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Can you help me to prove that in any trapezoid, which is not a parallelogram, the following points are collinear?

  • The midpoints of its bases.
  • The point of intersection of diagonals.
  • The point of intersection of the extended sides of trapezoid that are not parallel.

Thank you very much!

MvG
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wonderingdev
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2 Answers2

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These form a triangle:

  • A: The point of intersection of the extended sides of trapezoid that are not parallel.
  • B & C The endponts of the longest base.

And all the points are on the on the median from the point A.

So that is enough hint

GOOD LUCK

Willemien
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0

Without loss of generality you may assume the corners to have the following coordinates:

$$(a, 0)\qquad (b, 0)\qquad (c, h)\qquad (d, h)$$

Then the points you mention have coordinates

$$ \begin{array}{ccc} \left(\frac{a+b}2, 0\right) &\qquad& \left(\frac{c+d}2, h\right) \\ \left(\frac{ad-bc}{a-b-c+d}, \frac{(a-b)h}{a-b-c+d}\right) &\qquad& \left(\frac{ac-bd}{a-b+c-d}, \frac{(a-b)h}{a-b+c-d}\right) \end{array} $$

All of them are incident to the line

$$2h\,x + (a+b-c-d)y=(a+b)h$$

I actually computed these coordinates and the corresponding line using projective geometry, where both operations (joining points and intersecting lines) can be expressed easily using the cross product.

Community
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MvG
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