Hi Vladimir,

I will give you an analysis of my construction if you first tell me how you got your figure ( looks like it came from Sketchpad ) into a gif format.

Using your figure ( with B' as the midpoint of AC - don't know how to do subscripts ) here is an easier proof for this case:

Triangles BAP and SAB' are similar. Therefore, AS/AB' = AB/AP.

AS*AP = AB*AB' = AB*AC/2
AS*AP*sin(<SAP)/2 = (AB*AC*sin(<BAC)/2)/2
Area(SAP) = Area(BAC)/2

Bractals