## 1. Thales theorem - II

Given are parallel lines e1, e2, ... and three transversals to them a, b and c, passing through the same point O.
Then on each parallel the transversals cut two segments in the same proportion: AY1/Y1A' = BY2/Y2B' = ... .

The inverse is also true. If three transversals a, b and c cut on parallels e1, e2, ... segments in the same proportion (i.e. AY1/Y1A' = BY2/Y2B' = ...) then a, b and c pass through the same point O.

To show, for example that BY2/Y2B' = AY1/Y1A', draw parallels to a, b from Y2 and define points A*, A'' as shown. Apply then Thales theorem considering the parallels a and line A*Y2. According to that theorem AY1/BY2 = AY1/AA* = Y1O/Y2O. Analogously Y1A'/Y2B' = Y1O/Y2O.
From this follows AY1/BY2 = Y1A'/Y2B' or equivalently AY1/Y1A' = BY2/Y2B'.
The inverse is proved by reversing the argument and applying the inverse of Thales theorem.

## 2. Corollary of Thales theorem

Given are parallel lines e1, e2, ... and two transversals to them a, b. The intersection points A*, B* ... of the diagonals of any trapezium with parallel sides on two of the ei's and non parallel sides on a, b, lie along a line (c) passing through the intersection point O of lines a and b as well as passing from the middles of all sides EE', DD', ... .

In fact, line c is the median of all triangles OEE', ODD', ... . We see this by examining one of this triangles and applying the previous theorem. Preferably we draw a parallel from A* and show that A*A1 = A*A2.
For this subtract from triangles AA'B and AA'B' their common part AA'A* and see that area(AA*B) = area(A'A*B'). Then, area(AA*B) = dA*A1/2, area(A'A*B') = dA*A2/2 ==> A1A* = A*A2, d being the distance of the parallel lines e1, e2. Hence A* is the middle of A1A2.

## 3. Application of Thales theorem

Let F be the middle of side AB of triangle ABC. Take points {G,H} on side AB symmetric about F. Draw also a line e parallel to AB intersecting {CG,CH} at points {J,K} respectively. Show that lines {AJ,BK} intersect at a point I of the median CF.

## 4. A second application

Divide the sides of a triangle ABC each in three equal parts through points {D,E,H,I,F,G}. Join the division points on each side with the opposite vertex to build a hexagon KMLOJN as shown in the picture below. Show that the quotient of areas is area(ABC)/area(KMLOJN) = 10.

## 5. A third application

Divide the sides of triangle ABC each in three parts through points {A',A'',B',B'',C',C''}, such that
A'B/A'C=B'C/B'A=C'A/C'B=k και A''B/A''C=B''C/B''A=C''A/C''B=k' (*).
Join the dividing points with the opposite sides and construct through the intersections of the various lines the various triangles, quadrangles, pentagons and the central hexagon of the figure shown. Show that:
[1] The area of the resulting polygons remain unchanged if a vertex of ABC is translated along the parallel to the opposite side.
[2] Triangles with the same color have equal areas.
[3] The ratio of areas of each of the triangles, quadrangles, pentagons and central hexagon to the area of triangle ABC is independent of the shape of ABC.
In all these variations of the shape of ABC we assume that the ratios defining the dividing points {A',A'',B',B'',C',C''} remain constant (i.e. (*) remains valid all the time).

From [1] follows that for the proof of the other two properties we can restrict ourselves to an equilateral ABC. A calculation of the various areas shown, from which our claims can be deduced, is contained in the reference from Steiner given below.

Remark and problem From Pascal's theorem follows that hexagon DEFGHI can never have a circumscribing it conic. But for some ratios {k,k'} the corresponding lines {DG, EH, FI} do have a common point and consequently, by Brianchon's theorem, there is an inscribed conic. Find the relation of {k,k'} for which this happens.