Repeating the process with the arcs AC and CB, and continuing the repetition indefinitely, we divide up the required area and the remainder of the triangle **ATB** into corresponding elements, each element of the former being double the corresponding elements of the latter.

If we draw a line at right angles to TCV, meeting TCV produced in M and parallels through A and B in K and L, the area of the triangle **ATB** is KL.

To) than the area of the trapezium Kabl by two-thirds of the area of the triangle **ATB** (§ 34).