220 
Mr. Atwood's Disquisition on 
BO, as the sine of the angle AUB to radius. Let BG be taken 
a geometrical mean proportional between the lines BF and BE ; 
and from the point G, in the direction of the line GU, set off GZ 
equal to BF : join AZ ; and, through the point G, draw GS 
parallel to ZA, intersecting BA in the point S. Through S, 
draw the line CH parallel to BU : the area ASH will be equal 
to the area BSC. 
Since, in the triangles ASH, BSC, the angle ASH is equal 
to the angle BSC, the areas of the triangles will be in a ratio 
compounded of the ratios of the sides, including the equal 
angles ; that is, the area of the triangle ASH, will be to the 
area of the triangle BSC in the ratio of SA x SH to SB x 
SC. By the construction, the angle ASH = the angle ABU 
= OPO ; and the angle AHS = the angle AUB : also, by 
construction, 
BO : BD : : rad. : cos. ASH. 
Also BD : BE : : rad. : sin. SAH. 
And BF : BO : : sin. AHS : rad. 
Joining these ratios BF : BE : : sin. AHS x rad. : cos. ASH x sin. SAH. 
But, by the construction, and by the similarity of the triangles, 
BGS, BZA, BF or GZ : BE* : : BF a : BG 1 : : GZ 2 : BG 2 : : SA 2 : SB 2 : 
Wherefore SA 2 : SB 2 : : sin. AHS x rad. : cos. ASH x sin. SAH. 
And by trigonometry SH : SA : : sin. SAH : sin. AHS ; 
and SB : SC : : cos. ASH : rad. 
Joining these ratios SA x SH : SB x SC : : rad. : rad. 
But the area ASH is to the area BSC as SA x SH to SB x 
SC ; consequently, the area ASH is equal to the area BSC. 
To proceed with the construction of the second case. 
Through the point S, (fig. 4.) determined by the preceding 
construction, draw the line CH inclined to BA at the angle 
* Because the ratio of BE to BG is equal to the ratio of BG to BF, by the construc- 
tion, it will follow that the ratio of BE to BF is double the ratio of BE to BG. 
