Department of Mathematics Education

Exploration of A Triangle Ratios Problem

by

Jim Wilson

---

Given any triangle ABC and an arbitrary point M in its interior. Extend segments from each vertex through M to intersect the opposite sides, giving the segments AD, BE, and CF. Form the Ratios of segments

One approach

The approach here will be to construct auxilliary lines so that similar triangles can be found that allow substitution for the three ratios.


In a similar manner, construct a line parallel to BE to arrive at

and a line parallel to CF to arrive at

Finishing it.

These three ratios of segments on the sides of the triangle are constructed by the same orientation on each side. They are defined for any point M and we know from Ceva's Theorem that

Now, the difference of the product and the sum of the segments on the Cevians through M is

= [x(1 + z)y(1 + x)z(1 + y)] - [x(1 + z) + y(1 + x) + z(1 + y)]

= [xyz(1 + z)(1 + x)(1 + y) ] - [x + y + z + xy + yz + xz]

= [xyz(1 + x + y + z + xy + yz + xz + xyz) ] - [x + y + z + xy + yz + xz]

= [2 + x + y + z + xy + yz + xz ] - [x + y + z + xy + yz + xz]

= 2 +[x + y + z + xy + yz + xz ] - [x + y + z + xy + yz + xz]

= 2