‘roe 
Proceedings of the Royal Society 
3. The cases of certain irregular figures, in addition to those of 
isosceles triangles, are then deduced. 
If in fig. 1 a line he drawn from B at right angles to CB, and 
meet CG produced in J, the triangle OBJ is double of the triangle 
CBG. Hence the proposition follows in the case of equiangular 
right-angled triangles, one of the sides containing the right angle 
being a side of the given triangle j and hence also in the case of 
similar rectangles described on the sides. [E.) 
Again, it follows from (G), since the perpendicular from the 
vertex of an isosceles triangle on the base bisects the triangle, that 
the proposition is true in the case of equiangular right-angled 
triangles, the hypotenuses being respectively a side of the given 
triangle ; and also, by doubling the triangles in result (G), it follows 
in the case of rhombuses of the same angle, described on the sides. 
{F.) 
By combining these results we may show the proposition true in 
the case of any similar triangles described on the sides. 
For, since CJ, the hypotenuse of the triangle CBJ, is double of 
CG, and since CG, and corresponding lines HB, CP, have been 
shown to be the sides of a right-angled triangle equiangular to 
ABC, so also will CJ and its corresponding lines be the sides of a 
similar right-angled triangle. 
Again, the corresponding sides in the class of right-angled tri- 
angle in {F) are also the sides of a right-angled triangle. For, if 
GU be drawn at right angles to BJ-, GUB is such a triangle by 
what precedes. But GU is half of CB. Thus GU and corre- 
sponding lines are sides of a triangle similar to ABC, 
How, let AB (fig. 3) be the hypotenuse of the given triangle 
ABC, and let the triangle ADB, right- 
angled at D, be described upon it. By 
result (i^) the triangle ADB = sum of 
corresponding triangles on other sides 
of ABC, Produce D, and draw any 
line AE to meet it at E. Then, since 
AD and corresponding lines are the sides 
of a right-angled triangle, we have, by 
{E)^ the triangle ADE = sum of corre- 
the whole triangle ABE is equal to the 
