« ΠροηγούμενηΣυνέχεια »
But if one of the vertices, as D, be within the other triangle Book I. ACB; produce AC, AD to E, F; therefore, because AC is equal to AD in the
E triangle ACD, the angles ECD, FDC
F upon the other side of the base CD are equal a to one another, but the
a 5. 1. angle ECD is greater than the angle BCD; wherefore the angle FDC is likewise greater than BCD; much more then is the angle BDC greater than the angle BCD. Again, because CB is equal to DB, the angle BDCA is equal a to the angle BCD; but BDC has been proved to be 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 66
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 likewise their bases equal, the angle which is contained by the two sides of the one shall be equal to the angle contain. ed by the two sides equal to them of the other.
Let ABC, DEF be two triangles, having the two sides AB,
For, if the triangle ABC beapplied to DEF, so B
F that the point B be on E, and the straight line BC upon EF; the point C shall
Book I. also coincide with the point F; because BC is equal to EF;
therefore BC coinciding with EF, BA and 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 imposa 7. 1, sible a ; therefore, if the base BC coincides with the base EF,
the sides BA, AC cannot but coincide with the sides ED, DF;
wherefore likewise the angle BAC coincides with the angle b 8. Ax. EDF, and is equal o to it. Therefore w if two triangles,” &c. Q. E. D.
PROP. IX. PROB. To bisect a given rectilineal angle, that is, to divide it into two equal angles.
Let BAC be the given rectilineal angle, it is required to
bisect it. a 3. 1. Take any point D in AB, and from AC cut a off AE equal to AD; join DE, and upon it de
A bl. 1. scribe an equilateral triangle DEF;
then join AF; the straight line AF
the angle DAF is equal to the an-
PROP. X. PROB.
Let AB be the given straight line ; it is required to divide
it into two equal parts. a l. ). Describe a upon it an equilateral triangle ABC, and bisect b 9. 1. • the angle ACB by the straight line CD. AB is cut into
two 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.
Let AB be a given straight line, and C a point given in it; See N. it is required to draw a straight line from the point C at right angles to AB.
Take any point D in AC, and a make CE equal to CD, and a 3. 1. upon DE describe the equilateral triangle DFE, and join
F FC; the straight line FC drawn from the given point C is at right angles to the given straight line AB.
Because DC is equal to CE, and FC common to the two triangles DCF, ECF; the
AD С E B two sides DC, CF, are equal to the two EC, CF, each to each ; and the base DF is equal to the base EF; therefore the angle DCF is equal < to the c 8. 1. angle ECF; and they are adjacent angles.
djacent angles. But when the adjacent angles which one straight line makes with another straight line are equal to one another, each of them is called a right d angle; therefore each of the angles DCF, ECF, is a d 10. Def. right angle. Wherefore from the given point C, in the given 1. 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
Book I. line, the angle CBE is equal a to the angle EBA ; in the
E a 10. Def.
same manner, because ABD 1.
is a straight line, the angle
С have a common segment.
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
E centre C, at the distance CD, b 3. Post. describe b the circle EGF meeting AB in F, G; and bi
H c 10. 1. sect © FG in H, and join CF, AF
Because FH is equal to HG, and HC common to the two tri
angles FHC, GHC, the two sides FH, HC are equal to the d 15. Def. two GH, HC, each to each; and the base CF is equald to the
base CG; therefore the angle CHF is equal to the angle e 8. 1.
CHG; and they are adjacent angles; but when a straight line standing 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. Thie 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.
Let the straight line AB make with CD, upon one side of Book I. 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
C right a angle; but, if not, from the point B draw BE at right a Def. 10. angles to CD; therefore the angles CBE, EBD are two right b 11. 1. angles a ; and because CBE is equal to the two angles CBA, ABE together, add the angle EBD to each of these equals; therefore the angles CBE, EBD, are equal to the three angles c 2. Ax. 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 demonstrated to be equal to the same three angles; and things that are equal to the same are equal to one another; d 1. Ax. 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
А 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
E with CB. For, if BD be not in the
D same straight line with CB, let