OF SPACE. 21 



Let AB>AC, then Z.ACB>AI}C. For takinj; ^ 



ADzrAC, and joining CD, ZACDzzADC (87). 



But Z ADC > CBD (97), and ACB > ACD, therefore a D" 



much more /.ACB > CBD, or ABC, 



99. Theorem. Of two triangles on the same base, 

 the sides of the interior contain the greater angle. 

 Produce A B to C, then Z ABD > ACD (97), and E 



Z.ACD>AEC, therefore much more ABD> 

 AED. //^^ 



A D 



100. Problem. To make a triangle, having its sides 

 equal to three given right lines, every one of them being 

 less than the sum of the other two. 



Take AB equal to one of the lines, and on the 

 centres A and B describe two circles with radii 



equal to tlie other two lines; draw AC and BC to 



A B 



the intersection C, and ABC will be the triangle re- 

 quired. -" 



101. Problem. At a given point in a right line, to 

 make an angle equal to a given angle. 



In the lines forming the given angle ABC, take 

 any two points, A and C, join AC, and taking 

 DE=BC, make the triangle DEF, having DF= 

 BA and FE=AC (100), then Z.FDE=ABC 

 (89). 



102. Theorem. If two triangles have two angles and 

 a side respectively equal, the whole triangles are equal. 



Let the equal sides be AB and CD, inter- 

 vening between the equal angles, then if on 

 AB a triangle equal to CDE be supposed to 

 be constructed, the points A and B, and the 

 angles at A and B being the same in this tri- 

 angle and in ABF, the sides must coincide both in position and in 

 length; therefore ABFzzCDE. 



If the equal sides are AF and CE, opposite to equal angles, then 

 ABziCD, and the whole triangles are equal. For if AB is not equal 



