XX INTRODUCTION. 



indicated by OA, with a similar convention for OB. 

 Completing the parallelogram OACB, we shall prove 

 that a rotation around OC y through an angle propor- 

 tional to the length OC, will have precisely the same 

 effect as the two given rotations. 



Consider any point P of the body which lies in 

 the plane of the axes. The rotation OA will depress 

 the point P below the plane of the paper along 

 the normal to a small distance which is propor- 

 tional to the product of OA, and the perpendicular 

 PQ ; that is, proportional to the area of the triangle 

 POA. In the same way the rotation around OB will 

 raise P above the plane of the paper to a distance 

 which is proportional to the area viPOB. The joint effect 

 will be to raise P to a distance above the paper propor- 

 tional to the difference between the areas of the triangle 

 POB and POA. that is, to the area of POC; but this is 

 precisely the same effect as would be produced by a rota- 

 tion around OC through an angle proportional to OC. 



To prove that POC = POB - POA : draw PR 

 parallel to OA ; then OAR = GAP, and BRC^BPC, 

 whence POA + PBC = OBC ; also we have 



POA + POB + PBC = POA + POC + OBC, 



since each side represents the area of the figure 

 OAPBC ; therefore 



POB = POA + POC. 

 or 



POC = POB -POA. 



The rotation around OC must, therefore, produce 

 precisely the same effect on every point in the plane as 

 is produced by the joint effect of the rotations around 

 OA and OB. 



