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E to DF; and also the base BC equal to the base EF. The angle BAC is equal to the angle EDF.
For, if the triangle ABC be applied to the triangle DEF, so 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 AC shall coincide with ED, and DF; for, if BA, and CA do not coincide with ED, and FD,
but have a different situation as EG and FG; then, upon the same base EF, and upon the same side of it, there can be two triangles EDF. EGF, 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 (7. 1.); 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 EDF, and is equal (8. Ax.) to it. Therefore 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. • Take any point Din AB, and from AC cut (3. 1.) off AE equal to AD; join DE, and upon it describe
A (1. 1.) an equilateral Triangle DEF; then join AF; the straight line AF bisects the angle BAC.
Because AD is equal to AE, and 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; but the base DF is also equal to the base EF; therefore the angle DAF i. equal (8.1.) to the angle EAF: wherefore the given rectilineal angle BAC is bisected by the straight line AF. Which was to be B 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 (1. 1.) upon it an equilateral triangle ABC, and bisect (9. 1.) the angle ACB by the straight line CD. AB is cut into two equal parts in the point D.
Becayse AC is equal to CB, and CD common to the two triangles ACD, BCD: the two sides AC, CD, are equal to the two BC, CD, each to each ;. but the angle ACD is also equal to the angle BCD, therefore the base AD is equal to the base (4. 1.) DB, and the straight line AB is divided into two eqnal parts in the point D. Which was to be done.
PROP. XI. PROB. To draw a straight line at right angles to a given straight
line, from a given point in that line, Let AB be a giyen straight line, and C a point given in it; it is required to draw a straight line from the point C at right angles to AB.
Take any point D in AC, and (8. 1.) make CE equal to CD, and upon DE describe (1. 1.) the equilateral triangle DFE, and join FC; the straight line FC, drawn from the given point C, is at right angles to the given straight line AR,
Because DC is equal to CE, and FC common to the two tri. angles DCF, ECF, the two sides DČ, CF are equal to the two EC, CF, each to each; but the base: DF is also equal to the base EF; therefore the angle DCF is equal (8.1.) 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 is called a right (7. def.) angle; therefore each of 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 dope: .
f 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 a 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 perpendi
С cular to AB fron the point C. Take any point D
the other side of Å B, and from the centre.C, at the distance CD, describe (3. Post.) the circle EGF meeting AB in F, G; and А
B bisect (10. 1.) FG in H, and join CF, CH, CG; the straight
D 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; but the base CF is also equal (11. Def. 1.) to the base CG; therefore the angle CHF is equal (8. 1.) to the angle CHG; and they are adjacent angles; now 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 live AB. Which was to be done,
PROP. XIII. THEOR. The angles which one straight line makes with another upon 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 it the angles CBA, ABD; these are either two right angles, or are together equal to two right angles.
For, if the angle ČBA be equal to ABD, each of them is a right angle (Def. 7.); but, if not, from the point B draw BE at right anz
gles (11. 1.) to Ch; therefore the angles CBF, EBD are two right angles. Now, the angle CBE is equal to the two angles.CBA, ABE together; add the angle EBD to each of these equals, and the two angles CBE, EBD, will be equal (2. Ax.) to the three CBA, ABE, EBD. Again, the angle DBA is equal to the two angles DBE, EBA; add to each of these equals the angle ABC; then will the two angles DBA, ABC be 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 (1. Ax.) 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.
B therefore, because the straight line AB makes angles with the straight line CBE, upon one side of it, the angles ABC, ABE are together equal (13. 1.) to two right angles; but the angles ABC, ABD are likewise together equal to two right angles; therefore the angles CBA, ABE are equal to the angles CBA, ABD: Take away the common angle ABC, and the remaining angle ABE is equal (3. Ax.) to the remaining angle ABD, the less to the greater, which is impossible; therefore BE is not in the same straight line with BC. And in like manner, it may be demonstrated, that no other can be in the same straight line with it but BD, which therefore is in the same straight line with CB. Wherefore, if at a point, &c. Q. E. D.
} PROP. XV. THEOR. If two straight lines cut one another, the vertical, or op
posite angles are equal. Let the two straight lines AB, CD cut one another in the point E; the angle AEC shall be equal to the angle DEB, and CEB to AED.
For the angles CEA, AED, which the straight line AE makes with the straight line CD, are together equal (13. 1.) to two right angles; and the angles AED, DEB,
C which the straight line DE makes with the straight live AB, are also together equal (13. 1.) to two right angles ;
B therefore the two angles CEA, AED are equal to the two AED, DEB. Take a
common angle AED, and the remaining ang'e CEA is equal (3. Ax.) to the remaining angle DEB. In the same manner it may be demonstrated that the angles CEB, AED are equal. Therefore, if two straight lines, &c. Q E. D.
COR. 1. From this it is manifests that if two straight lines cut one another, the angles which they make at the point of their intersection, are together equal to four right angles.
Cor. 2. And hence, all the angles made by any number of straight lines meeting in one point, are together equal to four right angles.
PROP. XVI. THEOR. If one side of a triangle be produced, the exterior angle is greater than either of the interior, and opposite angles.
Let ABC be a triangle, and let its side BC be produced to D, the exterior angle ACD is greater than either of the interior opposite angles CBX, BAC.
Bisect (10. 1.) AC in E, join BE and produce it to F, and make EF equal to BE; join also FC, and produce AC to G.
Because AE is equal to EC, and BE to EF; AE, EB are equal to CE, EF, each to each; and the angle AEB is equal (15. 1.) to the angle CEF, because they are vertical angles; therefore the base AB is equal (4. 1.) to the base CF, and the triangle AEB to the triangle CEF, and the remaining angles to the remaining angles, each to each, to which the equal sides are opposite; wherefore the angle BAE is equal to the angle ECF; but the angle ECD is greater than the angle ECF; therefore the angle ECO, that is ACD, is greater than BAE: In the same manner, if the side