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Answer:

First question: Since this shape is a square, the midpoint of the two diagonals shall coincide with each other.

Second question: assume that by "midpoint" the question refers to the centroid of the triangle. The centroid of a triangle is on its median 2/3 the way from the corresponding vertice.

Step-by-step explanation:

First question

Midpoint of the diagonal between (x1, y1) and (x3, y3):

[tex]\displaystyle \left(\frac{x_1 + x_3}{2}, \frac{y_1 + y_3}{2}\right)[/tex].

Similarly, midpoint of the diagonal between (x2, y2) and (x4, y4):

[tex]\displaystyle \left(\frac{x_2 + x_4}{2}, \frac{y_2 + y_4}{2}\right)[/tex].

The two midpoints shall coincide. Therefore,

[tex]\displaystyle \frac{x_1 + x_3}{2} = \frac{x_2 + x_4}{2} \implies x_1 + x_3 = x_2 + x_4[/tex].

Similarly,

[tex]\displaystyle \frac{y_1 + y_3}{2} = \frac{y_2 + y_4}{2} \implies y_1 + y_3 = x_2 + x_4[/tex].

Second question

The centroid of a triangle divides all three of its medians at a 2:1 ratio. If the length of a median of the triangle is [tex]L[/tex], the centroid of that triangle is at a distance of [tex]2/3\;L[/tex] from the vertex on that median.

Start with the median that goes through the vertex [tex](x_1, y_1)[/tex]. That median also goes through the midpoint between [tex](x_2, y_2)[/tex] and [tex](x_3, y_3)[/tex].

  • Vertex: [tex](x_1, y_1)[/tex].
  • Midpoint of the opposite side: [tex]\displaystyle \left (\frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2}\right)[/tex]

The centroid will be located at

[tex]\displaystyle \left (\phantom{\frac{\phantom{x_{1}}}{2}}\right.x_1 + \frac{2}{3}\underbrace{\left(\frac{x_2 + x_3}{2} - x_1\right)}_{\begin{gathered}\text{Separation}\\[-0.5em]\text{in }x\text{-}\\[-0.5em]\text{direction}\end{gathered}}, \quad y_1 + \frac{2}{3}\underbrace{\left(\frac{y_2 + y_3}{2} - y_1\right)}_{{\begin{gathered}\text{Separation}\\[-0.5em]\text{in }y\text{-}\\[-0.5em]\text{direction}\end{gathered}}}\left.\phantom{\frac{\phantom{x_{1}}}{2}}\right)\right)[/tex].

Simplify this expression to obtain:

[tex]\displaystyle \left (\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 +y_2 + y_3}{3}\right)\right)[/tex].

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