OF SPACE. 



13 



Bdding ABE, the sum of all three is equal to ABE+EBC, 

 or to two right angles (94). 



109. Theorem. Right lines joining the 

 extremities of equal and parallel right lines, 

 are also equal and parallel. 



A j^ Let AB and CD be equal, and parallel. 



/ ^-""^ Then AC will be equal and parallel to 



L^:::__J BD. For, joining BC, aABC=BCD 



^' " (105), and the triangles ABC, DCB, are 



equal (86), and AC=DB; also aACB=:DBC, therefore 



AC||BD (104). 



110. Definition. A figure of which the 

 opposite sides are parallel, is called a paral- 

 lelogram . 



111. Definition. A straight line joining 

 the opposite angles of a parallelogram is 

 called its diagonal. 



112. Definition. A parallelogram, of 

 which the angles are right angles, is a rect- 

 angle. 



113. Definition. An equilateral rect- 

 angle is a square. 



1 14. Theorem. The diagonal of a pa- 

 rallelogram divides it into two equal triangles, 

 and its opposite sides are equal. 



^ B For ABC is equiangular with DCB 



\ ^-^ \ (105), and BC is common, therefore they 



Q- ^ are equal (102), and AB=CD,and AC=: 



BD. 



115. Theorem. Parallelograms on the 

 same base, and between the same parallels, 

 are equal. 



A B j^ P Since AB=CD, both being 



equal to EF, AC=BD(i5,or 16), 

 and the triangle AEC is equian- 

 gular (105) and equal (102) to 

 BFD j therefore deducting each 



of them from the figure AEFD, the remainder ED is equal 



to the remainder AF. 



116. TheoreSi. Parallelograms on equal 

 bases, and between the same parallels, are 

 equal. 



For each is equal to the paral- 

 lelogram formed by joining the 

 extremities of the base of the one, 

 and of the side opposite to the 

 base of the other (115). 



1 17. Theorem. Triangles on equal bases, 

 and between the same pan.llcls, are equal. 



Take AB and CD equal to the base 4 B C D 



EF or GH, and join BF and DM. 



Tiien F.B and GD are parallelograms 



between the same parallels (lOQ), 



and on equal bases, therefore they 



are equal (1 16), and their halves, the triangles AEF, CGVL 



(114), are also equal (1 8). 



118. Theorem. In anv right angled tri- 

 angle, (he square described on the hypote- 

 nuse is equal to the sum of the squares de- 

 scribed on the two other sides. 



Draw AB parallel to CD the side 

 of the square on the hypotenuse, then 

 the parallelogram CB is double any ' 

 triangle on the same base and be- 

 tween the same parallels (114, 117), 

 as ACD V but ACD=FCG, their 

 angles at C being each equal to ACG 

 increased by a right angle, FC to AC, 

 and GC to DC Again, GAH is a 

 right line (95), parallel to CF, therefore the triangle FCG 

 is half of the square CH on the same base, and CH:z:CB, 

 since they are the doubles of equal triangles. In the same 

 manner it maybe shown that GKr^GB ; therefore the 

 whole CDIG is equal to the sum of CH and GK. 



119. Problem. To find a common mea- 

 sure of any two quantities. 



Subtract the less continually from the greater, the re- 

 mainder from the less, the next remainder from the pre- 

 ceding one, as often as possible, and proceed till there be 

 no further remainder ; then the last remainder will be the. 

 common measure required. For since it measures the pre- 

 ceding remainder, it will measure the preceding quantities 

 in which that remainder was contained, and which, in- 

 creased at each step by the remainders, makes up the origi- 

 nal quantities. 



For example, if the numbers 54 and 21 be proposed, 

 J4 — 21 — 21 = 12, 21 — 12=9, 12 — 8=3, 9 — 3 — 3 — 3 



