356 Direct Solution of a Geometrical Problem. 



bisected, and the portions of the bisecting lines A E and B D 

 contained between the vertices A, B and the opposite sides B C 

 and AC are equal: to prove that the angles » and ft are also 

 equal. 



Find a point F, so that FA = AD and FE = AB, and draw 

 lineFB. 

 I. We have 

 AF = ALK 



p E=AB L : follows the equality of triangles F AE and 

 ADB. 



AE = DB 

 II. 



Therefore 

 But 



Therefore 



or 



Also 

 and 



ADB-f £=D</E, 

 BEA-J-^=D#E. 



ADB+g=BEA+|- 



ADB = FAE,| 

 FEA=| J" 



FAE-f-|=BEA + FEA, 



FAB=BEF. 

 FB = FB, 



(I) 



FE = AB. 



III. From which follows the equality of triangles F B A and 

 F B E, considering that 



angles FAB = FEB = BEA+f=D^E. . . (II.) 



But 



D^E = 180°-*-±^ 



« + /3<180°, 



and 

 therefore 



D#E or FAB^>90°. 



