CHAMBERS'S INFORMATION FOR THE PEOPLE. 



The converse of Prop. IV. namely, that, if two 

 angles of a triangle be equal to each other, the sides 

 opposite to them are also equal \s proved by sup- 

 posing the triangle turned over and laid on its 

 former position. The learner should try to work 

 out the demonstration for himself. 



Cor. If a triangle be equiangular, it is also 

 equilateral. 



PROP. V. If two sides of a triangle be unequal, 

 the greater side has the greater angle opposite to it, 

 or the less side has the less angle opposite to it. 



Let ABC be a triangle, having the side AB 

 greater than the side AC ; 

 it is required to prove that 

 the angle ACB, which is 

 opposite to AB, is greater 

 than the angle ABC, which 

 is opposite to AC. From 

 AB cut off the part AD 

 equal to AC. Join DC. 

 Then in the isosceles tri- 

 angle ADC we have the angle ACD equal to the 

 angle ADC ; but the exterior angle ADC is greater 

 than the angle at B, therefore the angle ACD is 

 also greater than the angle at B ; much greater, 

 then, will the angle ACB be than the angle ABC. 



The converse of Prop. V. is, that, if two angles 

 of a triangle be unequal, the greater angle has the 

 greater side opposite to it. This is proved by 

 shewing that a contradiction would follow from 

 supposing it not to be true. 



PROP. VI. Any two sides of a triangle taken 

 together are greater than the third side. 



Let ABC be a triangle ; it is required to prove 

 that the two sides AB and AC taken together are 

 greater than the third 

 side BC. Produce 

 AB through A to D, 

 making AD equal to 

 AC. Join DC. Then 

 since the triangle 

 ACD is isosceles, we B - 

 have the angle ACD 

 equal to the angle ADC, therefore the whole angle 

 BCD is greater than the angle at D ; therefore 

 the side BD is greater than the side BC. But 

 BD is equal to BA and AC together, therefore 

 BA and AC together are greater than BC. 



EQUALITY OF TRIANGLES. 



PROP. VII. Two triangles are equal in every 

 respect : 



(i.) If two sides and the angle between them in 

 the one triangle be respectively equal to two sides 

 and the angle between them in the other. 



(2.) If two angles and the side between them 

 in the one triangle be respectively equal to two 

 angles and the side between them in the other. 



(3.) If the three sides of the one be equal to the 

 three sides of the other. 



(4.) If two angles of the one be equal to two 

 angles of the other, and a side opposite one of 

 these angles in the one triangle be equal to a side 

 opposite the equal angle in the other. 



(5.) If two sides of the one be equal to two 

 sides of the other, and an angle opposite one of 

 these sides in the one triangle be equal to an 

 angle opposite to the equal side in the other, the 



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angles opposite the other equal sides being either 

 both acute or both obtuse. 



(i.) Let ABC, DEF be two triangles, having the 

 side AB in the one equal 

 to the side DE in the other, 

 and the side AC equal to 

 DF, and the angle at A 

 equal to the angle at D ; 

 it is required to prove that 

 the two triangles are equal 

 in every respect. For if 

 the triangle ABC be ap- 

 plied to the triangle DEF, so that A lies on D, 

 and so that AB lies along DE, then B shall lie on 

 E, since AB is equal to DE; and AB being on 

 DE, AC shall be on DF, since the angle at A 

 is equal to the angle at D ; and the point C shall 

 lie on F, since AC is equal to DF. But B lies on 

 E, and C on F, therefore the line BC shall lie on 

 EF, and be equal to it, and the one triangle shall 

 exactly coincide with the other, and be equal to it 

 in every respect ; that is, the two triangles shall be 

 equal in area, and the angle at B shall be equal to 

 the angle at E, and the angle at C to the angle at F. 



(2.) Let ABC, DEF be two triangles, having 

 the angle B equal to the angle E, and the angle 

 C equal to the angle F, and the side BC equal to 

 the side EF ; it is required to prove that the 

 triangles are equal in 

 every respect. For if 

 the triangle ABC be 

 applied to DEF, so 

 that B lies on E, and 

 BC on EF, then the 

 point C shall lie on F ; 

 and BC lying along 

 EF, BA shall lie along 

 ED, since the angle B is equal to the angle E ; 

 and CA shall lie on FD, since the angle C is equal 

 to the angle F ; and the point A, which lies both 

 on AB and on AC, shall lie both on DE and on 

 DF; that is, it must lie on D, the only point 

 common to both. Hence the triangles are equal 

 in every respect. 



(3.) Let ABC and DEF be two triangles, hav- 

 ing the three sides of the one equal respectively to 

 the three sides of the other ; it is required to prove 

 that the triangles are equal in every respect. For 

 if the triangle ABC be applied to DEF, so that B 

 shall lie on E, and BC on EF, then C shall lie on F, 

 since BC is equal to EF, and let the triangle ABC 

 take the position GEF. Join DG. Since ED 



equals EG, the angle EGD equals the angle EDG 

 (Prop. IV.) ; and since FD equals FG, the angle 

 FGD equals the angle FDG, therefore the whole 

 angle EGF equals the whole angle EDF ; that is, 

 the angle BAG is equal to the angle EDF ; then 

 in the two triangles ABC, DEF, we have two 

 sides in the one equal to two sides in the other, and 

 the contained angles equal ; therefore, by the first 



