From the trapezium ABCF take ▲ FDC, and from the same trapezium take ▲ EAB, and the remainders are equal, that is, ABCD= EBCF. (ax. 3) Therefore, parallelograms on the same, &c. Q. E. D. [In the latter part, ax. 3 is used in rather a peculiar manner. It will however be clear to the beginner that if we cut off a certain portion from a carpet and an equal portion from an equal carpet that the portions which remain are equal. Note carefully that the parallelograms are equal in area, but are not equal in all respects, nor identical in figure. The area of a four-sided field may be equal to that of a triangular field, and the fields may be said to be equal, but their figures are not equal. The word equivalent would better express equality of area, leaving the word equal to express equality of area and of figure also. Thus the parallelograms in this prop. are equivalent: the triangles in prop. 4 are equal. But the fact is there are few cases in which any ambiguity can arise, and a little care is all that is wanted.] Describe a parallelogram equal to a given square, and having an angle equal to half a right angle. PROPOSITION XXXVI. THEOREM. Parallelograms on equal bases, and between the same parallels, are equal to one another. Let ABCD, EFGH bes on = bases BC, FG, and between the same ||s AH, BG. = Dem. Then BC FG (hyp.), and FG EH, (i. 34) • : and these lines are ||, (hyp.) and joined towards the same parts by BE, CH; .. BE, CH are both = and ||; (i. 33) ABCD, on the same base and between the PROPOSITION XXXVII. THEOREM. Triangles on the same base, and between the same parallels, are equal to one another. Let the As ABC, DBC be on the same base BC, and between the same ||s AD, BC. Then shall ▲ ABC = ^ DBC. Constr. Produce AD both ways to E, F; through B draw BE || CA, and through C draw CF || BD. Dem. Then each of the figures EBCA, DBCF is and EBCA = DBCF, ... they are on the same base BC, and between the same ||s BC, EF. ; And (i. 35) the diameter AB bisects EBCA; (i. 34) .. AABC is half EBCA; also the diameter DC bisects DBCF, (i. 34) but the halves of equal things are equal; (ax. 7) .. A ABC = A DBC. Therefore, triangles, &c. Q.E.D. 1. Bisect a triangle by a line drawn through a given point in one of the sides. 2. Given an isosceles triangle and a point in one of its equal sides, draw a line from the point to the opposite side produced, which shall make with these sides a triangle equal to the given triangle. E PROPOSITION XXXVIII. THEOREM. Triangles on equal bases, and between the same parallels, are equal to one another. Let As ABC, DEF be on = bases BC, EF, and between the same ||s BF, AD. Then shall ▲ ABC = A DEF. Constr. Produce AD both ways to G, H; through B draw BG || CA, and through F draw FH (i. 31) || ED. Dem. Then each of the figures GBCA, DEFH is ; (def.) and they are equal, . they are on equal bases BC, EF, and between the same ||s BF, GH, And the diameter AB bisects (i. 36) GBCA, (i. 34) .. ▲ ABC is half — GBCA; also, the diameter DF bisects DEFH, (i. 34) but the halves of equal things are equal; (ax. 7) .. AABC = A DEF. Therefore, triangles, &c. Q. E. D. [The bases of the triangles are of course in the same straight line.] ABC, ABD are two equal base and on opposite sides of it. then CE, ED are also equal. triangles on the same If CD meets AB in E, PROPOSITION XXXIX. THEOREM. Equal triangles on the same base and on the same side of it are between the same parallels. Let the equal As ABC, DBC be on the same base BC, and on the same side of it. Then the As ABC, DBC shall be between the same ||s. Constr. Join AD; AD shall be || BC. For, if it is not, through A draw AE || BC, (i. 31) meeting BD or BD produced in E, and join EC. Dem. Then ▲ ABC: = AEBC, they are on the same base, and between the same ||s BC, AE; (i. 37) but ABC = DBC; (hyp.) .. also ▲ DBC= ^ EBC the greater the less, which is impossible: = therefore AE is not parallel to BC. In the same way it can be shown, that no other line through A but AD is || BC; .. AD is || BC. Therefore, equal triangles, &c. Q. E. D. |