PHYSICAL AND LITERARY. 115- 



the triangles, GDK, GFK are equal, and the 

 fpace GHK common to both, the fedor KDH 

 "Vvill be equal to the fum of the fpaces KEFK, 

 GFHG ; therefore the fpdor KDE is equal 

 to the fpace KEFK. And, becaufe the tri- 

 angle KDC is equal to the triangle KFC, the 

 fpace KCDE will be equal to the fedor KCF; 

 therefore (adding the fedor BCK to both) 

 the fedor BDE will be equal to the fedor 

 BCF. Again, becaufe the arc AFB is to the 

 arc BE as/* is to 5-, the femicircle will be to 

 the fedor BCF as /> is to 5^ j therefore the 

 femicircle will be to the fedor BDE as p is 

 to^'. Qi^E. D. 



' But, as this would require a good deal of 

 trigonometrical calculation, the following 

 method may be ufed ; which will give the 

 point fought very nearly, when this problerrj 

 is of ufe in the planetary fyftem. 



PROP. V. Probl. 3. Fig. 5. 



Let there be a femicircle whofe dia- 

 meter is AB and centre C, and let 

 D be a point in the diameter not 



very 



