874 On a Solution of the Theory of Parallels, 



equal. Therefore BDA is greater than a right angle. Therefore 

 in the triangle ADC there is one angle greater than two right 

 angles, which is impossible (XVII.). Nor can the triangle ED C 

 lie above or below AB and AC (as the dotted lines in the figure), 

 for the angles at A and D are equal (XXI.) ; therefore ED must 

 fall on AB, and therefore the triangles are, in every respect, equal. 



Proposition II. 



The three interior angles of every right-angled triangle are together 



equal to two right angles. 



Let ABC be a right-angled tri- p 

 angle ; produce AB till AD is equal 

 to AC, and take, in AC, AE equal to 

 AB. Draw DC and BE. At the 

 point B draw BE perpendicular to AD. 



As the triangle DAC is an isosceles 

 right-angled triangle, the angles at 

 D and C are equal to half right 

 angles (Prop. I.) . The triangle BDF 

 is isosceles, because one angle, DBF, 

 is a right angle, and another, BDF, 

 is half a right angle (Lemma I.). The angle CFB is greater 

 than a right angle, and, being the supplement of half a right 

 angle, is equal to CEB, for the angle AEB is equal to half a 

 right angle (Prop. I.) . BE is equal to BD ; BD is equal to EC ; 

 therefore BF and EC are equal. 



In the triangles BFC and CEB, two sides, and an angle greater 

 than a right angle, are respectively equal ; therefore the triangles 

 are equal in every respect (Lemma II.) ; therefore the angles 

 EBC and ECB are equal to the angles FCB and FBC. Now 

 the angles FBE and FCE are each half right angles ; therefore 

 the angles EBC and ECB are together equal to half a right 

 angle. To these add ABE, which is half a right angle, and 

 BAE, which is a right angle ; then the three interior angles of the 

 right-angled triangle ABC are together equal to two right angles. 



Proposition III. 



The three interior angles of every triangle are together equal to 

 two right angles. 



Let fall upon the greater side 

 AC the perpendicular BD. 



In each of the right-angled tri- 

 angles thus formed there are two 

 right angles (Prop. II.). Taking 

 away the two right angles at D, 

 there remains the three angles of 

 the triangle ABC, together equal to two right angles. 



