of Edinhurgli, Session 1883-84, 
705 
and thus the equal sides of the isosceles triangles are also the sides 
of a right-angled triangle. Now draAv a line from B at right angles 
to CB, and let it meet CG produced in J ; then the triangle BGJ 
is isosceles, BG being equal to GJ, and is equal to the triangle GCB. 
The same being true in the case of corresponding triangles BHO 
and CPQ described on the other side, it follows that 
ABHO= ABGJ+ A CPQ, 
But HB, BG, and CP are the sides of a right-angled triangle ; and 
this triangle being equiangular to the triangle ABO, the triangle 
BGJ is any isosceles triangle described on BG. Hence the propo- 
sition follows in the case of equiangular isosceles triangles^ one of 
their equal sides being respectively a sid.eof the right-a.ngled triangle, 
(C.) 
The case of similar segments of circles described on the sides 
may be deduced either by the application of result {A) or of (G), 
The latter may be employed by inscribing in the circles, of which 
the described segments are parts, regular polygons of the same 
number of sides, and joining their centres to the angular points of 
the polygons which are external to the sides of the triangle. These 
joining lines being the sides of a right-angled triangle, it may be, 
shown by (G) that the figure external to AB is equal to the sum of 
the corresponding figures for BG and CA, and therefore, in the 
limit, the proposition follows in the case of the segments. 
It may, however, be more simply shown by the application of (A) 
directly. Let AB (fig. 2) be the hypotenuse 
of the right-angled triangle ABC^ and ADEB 
any segment of a circle described on it. If 
D be the middle point of the arc, the tri- 
angle ADB = sum of the corresponding tri- 
angles by (A). Again, if E be the middle 
point of arc DEB, since DB and the cor- 
responding lines are sides of a right-angled 
triangle, the triangle DEB = sum of corresponding triangles. Pro- 
ceeding in like manner with the arcs DE and EB, it follows finally 
that the segment DEB = sum of corresponding segments. The 
same being true of the segment AFD, therefore the whole segment 
AEDEB = sum of similar segments on CA and BC. (D.) 
VOL, XII. 2 z 
D 
