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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.

Theorem:

Theorem:

MN || AB || DC
AM = MD; BN = NC

MN central median, AB and CD are bases, AD and BC are legs

Theorem:

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.
Solution:
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.