PARTKULAELV APPUCAB.E TO SOME CEODET.CAL PROBLEMS^ ■„<, 



THEOREM. 



.he.e ,.. Wng either r„^ ;tl'i7;V„Tlr'?/-'f> ^^"' 



vngs. 1, 2.) or both without the angle 



Fig. 2. 



i?^C (% 3). Let the lines be sucli that the rectangle BA Ad i. 

 equal to the rectangle BA.AC, draw Hues fro. B Ja t^BaMC- 



"i>r:^; a^Lfaf >'' "- '--'' -- ^^-^- - ^^^ --.'e.- 



Demonstration. Because by hypothesis AD.Ad=AB.AC, there- 

 fore^^ : ^Z>=^^ : ^C. Now the angles BAD, dAC ar eql, 

 by hypothesis, therefore the triangles BAD, dAC are similar. 



Also because ^^ : Ad=AD : AC, and the angles BAd DAC are 

 equal, for they are the sums or differences of the equal angles BAD, 



