266 BRADLEY M. PATTEN 



deflections' already ascertained. The magnitude of the angle 

 may bear no direct relation to the actual angle at which the sen- 

 sitive areas are located in the body of the animal, because of the 

 many factors which may modify the direction of the rays before 

 they fall on the sensitive surfaces. The significant test of the 

 hypothesis would be the constancy of the angle when computed 

 from experimental data obtained under varying conditions. 



The method of constructing such an angle is shown in figure 23, 

 in which the opposing lights are assumed to be of a two-to-one 

 ratio of intensity. The line AB is drawn perpendicular to the 

 direction of the rays of light. On the line AB, construct angle 

 BOC equal to the actual average angular deflection of the larvae 

 at a two-to-one ratio of lights (p. 245). The problem now re- 

 solves itseK into the construction of an angle about OC as a 

 bisector, which shall be of such a magnitude that equal distances 

 on its opposite sides shall have projections on the line AB of the 

 ratio of two to one. 



Construction: From a point D on the line OC draw Dh perpen- 

 dicular to AB. Lay off on AB distances hx and hy, such that 

 hy = 2hx. From x and?/ erect lines perpendicular to AB, they 

 will intersect OC at / and e respectively. Bisect the line ef, 

 and at its middle point, g, construct a line kl perpendicular to 

 OC. From the point of intersection of kl and yy' (M), draw^ a 

 line to D. From the intersection of kl and xx' (N), draw a line 

 toD. 



The angle MDN^ is the desired angle 

 Proof: eg=gf (construction) 

 Angle egM = angle fgN (construction) 

 Angle Meg = angle Nfg (alternate int. angles of parallel 

 lines, yy' and xx' being parallel by construction) 



Therefore triangle Meg = triangle. iVgff (side and two adjacent 

 angles being equal) 



Ng = gM (similar sides of equal triangles) 

 gD = gD (identical) 



Therefore triangle NgD = triangle Mgd (rt. triangles, altitude 

 and base equal) 



Therefore angle gDM = angle gDN and side DM = side DN. 



