RESPECTING THE THEORY OF PARALLELS. 445 



We may advance yet a stage further. It will suffice if the needed postulate 

 read thus — 



Postulate 3. Given a finite triangle and a finite number, it is possible to find 

 a triangle which shall exceed the multiple of the given triangle by the given 

 number. 



If this postulate be admitted, we can prove that the angles of a triangle are 

 together equal to two right angles as follows : — 



Let ABC be a finite triangle : ^ABC will be a finite angle, however small ; 

 and the multiplier m, which applied to ^ABC shall make it greater than two 

 right angles, will be finite. By the postulate a triangle can be found the area of 

 which shall exceed m times the area ABC. Let DEF be this triangle ; then 

 5DEF : ^ABC : : area DEF : area ABC 



* :p^m : 1 



Therefore ^DEF:^w^ ^ABC 



:^two right angles ; which is absurd 

 Hence, the angles of a triangle cannot fall short of two right angles. 



It must be remembered that this is not put forth as a demonstration : it is 

 merely exhibited to shew what more will suffice to render the demonstration 

 possible. 



Cor. It is evident that the area of a quadrilateral is proportional to its angular 

 defect from four right angles ; and that, with this understanding, everything 

 which has been here demonstrated of triangles is applicable to quadrilaterals. 



Prop. XVIII. If from two points in one straight line the perijendiculars drawn to 

 another straight line are equal., so that the lines are parallel (Prop. VI.), a7id a 

 straight line he drawn from one of the given poijits to a point in the given line 

 produced on the side towards which the point lies., the exterior angle which 

 this line makes with the line joining the points exceeds the interior opposite 

 angle which it makes with the other line, hy the angular defect of the quadri- 

 lateral formed hy these three lines and the common perpendicular. 



Let CE=DF in the figure of Prop. VI., and let any point B in EF produced be 

 joined with D and let BD be produced to L ; -===:CDL exceeds FBD by the angular 

 defect of KHBD. For 90°- CDF = |^CEFD (Prop. XVIL, Cor.) 



= 5KHFD (Prop. VI., Cor. 2), 

 and CDL + CDF + FDB = 180° ; 

 therefore CDL-FBD = 90°-CDF + 90°-(FBD + FDB) 

 =^KHFD+^DFB 

 = 5KHBD. 



VOL. XXIII. PART III. 6 E 



