Euclid's Elements Book I

23 definitions · 5 postulates · 5 common notions · 48 propositions

Proposition I.41 · Theorem

If a parallelogram have the same base with a triangle and be in the same parallels, the parallelogram is double of the triangle.

Drag the points circled in red. The figure changes, and every step of the proof still holds.
  1. Let the parallelogram ABCD have the same base BC with the triangle EBC, and let it be in the same parallels BC, AE. The claim: the parallelogram ABCD is double of the triangle BEC. (Try dragging A, B, C or E.)
  2. Let AC be joined.
    Postulate 1
    Postulate 1: To draw a straight line from any point to any point.
  3. The triangle ABC is equal to the triangle EBC, for it is on the same base BC with it and in the same parallels BC, AE.
    Proposition I.37
    Proposition I.37: Triangles which are on the same base and in the same parallels are equal to one another.
  4. But the diameter AC bisects the parallelogram ABCD, so ABCD is double of the triangle ABC; therefore it is also double of the triangle EBC.
    Proposition I.34
    Proposition I.34: In parallelogrammic areas the opposite sides and angles are equal to one another, and the diameter bisects the areas.