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AEt, BCF, muft be equal; wherefore the lides are pro- 
portional, and BC and ae, fubtending equal angles muft 
be homologous; but bc is equal to ac, which is greater 
thaiiAE; confeqviently the triangle bfc is greater than 
AFE, and fo the equilateral triangle abc is greater than 
the triangle abe. In the fame manner, the triangle 
ABE may be proved greater than ade;. for ahe is com- 
mon, and the two Mangles adh,‘bhe, are ITmilar, and 
their lides proportional; and ad and be, fubtending 
equal angles, muft bd homologous;* but be is greater 
than BC, which is equal to ab,. and that again greater 
than AD confequently be is greater than ad, and the 
whole triangle aeb greater than aed ; and fo the equi-? 
lateral triangle muft, a fortiori^ be greater, thhi. aed-^ 
. hfext, let the triangle ade be fuppoftd greater than 
the equilateral triangle abc, and let the angle ade fall 
fomewhere in the fegment bdc, (fee fig. 2.) fo that the 
fegment ec may be greater than bd ;, for if it were not, 
the angle aed being applied to any of the angles of the 
equilateral triangle, the demonftration would become the 
fame as in the firft cafe: wherefore the fegments aeg, 
BDC,. being equal, and bd being lefs tlian ec, ae muft be 
left than dc. Draw the right line dc, and then in the 
two triangles adc,. ade, the triangle afd is common, 
and the two. triangles afe, dfc, are equiangular and 
fimilar, and the fides ae, dc, fubtending equal angles, 
are homologous ; but dc is greater than ae ; fo the trian- 
gle. DFC is greater than the triangle afe, and the whole 
S f 2 triangle 
O 
