Area Theorems

Two figures are said to be between the same parallels when both lie inside the strip formed by one pair of parallel lines, with one side resting on the same base line. Because the two parallel lines never get closer or further apart, the perpendicular distance between them (the height) is the same everywhere. Area of a parallelogram depends only on its base and this height, so the strip fixes the height and the common base fixes the width. That is why two very differently shaped figures in the same strip can still share the same area.

The first main theorem states that parallelograms on the same base and between the same parallels are equal in area. The proof slides one parallelogram into the other: triangles cut off at one end exactly fill the gap at the other end, because opposite sides of a parallelogram are equal and parallel. So area ABCD = base × height = area ABEF, even though one may look long and slanted and the other short and upright. The shape changed, the area did not.

A triangle on the same base and between the same parallels as a parallelogram has exactly half its area. The reason is simple: complete the triangle into a parallelogram by adding an equal triangle, and a diagonal always splits a parallelogram into two equal triangles. So area of triangle = half × base × height. From this follows the matching theorem: triangles on the same base and between the same parallels are equal in area, because each is half of the same parallelogram.

These theorems give two quick everyday results. A median of a triangle joins a vertex to the midpoint of the opposite side, so it creates two smaller triangles with equal bases and the same height (measured from that vertex) - hence equal areas. A diagonal of a parallelogram splits it into two congruent triangles, again of equal area. Many Selina problems chain these facts together to prove that two seemingly unrelated triangles in a figure have equal areas.