the diagonals of a parallelogram bisect each other. Here are some parallelograms: The opposite sides of a parallelogram are equal and parallel. The diagonals of a parallelogram always bisect each other. (We may prove this using Euclid 1:34, etc. )

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**In the triangle above, if we bisect AC and AB… **

We find this theorem in a slightly different form in Euclid 6: 2. Parallel to AB, and equal to half of AB. A A A A B B Our second assumption is that, if we bisect two sides of any triangle, then draw a line between those two points, that line is parallel to the third side of the triangle, and ½ of its length. In the triangle above, if we bisect AC and AB… Then draw a line between those points… Then that line is parallel to AB, and ½ of the length of AB.

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**Proof that the three medians of any triangle meet at one point.**

B C Proof that the three medians of any triangle meet at one point.

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**Draw the median to point A.**

B C E D F Find the midpoint of CB. Draw the median to point A. Find the midpoint of AC, and draw the median to point B. The medians meet at a point, F.

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**Line ED is parallel to line AB, and is ½ the length of AB.**

C E D F Draw line ED. Line ED is parallel to line AB, and is ½ the length of AB. Point E splits side AC in half. Point D cuts side CB in half. Therefore ED is parallel to AB and is ½ as long as AB.

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**Why? C E D A B Locate the midpoint of BF. Locate the midpoint of AF.**

G Locate the midpoint of BF. Locate the midpoint of AF. Draw line HG. Line HG is parallel to line AB, and is ½ the length of AB. On triangle ABF, point G cuts FB in half. Point H cuts AF in half. There fore HG is parallel to AB and is one half as long. Why?

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A B C E D F H G Line HG is parallel to line ED, and is the same length. Why? (ED and HG are both parallel to AB, and both equal ½ of AB) The rectilinear figure HGED is a parallelogram. (Opposite sides are equal and parallel.) Why?

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**Why? C E D A B Therefore HF = FD. And EF = FG.**

The rectilinear figure HGED is a parallelogram. The diagonals of a parallelogram bisect one another. Why?

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**But we already know that AH = FH…. And that BG = FG.**

C E D F H G HF = FD and EF = FG. But we already know that AH = FH…. And that BG = FG. Thus AH = HF = FD. And BG = GF = FE.

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A B C E D F H G Hence medians AD and medians BE meet at point F, which is located at 2/3 of the distance between each vertex and the median of the opposite side. Because we have shown that this is true for any two sides and vertices, we know that it must be true for the third side and vertex.

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A B C E D F H G Therefore, the three medians of any triangle meet at a common point, known as the centroid, that cuts each median into a ratio of 2:1. We have not only proved our theorem, but we have learned something about where this point is located.

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Therefore, the three medians of any triangle meet at a common point, known as the centroid, that cuts each median into a ratio of 2:1. We have not only proved our theorem, but we have learned something about where this point is located.

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