THEORY OF POLYCONIC PROJECTIONS. 



61 



sphere at 0. The triangle OFF' is isosceles; therefore, the 

 line F'F' is equally inclined to the planes of the great 

 circles, since it is equally inclined to their perpendiculars 

 OF and OF' . Produce FF' in both directions to intersect 

 the planes of the circles, the one at Q and the other at Q'. 

 The triangle OPG = the triangle OF'Q', since OF = OF', 

 10PQ= lOF'Q', and Z.FOQ= IF'OQ' . Therefore, 

 QO = Q'0 and QD = Q'C'. Pass a plane through FF' and 

 let QGHG' be its trace on the plane of BED' and let 

 Q'F'HFhe the trace on the plane of CEC. Then Z OQH = 

 /.OQ'H, since the corresponding right triangles are equal. 

 The arc DG will therefore equal the arc C'F', and the arc 

 G'D^ will equal the arc OF, since Q and Q' are the same 

 distance from their respective great circles. But the arc 

 GEG' = Tr-(DG + D'G') and the arc FEF' = tt - (F' 6" + CF) . 

 Therefore, the arc GEG' is equal to the arc FEF', and the 

 proposition is proved. 



Problem 8. — To determine the shortest distance between 

 two points whose projections P and Q are given; that is, 

 to determine the arc of a great circle between them: 



Fig. 22. — Great circle arc between two points on stereographic projection. 



