You will need
- The sides of the triangle, the angles of a triangle
Instruction
1
Let in triangle ABC, MN - an average line connecting the midpoints of the legs AB (point M) and AC (point N).
By property, the middle line of the triangle connecting the centers of two sides parallel to the third side and equal to its half. So the middle line MN is parallel to side BC and equal to BC/2.
Therefore, to determine the length of the middle line of the triangle, it is sufficient to know the side length of this third side.
By property, the middle line of the triangle connecting the centers of two sides parallel to the third side and equal to its half. So the middle line MN is parallel to side BC and equal to BC/2.
Therefore, to determine the length of the middle line of the triangle, it is sufficient to know the side length of this third side.
2
Now let the known sides, means which connects the middle line MN, that is, AB and AC and the angle BAC between them. Since MN is the middle line, then AM = AB/2 and AN = AC/2.
Then the cosine theorem is true: MN^2 = (AM^2)+(AN^2)-2*AM*AN*cos(BAC) = (AB^2/4)+(AC^2/4)-AB*AC*cos(BAC)/2. Hence, MN = sqrt((AB^2/4)+(AC^2/4)-AB*AC*cos(BAC)/2).
Then the cosine theorem is true: MN^2 = (AM^2)+(AN^2)-2*AM*AN*cos(BAC) = (AB^2/4)+(AC^2/4)-AB*AC*cos(BAC)/2. Hence, MN = sqrt((AB^2/4)+(AC^2/4)-AB*AC*cos(BAC)/2).
3
If we know the sides AB and AC, then the middle line MN can be found knowing the angle ABC or ACB. Suppose, for example, known angle ABC. Since the property of the middle line MN is parallel to BC, the angles ABC and AMN are corresponding, and therefore ABC = AMN. Then the cosine theorem: AN^2 = AC^2/4 = (AM^2)+(MN^2)-2*AM*MN*cos(AMN). Therefore, the side MN can be found from the quadratic equation (MN^2)-AB*MN*cos(ABC)-(AC^2/4) = 0.
