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Book I. But if one of the vertices, as D, be within the other triangle ACB; produce AC, AD to E, F; there
ase CD are a 5. 1. equal a to one another, but the angle
ECD is greater than the angle BCD;
B greater than the same BCD; which is impossible. The case in which the vertex of one triangle is upon a side of the other, needs no demonstration.
Therefore upon the same base, and on the same side of it, there cannot be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise those which are terminated in the other extremity, Q. E. D.
PROP. VIII. THEOR.
IF two triangles have two sides of the one equal to two sides of the other, each to each, and have like. wise their bases equal; the angle which is contained by the two sides of the one shall be equal to the angle contained by the two sides equal to them, of the other.
Let ABC, DEF be two triangles, having the two sides AB,
For, if the tri-
F that the point B be on E, and the straight line BC upon EF: the point C shall also coincide with the point F. Because
BC is equal to EF; therefore BC coinciding with EF, BA and Book I. AC shall coincide with ED and DF; for, if the base BC coincides with the base EF, but the sides BA, CA do not coincide with the sides ED, FD, but have a different situation, as EG, FG; then, upon the same base EF, and upon the same side of it, there can be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise their sides terminated in the other extremity; but this is impossible a ; therefore, if the base BC coincides with the a 7. 1. base EF, the sides BA, AC cannot but coincide with the sides ED, DF; wherefore, likewise, the angle BAC coincides with the angle EDF, and is equal b to it. Therefore, if two triangles, b 8. Ax. &c.
Q. E. D.
TO bisect a given rectilineal angle, that is, to di. vide it into two equal angles.
Let BAC be the given rectilineal angle; it is required to bisect it.
Take any point D in AB, and from AC cut a off AE equal to a 3. 1. AD; join DE, and upon it describe b
b 1. 1. an equilateral triangle DEF; then join AF; the straight line AF bisects the angle BAC. Because AD is equal to AE, and
D AF is common to the, two triangles DAF, EAF; the two sides DA, AF are equal to the two sides EA, AF, each to each; and the base DF is equal to the base EF; therefore the
F angle DAF is equal < to the angle B
Cc 8.1 EAF; wherefore the given rectilineal angle BAC is bisected by the straight line AF. Which was to be done.
PROP. X. PROB.
TO bisect a given finite straight line, that is, to divide it into two equal parts.
Let AB be the given straight line ; it is required to divide it into two equal parts.
Describe a upon it an equilateral triangle ABC, and bisect b a 1. 1. the angle ACB by the straight line CD. AB is cut into two b 9. 1. equal parts in the point D.
TO draw a straight line at right angles to a given straight line, from a given point in the same.
See N. Let AB be a given straight line, and C a point given in it; it
is required to draw a straight line from the point C at right an.
gles to AB. a 3. 1. Take any point D in AC, and a make CE equal to CD, and bl. 1. upon DE describe b the equi
Because DC is equal to CE,
the base DF is equal to the base EF; therefore the angle DCF 8.1. is equal to the angle ECF; and they are adjacent angles.
But, when the adjacent angles which one straight line makes
with another straight line are equal to one another, each of them d10. Def. is called a right d angle; therefore each of the angles DCF, 1. ECF is a right angle. Wherefore, from the given point C, in
the given straight line AB, FC has been drawn at right angles to AB.
Which was to be done. Cor. By help of this problem, it may be demonstrated, that two straight lines cannot have a common segment.
If it be possible, let the two straight lines ABC, ABD have the segment AB common to both of them. From the point B draw BE at right angles to AB; and because ABC is a straight
PROP. XII. PROB. TO draw a straight line perpendicular to a given straight line of an unlimited length, from a given point without it.
Let AB be the given straight line, which may be produced to any length both ways, and let C be a point without it; it is required to draw a straight line
C perpendicular to AB from the point C.
Take any point D upon the other side of AB, and from the centre C, at the distance
H CD, describe b the circle EGF
b 3. Post. A F
GB meeting AB in F, G; and bi
D sectc FG in H, and join CF,
c 10. 1. CH, CG ; the straight line CH, drawn from the given point C, is perpendicular to the given straight line AB.
Because FH is equal to HG, and HC common to the two triangles FHC, GHC, the two sides FH, HC are equal to the two GH, HC, each to each ; and the base CF is equal d to the d15. Def. base CG; therefore the angle CHF is equal e to the angle CHG;
1. and they are adjacent angles ; but when a straight line standing e 8. 1. on a straight line makes the adjacent angles equal to one another, each of them is a right angle, and the straight line which stands upon the other is called a perpendicular to it; therefore from the given point C a perpendicular CH has been drawn to the given straight line AB. Which was to be done.
PROP. XIII. THEOR.
THE angles which one straight line makes with another upon the one side of it, are either two right angles, or are together equal to two right angles.
Book I. Let the straight line AB make with CD, upon one side of it,
the angles CBA, ABD; these are either two right angles, or are together equal to two right angles.
For, if the angle CBA be equal to ABD each of them is a a Def.10. right a angle; but if not, from the point B draw BE at right A
с b 11. 1. angles b to CD ; therefore the angles CBE, EBD are two right
angles a ; and because CBE is equal to the two angles CBA, ABE
together, add the angle EBD to each of these equals ; therec 2. Ax.
fore the angles CBE, EBD are equal to the three angles CBA, ABE, EBD. Again, because the angle DBA is equal to the two angles DBE, EBA, add to these equals the angle ABC; therefore the angles DBA, ABC are equal to the three angles DBE, EBA, ABC; but the angles CBE, EBD have been de
monstrated to be equal to the same three angles ; and things d 1. Ax. that are equal to the same are equal d to one another; therefore
the angles CBE, EBD are equal to the angles DBA, ABC; but CBE, EBD are two right angles; therefore DBA, ABC are together equal to two right angles. Wherefore, when a straight line, &c. Q. E. D.
PROP. XIV. THEOR.
IF, at a point in a straight line, two other straight lines, upon the opposite sides of it, make the adjacent angles together equal to two right angles, these two straight lines shall be in one and the same straight line.
At the point B, in the straight line AB, let the two straight lines BC, BD, upon the opposite sides of AB, make the adjacent angles ABC, ABD equal together to two right angles; BD is in the same straight line with CB.
For, if BD be not in the same straight line with CB, let BE be C