704 
Proceedings of the Boycd Society 
For a circle may be described about the figure HFKB ; therefore 
the angle KHB is equal to the angle KFB. But the angle HBK is 
equal to the angle FBM, since the isosceles triangles are equi- 
angular; therefore the remaining angle FMB is equal to the 
remaining angle HKB, and is a right angle. Thus K is the ortho- 
centre of the triangle GFB, and the line GKL is perpendicular to 
FB. 
Hence the figure FGKH is a parallelogram, and FG = KH, and 
therefore the triangle GFB is equal to the triangle BHD, since the 
rectangle whose sides are FG and DB is equal to the rectangle 
whose sides are HK and DB. 
Thus the whole figure GCFB is equal to the whole triangle HOB ; 
that is, denoting the isosceles triangles by A, C respectively, 
A -f GFB = HHB + CHF + GFB , 
or ^=HNB^-GHF. 
Similarly B + CFA==HNA + CNA, 
or i5-t-GHF = HNA. 
Thus A + B=0. 
Now let regular polygons of the same number of sides be described 
on the sides of the right-angled triangle, and suppose B, S, T to be 
the centres of the polygons. By joining K, S, T to the angular 
points of the respective figures of which they are centres, the poly- 
gons will be divided into the same number of isosceles triangles 
which are equal to one another in each polygon. But, since these 
triangles have the same vertical angle, by what precedes 
ATAB= ARBG+ ASGA. 
Hence the regular polygon described on AB is equal to the sum of 
the regular polygons of same number of sides described on BG and 
GA. (B.) 
2. Since the sides of the triangle GDK or of HFI are equal to 
FH, DG and PE respectively, it follows that 
HF2 = DG2 + PE2, 
But FB2 =DB2-fGE2, 
Therefore HB2 = BG^ -i- PC^ , 
