GEOMETRY. 



(2.) 7^i? exterior angle is equal to the interior 

 and remote angle. 



(3.) The two interior angles are together equal 

 to two right angles. 



(i.) Let AB and CD be two parallel lines inter- 

 sected by the line EG; it is required to prove 

 that the angle AFG is 

 equal to the alternate 

 angle FGD. If these 

 angles are not equal, at 

 the point F in the line 

 FG, make the angle HFG 

 equal to the angle FGD. 

 Then since these angles 

 are equal, the two lines 

 HK, CD are parallel ; but AB is parallel to CD 

 by hypothesis, therefore AB and HK, two inter- 

 secting lines, are parallel to the same line, which 

 is absurd ; therefore the angle AFG must be 

 equal to the angle FGD. 



(2.) Since AFG is equal to FGD, and AFG 

 also equal to EFB ; therefore EFB is equal to 

 FGD. 



(3.) To each of the equal angles AFG, FGD 

 add the angle BFG, then we have the two angles 

 AFG, BFG equal to the two BFG, FGD ; but 

 AFG, BFG make two right angles, therefore 

 BFG, FGD also make two right angles. 



Cor. It follows at once that lines which are 

 parallel to the same line are parallel to each 

 other; for if they are not parallel, they must 

 meet. Then through the same point we shall 

 have two lines drawn parallel to the same line, 

 which is absurd. 



THE PARALLELOGRAM. 



PROP. X. In a parallelogram : 



(i.) The opposite sides are equal. 



(2.) The opposite angles are eq^lal. 



(3.) Each diagonal bisects the figure. 



Let ABCD be a parallelogram ; it is required 

 to prove that its opposite sides AB and CD are 

 equal, also that AD and 

 BC are equal ; the angle 

 A equal to the angle C, 

 the angle B equal to the 

 angle D, and the triangle 

 ABC equal to the triangle 



ADC. In the two triangles ABC and ADC we 

 have two angles BAC, ACB in the one equal 

 respectively to the two ACD, DAC in the other 

 (Prop. IX.), and the side AC common to both; 

 therefore (Prop. VII. 4) the triangles are equal 

 in every respect that is, AB is equal to DC, 

 BC to AD, the angle B equal to the angle D, 

 and the triangle ABC equal in area to ADC. 



Cor. If through a point P in the diagonal of a 

 parallelogram lines be drawn parallel to the sides, 

 of the four parallelo- 

 grams thus formed, those 

 through which the dia- 

 gonal does not pass, and 

 which are called the com- 

 plements, are equal. 



Since the triangle ABC 



is equal to ADC, ASP to AQP, PVC to PTC ; 

 therefore SV equals PB. 



Conversely, a four-sided figure is a parallel- 

 ogram : 



. 

 / 

 / 



(r.) If the opposite sides are equal. 

 (2.) If the opposite angles are equal. 

 (3.) If two opposite sides are both equal and 

 Parallel. 



(i.) If AB equals CD, and AD equals BC 

 Join AC. Then we have two triangles ABC, ADC 

 having the three sides of 

 the one equal to the three 

 sides of the other, and 

 therefore (Prop. VII. 3) 

 their angles are equal 

 namely, angle BAC equal 

 to angle ACD ; but these 

 are alternate angles, therefore (Prop. IX. i) AB 

 is parallel to CD ; and the angle ACB equals the 

 angle CAD, therefore AD is parallel to CB. The 

 figure is consequently a parallelogram. 



(2.) If the opposite angles are equal namely, 

 angle BAD equal to angle BCD, and angle ABC 

 equal to angle ADC. Since 



the four angles of the figure A B 



together make up four right 7~ / 



angles, angle B and angle / / 



C together make two right / / 



angles ; therefore (Prop. IX. 



3) AB and CD are parallel. 



Again, angle D and angle C together make two 

 right angles, therefore AD and BC are parallel ; 

 therefore ABCD is a parallelogram. 



(3.) The learner will readily demonstrate this 

 for himself. 



PROP. XI. Parallelograms are equal in area: 



(i.) If they are on the same base, and between 

 the same parallels. 



(2.) If they are on equal bases, and between the 

 same parallels. 



(i.) Let ABCD be a D c 



parallelogram ; it is / / 



equal in area to another / / 



which stands on the A Z / 



same base AB and 



between the same par- E p c 



allels. 



First, Let the two 

 parallelograms ABCD, 

 ABDE stand as in the 



iigure. Then each of 



them is double of the triangle ABD, and con- 

 sequently equal to each other. 



Second, Let ABCD, ABEF stand as in the 

 figure, then (Prop. VII. i) the triangles EEC, AFD 

 are equal in area. 

 Take away the 

 common part 

 EOD, and there 

 remains the four- 

 s ided figure 



DOBC equal to the four-sided figure EOAF. 

 Add AOB to each ; therefore the parallelogram 

 ABCD is equal to the parallelogram ABEF. 



Third, Let ABCD, 

 ABEF stand as in the 

 figure, then (Prop. VII. i) 

 the triangles EEC, ADF 

 are equal in area. Add 

 o each the four-sided 

 figure ABED, and we have ABCD equal to ABEF. 



(2.) The parallelograms ABGH, CDEF are 

 equal, if they stand upon the equal bases AB and 



615 



