« ForrigeFortsett »
Let ABC be a triangle having the angle ABC equal to the Book I. angle ACB; the side AB is also equal to the side AC.
For if AB be not equal to AC, one of them is greater than the other. let AB be the greater, and from it cut'off DB equal to AC,
a. 3. 1. the less, and join DC. therefore because in the triangles DBC, ACB, DB is equal to AC, and BC common to both, the two
D fides DB, BC are equal to the two AC, CB, each to each; and the angle DBC is equal to the angle ACB ; therefore the bafe DC is equal to the bafe AB, and the triangle DBC is equal to the triangle - ACB, the less to the greater ;
b. 4. I. which is absurd. Therefore AB is not unequal to AC, that is, it is equal to it. Wherefore if two angles, &c.
Q. E. D.
PRO P. VII. THEO R.
there cannot be two triangles that have their lides which are terminated in one extremity of the base equal to one another, and likewise those which are terminated in the other extremity.
If it be possible, let there be two triangles ACB, ADB upon the fame bafe AB, and upon the same side of it, which have their fides CA, DA, terminated in the extremity A of the base, equal to one another, and likewise their fides CB, DB that are terminated in B.
Join CD; then, in the case in which the Vertex of each of the triangles is without the other triangle, because AC is equal to AD, the angle ACD is equal to the angle ADC. but the angle ACD is greater than the angle BCD, thereforethe angle ADCis greater also than BCD; much more then ist
B the angle BDC greater than the angle BCD. again, because CB is equal to DB, the angle BDC is equal to the angle BCD, but it has been demonstrated to be greater than it; which is impossible.
a. s. I.
Book I. But if one of the Vertices, as D, be within the other triangle ACB ; produce AC, AD to E, F. therefore
E because AC is equal to AD in the triangle
one another; but the angle ECD is greater
B the angle BCD; but BDC has been proved to be greater than the fame 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 fame 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.
IF two triangles have two sides of the one equal to
two sides of the other, each to cach, and have likewise their bases cqual; the angle which is contained by the two sides of the one shall be cqual 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, AC
For if the trian-
F point B be on E, and the straight line BC upon EF; the point C Thall also coincide with the point F, because BC is equal to EF.
therefore BC coinciding with EF,BA and AC shall coincide with ED Book I. and DF. for if the base BC coincides with the bafe EF, but the sides BA, CA do not coincide with the fides ED, FD, but have a different situation,as EG, FG; then upon the fame base EF and upon the same Side of it there can be two triangles that have their fides which are terminated in one extremity of the base equal to one another, and likewise their fides terminated in the other extremity. but this is impossible“. therefore if the base BC coincides with the base EF, a. 7. s. the sides BA, AC cannot but coincide with the sides ED, DF; wherefore like wise the angle BAC coincides with the angle EDF, and is equal o to it. therefore if two triangles, &c. Q. E. D. b. 8. Ar.
PROP. IX. PROB.
Take any point D in AB, and from AC cut à off AE equal to 2. 3. 1.' AD; join DE, and upon it describe b
b. l. 1. an equilateral triangle DEF, then join AF. the straight line AF bifects the angle BAC.
Because AD is equal to AE, and AF D E 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 bafé B/
F EF; therefore the angle DAF is equal
to the angle EAF. wherefore the given rectilineal angle BAC c. 8. 1. is bifected by the straight line AF. Wnich was to be done.
To bisect a given finite straight line, that is, to di
vide Let AB be the given straight line ; it is required to divide it into two equal parts. Describe * upon it an equilateral triangle ABC, and bisect ba. 1. 1.
b. 9.1. the angle, ACB by the straight line CD. AB is cut into two equal parts in the point D.
Book 1. Because AC is equal to CB, and CD
common to the two triangles ACD,
fore the base AD is equal to the base C. 4. I.
DB, and the straight line AB is divided
b. I. 3.
PRO P. XI. PROB.
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. 2. 3. 1.
Take any point D in AC, and make * CE equal to CD, and
Because DC is equal to CE,
therefore the angle DCF is equal o to the angle ECF; and they c. 8. 1.
are adjacent angles. but when the adjacent angles which one
straight line makes with another straight line are equal to one d. 10. Def. another, each of them is called a right d angle; therefore each of 1. the angles DCF, 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 poflible, 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 line, the
a.so. Def. 1.
angle CBE is equal to the angle
E EBA; in the same manner, because ABD is a straight line, the angle DBE is equal to the angle EBA. wherefore the angle DBE is equal to the angle CBE, the less to the greater; which isimposible. therefore two straight lines cannot have
A B a common segment.
To , a
PRO P. XU. PRO B.
C. 1O. I.
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 perpendicular to AB from the point C. Take any point D upon the
C other side of AB, and from the center C, at the distance CD, describe b the circle EGF meet
b. 3. Polt. ing AB in F, G; and bifect
H + FG in H, and join CF,CH, CG. A F
B the straight line CH drawn from
D 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 to the base CG; therefore d. 15. Del.s. the angle CHF is equal to the angle CHG; and they are adjacente. 8. s. 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 pointCa perpendicular CH has been drawn to the given straight line AB. Which was to be done.
PRO P. XIII. THEO R.
another upon one side of it, are either two right . angles, or are together equal to two right angles.