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From the trapezium ABCF take A FDC, and from the
same trapezium take A EAB, and the remainders are equal,
(ax. 3) ABCD=OEBCF. Therefore, parallelograms on the same, &c.
Q. E. D.
[In the latter part, ax. 3 is used in rather a peculiar
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 inay 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 be os on=bases BC, FG, and between the same ||s AH, BG.
Then shall O ABCD=DEFGH.
Constr. Join BE, CH. Dem. Then :: BC =FG (hyp.), and FG = EH, (i. 34) .. BC=EH;
(ax. 1) and these lines are llı (hyp.) and joined towards the same parts by BE, CH;
.. BE, CH are both = and II; (i. 33) .. EBCH is a o, (def. 36) and it ABCD, on the same base and between the same ||s;
(i. 35) For the same reason,
EFGH = EBCH; ..O ABCD = =O EFGH. (ax. 1) Therefore, parallelograms on equal, &c. Q. E. D.
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 || AD, BC.
Then shall A ABC= A 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 || BC, EF. (i. 35) And :: the diameter AB bisects DEBCA; (i. 34)
.. ABC is half EBCA; also :: the diameter DC bisects ODBCF, (i. 34)
.. DBC is half ODBCF, but the halves of equal things are equal; (ax. 7)
.:. Δ ABC = Δ 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.
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 || BF, AD.
Then shall A ABC= A DEF.
Constr. Produce AD both ways to G, H; through B draw BG || CA, and through F draw FH
(i. 31) Dem. Then each of the figures GBCA, DEFH is O;
(def.) and they are equal, they are on equal bases BC, EF, and between the same Is BF, GH.
(i. 36) And :: the diameter AB bisects o GBCA, (i. 34)
.: A ABC is half o GBCA; also, :: the diameter DF bisects O DEFH, (i. 34)
..) DEF is half O DEFH; but the halves of equal things are equal; (ax. 7)
.:. Δ ABC=Δ 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 triangles on the same base and on opposite sides of it. If CD meets AB in E, then CE, ED are also equal.
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 Is.
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 A ABC= A EBC, :;: they are on the same
base, and between the same || BC, AE; but a ABC=JDBC; (hyp.) .. also A DBC=A 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.