trianglerecthalf

A triangle and a rectangle with the same base and half its height

The hinged figure suggests that the area of a triangle is the same as the area of a rectangle with the same base and half the height.

Proof with rotations. When triangle BGE is rotated 180˚ around the midpoint E of side BA, the image of segment BG will be a perpendicular segment to the base at A. And GE and its image EH will be on the same line. In the same way, when triangle BDG is rotated 180˚ around the midpoint D of the other side BC, the image of BG will be a perpendicular segment to the base at C, and DG and its image will be on the same line. Therefore, CAHI will be a rectangle.

Synthetic proof. Let D and E be the midpoints of two sides of triangle ABC, and let BF be the height of the triangle. The line through the midpoints will be parallel to the base CA. To prove that the area of the rectangle CAHI has the same area as the original triangle ABC we need to prove that the triangles BGE and AHE are congruent (and also triangles BGD and CID are congruent).

Convince yourself of the following statements and give a reason why:
BE congruent to EA
BG congruent to HA
Angle α congruent to angle β
The previous facts are enough to guarantee the congruency of triangles BGE and AHE
In a similar way you can convince your self that triangle BGD is congruent to triangle CID.

If h is the height of the triangle, the height of the rectangle will be h/2. The area of the rectangle is its base times its height, b × h/2, so a formula for the area of the triangle is base times half the height,

Notice also the difference in the two proofs. In the proof by rotations, the assumption is that the pieces are congruent, so that the trapezoid and the shape obtained have the same area, and it is then shown that the shape obtained is indeed a rectangle. In the synthetic proof we start with a rectangle, and show that triangles are congruent so that the area of the rectangle and the area of the trapezoid are the same.