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a. 2. Cor. 19. I.
Book 1. to each of its angles. And, by the preceeding proposition, all
the angles of these triangles are equal to twice as many right angles as there are triangles, that is, as there are sides of the figure; and the same angles are equal to the angles of the figure, together with the angles at the point F, which is the common vertex of the triangles; that is, together with four right angles. Therefore all the angles of the figure, together with four right angles, are equal to twice as many right angles as the figure has fides.
Cor, 2. All the exterior angles of any rectilineal figure, are together equal to four right angles.
Because every interior angle
ABC,'with its adjacent exterior b 13. 1. ABD, is equal b to two right
angles; therefore all the interi-
equal and parallel straight lines, towards the fame
Join BC; and because AB is pa-
D the alternate angles ABC, BCD are equal“; and becaule AB is equal to CD, and BC common to the two triangles ABC, DCB, the two fides AB, BC are e. qual to the two DC, CB; and the angle ABC is equal to the angle ECD; therefore the base AC is equal to the base BD, and the triangle ABC to the triangle BCD, and the other angles
2 29. I.
b 4. I.
to the other angles , each to each, to which the equal fides are
equal to one another, and the diameter bisects them, that is, divides them in two equal parts.
N. B. A parallelogram is a four sided figure, of which the opposite sides are parallel ; and the diameter is the straight line joining two of its opposite angles.
Let ACDB be a parallelogram, of which BC is a diameter; the opposite sides and angles of the figure are equal to one another; and the diameter BC bifects it. Because AB is parallel to CD,
B and BC meets them, the alternate angles ABC, BCD are equal · to one another; and be
a 29. 1 caule AC is parallel to BD, and BC meets them, the alternate angles ACB, CBD, are equal a C
D to one another; wherefore the two triangles ABC, CBD have two angles ABC, BCA in one, equal to two angles BCD, CBD in the other, each to each, and one side BC common to the two triangles, which is adjacent to their equal angles; there. fore their other fides shall be equal, each to each, and the third angle of the one to the third angle of the other b, viz. b 26, I. the fide AB to the side CD, and AC to BD, and the angle BAC equal to the angle BDC: And because the angle ABC is equal to the angle BCD, and the angle CBD to the angle ACB; the whole angle ABD is equal to the whole angle ACD: And the angle BAC has been shown to be equal to the angle BDC; therefore the opposite sides and angles of parallelograms are equal to one another ; also, their diameter bisects them; for
l AB being equal to CD, and BC common, the two AB, BC are equal to the two DC, CB, each to each; and the angle ABC
Book 1. is equal to the angle BCD; therefore the triangle ABC is e mqual to the triangle BCD, and the diameter BC divides the
parallelogram ACDB into two equal parts. Q. E. D.
ARALLELOGRAMS upon the same base and between the same parallels, are equal to one another.
a 34. I.
See the ad Let the parallelograms ABCD, EBCF, be upon the same and 3d fi- base BC, and between the fame parallels AF, BC; the parallegures.
logram ABCD Mall be equal to the parallelogram EBCF.
If the fides AD, DF of the pa.
But, if the fides AD, EF, opposite B
А. DE F A E DF
b 1. Ax.
c 2. or 3. Ax.
d 29. 1.
e 4: I.
f 3. Ax.
ARALLELOGRAMS upon equal bases and between the
Let ABCDEFGH be
DE parallelograms upon e
H qual bases BC, FG, and between the same parallels AH, BG; the paral. lelogram ABCD is equal to EFGH.
Join BE, CH; and because BC is equal to
C F G FG, and FG to * EH, BC is equal to EH; and they are pa- a 34. I. rallels, and joined towards the same parts by the straight lines BE, CH : But straight lines which join equal and parallel straight lines towards the same parts, are themselves equal and parallel 5; therefore EB, CH are both equal and parallel, and 5 33. . EBCH is a parallelogram; and it is equal to ABCD, because c 35.1. it is upon the same bafe BC, and between the fame parallels BC, AD: For the like reason, the parallelogram EFGÀ is equal to the same EBCH: Therefore also the parallelogram ABCD is equal to EFGH. Wherefore parallelograms, &c. Q. E. D.
RIANGLES upon the same base, and between the
fame parallels, are equal to one another.
Let the triangles ABC, DBC be upon the same base BC and between the fame parallels
E А D F
Produce AD both ways to the points E, F, and thro' Bdraw - BE parallel to CA;
a 31. 8, and thro' C draw CF paral
B Icl to BD: Therefore each
C of the figures EBCA, DBCF is a parallelogram ; and EBCA is equal o to DBCF, because they are upon the same base BC, and b 35. I. between the same parallels BC, EF; and the triangle ABC is
Book 1. the half of the parallelogram EBCA, because the diameter AB
bisects Cit; and the triangle DBC is the half of the parallelo
gram DBCF, because the diameter DC bifects it. But the d 7. Ax.
halves of equal things are equald; therefore the triangle ABC is equal to the triangle DBC. Wherefore triangles, &c. Q. E. D.
C 34. I.
a 31. I.
PRO P. XXXVII. THEOR. TRIANGLEs upon equal bases, and between the fame
parallels, are equal to one another. Let the triangles ABC, DEF be upon equal bases BC, EF, and between the fame parallels BF, AD: The triangle ABC is equal to the triangle DEF.
Produce AD both ways to the points G, H, and through B draw BG parallela to CA, and through F draw FH parallel to ED: Then each of
G A D Η.
b 36. I.
C 34. 1.
PRO P. XXXIX. THE O R.
lame side of it, are between the fame parallels.
Join AD; AD is parallel to BC; for, if it is not, through the point A draw. AE parallel to BC, and join EC : The tri