be the area of the triangle after the translation. The total area in Figure 1(a), , is equal to plus . On the other hand, the same area is equal to plus . So we get and, since , we get which is the statement of the sliding argument. The same proof also works when the final triangle does not overlap the initial triangle, or it is a consequence of the previous argument by decomposing the translation into several small translations. AT′ = A′B′C′ AB′C′C AL AT′ AT AS AL+AT′ = AT + AS AT′ = AT AL = AS Of course, the area of is equal to the area of its reflection through the leg of the triangle, so the sliding argument might be interpreted as giving equality between the two areas and in Figure 1(b). We shall use this interpretation in what follows. We shall also use the corresponding version of the sliding argument when the triangle is translated parallel to the other leg .