Trapezium, Central Median of Trapezium, Triangle
The sides of the trapezium that are parallel, are called bases and those that are not parallel are called legs. If the legs are equal in length, then this is an isosceles trapezoid. The distance between the bases is called height of trapezoid.
Central Median of Trapezium
This is the line segment joining the middles of the two sides of a trapezium that are not parallel. The central median of a trapezium is parallel to its sides.
AM = MD; BN = NC
MN central median, AB and CD are bases, AD and BC are legs
Basic assignment: Prove that the central median of a trapezium halves each segment the ends of which lie on the two bases.
Central Median of a Triangle
The line segment joining the middles of two of the sides of a triangle is called central median of a triangle. It is parallel to the third side and its length is half the length of the third side.
Theorem: If a line segment crosses the middle of one side of a triangle and is parallel to another side of the same triangle, then this line segment halves the third side.
AM = MC and BN = NC =>
Application of the properties of the central medians in a trapezium and triangle
Division of the segment into a given number of equal parts
Assignment: Divide the segment AB given into 5 equal parts.
Let p be an arbitrary ray with A being the beginning, which does not lie on the AB straight line. We draw consecutively five equal segments on p AA1 = A1A2 = A2A3 = A3A4 = A4A5
We connect A5 with B and draw lines through A4, A3, A2 and A1 that are parallel to A5B. They cross AB respectively in the points B4, B3, B2 and B1. These points divide the segment AB into five equal parts. Indeed, from the trapezium BB3A3A5 we see that BB4 = B4B3. In the same way, from the trapezium B4B2A2A4, we obtain B4B3 = B3B2
While from the trapezium B3B1A1A3, B3B2 = B2B1.
Then, from B2AA2, it follows that B2B1 = B1A. We finally obtain :
AB1 = B1B2 = B2B3 = B3B4 = B4B
It is clear that if AB should be divided into another number of equal parts, we should project the same number of equal segments on p. Then we proceed in the way above-described.