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# 8.3: Medians and centroid

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A median of a triangle is the segment joining a vertex to the midpoint of the opposing side.

Theorem $$\PageIndex{1}$$

The three medians of any nondegenerate triangle intersect in a single point. Moreover, the point of intersection divides each median in the ratio 2:1.

The point of intersection of medians is called the centroid of the tri- angle; it is usually denoted by M. In the proof we will apply Exercise 3.4.3 and Exercise7.3.1; their complete solutions are given in the hits.

Proof

Consider a nondegenerate triangle $$ABC$$. Let $$[AA']$$ and $$[BB']$$ be its medians. According to Exercise 3.4.3, $$[AA']$$ and $$[BB']$$ have a point of intersection; denote it by $$M$$.

Draw a line $$\ell$$ thru $$A'$$ parallel to $$(BB')$$. Applying Exercise7.3.1 for $$\triangle BB'C$$ and $$\ell$$, we get that $$\ell$$ cross $$[B'C]$$ at some point $$X$$ and

$$\dfrac{CX}{CB'} = \dfrac{CA'}{CB} = \dfrac{1}{2};$$

that is, $$X$$ is the midpoint of $$[CB']$$.

Since $$B'$$ is the midpoint of $$[AC]$$ and $$X$$ is the midpoint of $$[B'C]$$, we get that

$$\dfrac{AB'}{AX} = \dfrac{2}{3}.$$

Applying Exercise 7.3.1 for $$\triangle XA'A$$ and the line $$(BB')$$, we get that

$\dfrac{AM}{AA'} = \dfrac{AB'}{AX} = \dfrac{2}{3};$

that is, $$M$$ divides $$[AA']$$ in the ratio 2:1.

Note that 8.3.1 uniquely defines $$M$$ on $$[AA']$$. Repeating the same argument for medians $$[AA']$$ and $$[CC']$$, we get that they intersect at $$M$$ as well, hence the result.

Exercise $$\PageIndex{1}$$

Let $$\square ABCD$$ be a nondegenerate quadrangle and $$X, Y, V, W$$ be the midpoints of its sides $$[AB], [BC], [CD]$$, and $$[DA]$$. Show that $$\square XYVW$$ is a parallelogram.

Hint

Use the idea from the proof of Theorem $$\PageIndex{1}$$ to show that $$(XY) \parallel (AC) \parallel (VW)$$ and $$(XV) \parallel (BD) \parallel (YW)$$.