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Ceva using vectors--special case

Some time ago we looked at Ceva's theorem (post). I'm starting on a book about Vector Calculus, and saw this question early in Chapter 1.

Using vectors, prove that the lines from each vertex of a triangle to the midpoint of the opposite side cross at a single point. The picture is as shown below:

We solve this by constructing parametric equations for the midpoint lines. The first one starts at O and travels along the vector **a** + (**b**-**a**)/2, which is the diagonal of the parallelogram formed by **a** and **b**. The equation is:

`u[`**a** + (**b**-**a**)/2]

where u is the parameter. Similarly, construct the line extending from A to its opposing side. It starts from A and travels along the vector **b**/2 - **a**. The equation is:

**a** + v(**b**/2 - **a**)

At the point where the vectors cross, these are equal:

`u[`**a** + (**b**-**a**)/2] = **a** + v(**b**/2 - **a**)

I had a little trouble at this point, and I must confess I peeked at the answer. We can solve this by considering that both the **a** and the **b** terms must balance. (Even though **a** isn't perpendicular to **b**, the part of **b** which is perpendicular to **a** is a constant fraction k of the whole).

This leads to:

`½ k u`**b** = ½ k v**b**

u = v = ⅔

`u`**a** - ½ u**a** = **a** - v**a**

½ u + v = 1

u = v = ⅔

Given this suggested solution u = v = ⅔ , we can easily verify that

`u[`**a** + (**b**-**a**)/2]

`⅔[`**a** + (**b**-**a**)/2]

` = ⅔ `**a** + ⅓ **b** - ⅓ **a** = ⅓ (**a** + **b**)

**a** + v(**b**/2 - **a**)

**a** + ⅔(**b**/2 - **a**)

` = `**a** + ⅓ **b** -⅔ **a** = ⅓ (**a** + **b**)

Now consider the third side. The midpoint line starts from B and travels along the vector **a**/2 - **b**. The equation is:

**b** + w(**a**/2 - **b**)

We observe that for w = ⅔ this becomes:

**b** + w(**a**/2 - **b**)

**b** + ⅔(**a**/2 - **b**)

**b** + ⅓ **a** - ⅔ **b** = ⅓ (**a** + **b**)

Thus, all three lines intersect at the same point.

I haven't got an extension to the general case just yet (i.e. any point, not just the midpoint).
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