There are two triangles ABC and XYZ, ABC congruent to XYZ. XYZ is out side of ABC. Suppose D is mid point AX, E midpoint BY, and F midpoint CZ. Prove D, E, F is colinear.

It is because I can form a parallelogram which can be proven easier... If one of the triangles is rotated other than 180 degrees then the contradicting proof is done by the distance formula for the midpoints in which they are not collinear when rotated other than 180 degrees.

Without loss of generality, let triangle XYZ be rotated 180 degrees. We will be able to form a parallelogram AYXB for easier proving.
(This also work with other figures as long as one of the traingles is rotated 180 degrees.)

By coordinates:
Let B = (0,0), A = ((x(1), y(1)), C = (x(2), 0), Z = (x(3), y(1)), X = (x(4),0), and Y = (x(5), y(1)).

Suppose we find D, E, and F.
D = ([x(1) + x(4)]/2, y(1)/2)
E = (x(5)/2, y(1)/2)
F = ([x(3)+x(2)]/2, y(1)/2)

By distance formula:
DE = [x(1) + x(4) - x(5)]/2
EF = [x(5) - x(3) - x(2)]/2
DF = [x(3) + x(2) - x(1) - x(4)]/2

Note that the points D, E, and F are coinciding because the diagonals AX and BY share the same midpoint as well as the segments AX and CZ which implies that DE + EF + DF = 2DF = 0. By segment addition postulate, the points are collinear and at the same time coinciding.

P.S. I don't know how to type Latex. Also, I used this time the right correspondences compared to the earlier post.

## Comments

Sort by:

TopNewestWhy you W.L.O.G rotated \(180^{0}\)?

Log in to reply

It is because I can form a parallelogram which can be proven easier... If one of the triangles is rotated other than 180 degrees then the contradicting proof is done by the distance formula for the midpoints in which they are not collinear when rotated other than 180 degrees.

Log in to reply

Without loss of generality, let triangle XYZ be rotated 180 degrees. We will be able to form a parallelogram AYXB for easier proving. (This also work with other figures as long as one of the traingles is rotated 180 degrees.)

By coordinates: Let B = (0,0), A = ((x(1), y(1)), C = (x(2), 0), Z = (x(3), y(1)), X = (x(4),0), and Y = (x(5), y(1)).

Suppose we find D, E, and F. D = ([x(1) + x(4)]/2, y(1)/2) E = (x(5)/2, y(1)/2) F = ([x(3)+x(2)]/2, y(1)/2)

By distance formula: DE = [x(1) + x(4) - x(5)]/2 EF = [x(5) - x(3) - x(2)]/2 DF = [x(3) + x(2) - x(1) - x(4)]/2

Note that the points D, E, and F are coinciding because the diagonals AX and BY share the same midpoint as well as the segments AX and CZ which implies that DE + EF + DF = 2DF = 0. By segment addition postulate, the points are collinear and at the same time coinciding.

P.S. I don't know how to type Latex. Also, I used this time the right correspondences compared to the earlier post.

Log in to reply

For other rotations except for 0 and 180 degree rotation (e.g. 90 degree rotation), this can be checked by distance formula.

Log in to reply