Two vertices of a triangle lie on some line. The third vertex belongs to the line that is parallel to the first one. Prove the area of that triangle does not depend on the position of the third vertex on the line.
the altitude of the triangle is the distance between the lines.
A = 1/2 bh
b is constant, h is constant.
To prove that the area of the triangle in question does not depend on the position of the third vertex on the line, we can use geometry and the properties of parallel lines.
Let's assume that the two given vertices of the triangle lie on a line, which we'll call line AB. The third vertex lies on a line parallel to AB. Let's call this line CD.
Now, let's consider two different positions for the third vertex, say C and C'. Without loss of generality, assume that C is closer to A than C'.
Since CD is parallel to AB, we can draw a transversal line through A intersecting CD at a point E, as shown in the diagram below:
```
A____________B
/. /
/ . /
/ . /
/ . /
/ . /
/____.EE.__/
C C'
```
Now, we can observe that triangle ABC and triangle AB'C' have the same height, which is the perpendicular distance between lines AB and CD (marked by the line EE in the diagram).
The bases of the two triangles, AC and AC', are both line segments on line CD. Since CD is parallel to AB, AC and AC' are parallel as well.
By the properties of triangles with the same height and parallel bases, we know that these two triangles have the same area. Therefore, the area of the triangle does not depend on the position of the third vertex C on line CD.
This can be generalized to any position of the third vertex on the line parallel to AB. The triangle's area will remain the same because the height and the bases remain constant.
Thus, we have proven that the area of the triangle in question does not depend on the position of the third vertex on the line.