Fine area of hexagon traced by cevians of trisecting points of triangle ABC

Question

In any triangle ABC, two points are marked on each side, in order to divide the side into three equal parts. When tracing the cevians to each of the six marked points there will be a hexagon in the center of the figure. Give the ratio between the area of ​​the hexagon and the area of ​​the triangle ABC.

Answer 1

Let [] denote areas. For the trisecting points D, E and F on the respective sides of the triangle ABC, it is well-known that

$$[XYZ] = \frac17[ABC]$$

It can also be shown that

$$\frac{XU}{XZ} = \frac14,\>\>\>\>\>\frac{XV}{XY} = \frac25$$

As a result,

$$[XVU] = \frac14\cdot\frac25 [XYZ] = \frac1{10}[XYZ]$$

Similarly, $$[YTS] = [ZWH] =\frac1{10}[XYZ]$$

Thus, the area of the hexagon is

$$[UVSTHW] = [XYZ] - [XVU] - [YTS] - [ZWH] $$ $$= (1-\frac3{10})[XYZ]=\frac7{10}[XYZ]=\frac1{10}[ABC]$$

Thus, the are ratio between ​​the hexagon and the triangle ABC is $\frac1{10}$.