398 



PROCEEDINGS OF THE AMERICAN ACADEMY. 



As an extension of the idea of similarity for triangles, we may say 

 that any two polygons which have their corresponding sides parallel 



and in proportion are similar. It fol- 

 lows that if any two corresponding 

 lines are drawn in the polygons, these 

 lines must be parallel. 



XI. If on two sides of a triangle 

 similar parallelograms be constructed, 

 and on the third side a parallelogram 

 with diagonals parallel to the diagonals 

 of the other parallelograms, the area 

 of this parallelogram will be equal to 

 the difference of the areas of the other 

 two. 



The areas (Figure 7) of the paral- 

 lelograms on AB, CA, BC are respec- 

 tively four times the areas of the 

 triangles ABF, CAE, BCD. If we take the unit parallelogram with 

 sides parallel to the diagonals, it will suffice to prove that 



Figure 6. 



Figure 7. 



FB X AF = AEX EC — BD X CD, 



for each of these areas is twice the area of the corresponding triangle. 

 In the similar triangles ACE and GCD, 



EC -.CD: -.AE-.DG. 



