PROP. VII. (THEOREM.)-Upon the same base (AB), and on the same side of it, there cannot be two triangles (ACB, ABD) having their sides (CA, DA) terminated in one extremity of the base (A), equal to one another, and likewise those (CB, DB) terminated in the other extremity (B).

C D

Join CD. First, let the vertex of each triangle be without the other triangle. Because AC is equal (Hyp.) to AD in the triangle ACD, the angle ACD is equal (I. 5) to the angle ADC. But the angle ACD is greater (Ax. 9) than the angle BCD. Therefore the angle ADC is also greater than the angle BCD. Much more, then, is the angle BDC greater than the angle BCD. Again, because CB is equal (Hyp.) to DB, the angle BDC is equal (1.5) to the angle BCD. demonstrated that the angle BDC is greater than the angle BCD. Therefore, the angle BDC is both equal to, and greater than the angle BCD; which is impossible.

A

B

But it has been

E

Secondly, let the vertex of one of the triangles be within the other triangle. Produce AC and AD to E and F. Because AC is equal (Hyp.) to AD in the triangle ACD, the angles ECD, FDC upon the other side of the base CD are equal (I. 5) to one another. But the angle ECD is greater (4x. 9) than the angle BCD. Therefore 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 (Hyp.) to DB, the angle BDC is equal (1.5) to the angle BCD. But it has been proved that the angle BDC is greater than the angle BCD. Therefore, the angle BDC is both equal to, and greater than the angle BCD, which is impossible

A

The third 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, &c. Q. E. D.

PROP. VIII. (THEOREM.)—If two triangles (ABC, DEF) have two sides (AB, AC) of the one equal to two sides (D E, DF) of the other each to each (viz., A B to D E, and AC to DF) and have likewise their bases (BC, EF) equal; the angle (BAC) which is contained by the two sides of the one is equal to the angle (EDF) contained by the two sides equal to them, of the other.

For, if the triangle ABC be applied to the triangle DEF, so that the A point B shall be on E, and the base BC upon the base EF; then, the point C shall coincide with the point F, because BC is equal (Hyp.) to EF. And BC coinciding with EF, BA and AC shall coincide with ED and DF. For, if the base BC coincides with the base

D G

B

CE

EF, and the sides BA, AC do not coincide with the sides ED, df, but have a different situation as EG, GF; then, upon the same base E F, and upon the same side of it, there can be two triangles having their sides terminated in one extremity of the base equal to one another, and likewise those terminated in the other extremity. But this is (I. 7) impossible. Wherefore, if the base BC coincides with the base EF, the sides BA, AC cannot but coincide with the sides ED, DF. Therefore, the angle BAC coincides with the angle EDF, and is equal (Ax. 8) to it. Therefore if two triangles, &c. Q. E. D.

PROP. IX. (PROBLEM.)—To bisect a given rectilineal angle (BAC); that is, to divide it into two equal angles.

Take any point D in AB, and from AC cut off AE (I. 3) equal to AD. Join DE. Upon DE opposite to the triangle DAE, describe (I. 1) an equilateral triangle DEF. Join AF.

The straight line AF bisects the angle BAC.

Because AD is equal (Const.) 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; and the base DF is equal (Const.) to the base EF. Therefore the angle DAF is equal (I. 8) to the angle EAF. Wherefore the given rectilinear angle BAC is bisected by the straight line AF. Q. E. F.

A

F

B

E

PROP. X. (PROBLEM.)-To bisect a given finite straight line (AB); that is, to divide it into two equal parts.

Describe upon AB (I. 1) an equilateral triangle ABC, and bisect (1.9) the angle ACB by the straight line CD, meeting AB in the point D. Then, AB is divided into two equal parts

at the point D.

Because AC is equal (Const.) to CB, and CD common to the two triangles ACD, BCD, the two sides AC, CD, are equal to the two sides BC, CD, each to each; and the angle ACD is equal (Const.) to the angle BCD. Therefore the base AD is equal to the base (I. 4) DB, and the straight line AB is divided into two equal parts at the point D. Q. E. F.

A D

PROP. XI. (PROBLEM.)-To draw a straight line at right angles to a given straight line (AB), from a given point (C) in the same.

Take any point D in AC, and make (I. 3) CE equal to CD. Upon DE describe (I. 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 AB.

E B

Because DC is equal (Const.) to CE, and FC common to the two triangles DCF, ECF, the two sides DC, CF, are equal to the two sides EC, CF, each to each; and the base DF is equal (Const.) to the base EF. Therefore the angle DCF is equal (I. 8) to AD the angle ECF: and they are adjacent angles. But when the adjacent angles which one straight line inakes with another straight line are equal to one another, each of them is called a right (Def. 10) angle. Therefore each of the angles DCF, ECF, is a right angle. Wherefore, from the given point C, in the given straight line AB, a straight line FC has been drawn at right angles to AB. Q. E. F.

Cor.-Two straight lines cannot have a common segment; that is, they cannot coincide in part without coinciding altogether.

If it be possible, let the two straight lines ABC, ABD, have the segment AB common to both of them.

E

From the point B draw (I. 11) BE at right angles to AB. Because ABC is a straight line, the angle CBE is equal (Def. 10) to the angle EBA. Because ABD is a straight line, the angle DBE is equal to the angle EBA. Therefore (4x. 1) the angle DBE is equal to the angle CBE, the less to the greater, which is impossible. Therefore two straight lines cannot have a common segment. Q. E. D.

A

B

D

PROP. XII. (PROBLEM.)-To draw a straight line perpendicular to a given straight line (AB) of unlimited length, from a given point (C) without it. Take any point D upon the other side of AB, and from the centre C, at the distance CD, describe (Post. 3) the circle EGF meeting AB in F and G. Bisect (I. 10) FG in H,

and join CH. The straight line CH, drawn from the given point C, is perpendicular to the given straight line AB.

A F

H

D

Join CF, CG. Because FH is equal (Const.) to HG, and HC common to the two triangles FHC, GHC, the two sides FH, HC, are equal to the two sides GH, HC, each to each; and the base CF is equal (Def. 15) to the base CG. Therefore the angle CHF is equal (I. 8) to the angle CHG; and they are adjacent angles. But when a straight line standing on another 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 (Def. 10) to it. Therefore from the given point C a perpendicular CH has been drawn to the given straight line AB. Q. Ë. D.

PROP. XIII. (THEOREM.)-The angles (CBA, ABD) which one straight line (AB) makes with another (CD) upon one side of it, are either two right angles, or are together equal to two right angles.

For if the angle CBA be equal to the angle ABD, each of them

A.

E

A

B

is a right angle (Def. 10). But if the angles CBA, ABD, be unequal, from the point B draw BE at right angles (I. 11) to CD. Therefore, the angles CBE, EBD, (Def 10), are two right angles. But the angle

CBE is equal to the two angles D
CBA, ABE, together. To

each of these equals add the

angle EBD. Therefore the two angles CBE, EBD, are equal (Ax. 2) to the three angles CBA, ABE, EBD. Again, the angle DBA is equal to the two angles DBE, EBA. To each of these equals add the angle ABC. Therefore the two angles DBA, ABC, are equal (4x. 2) to the three angles DBE, EBA, ABC. But the two angles CBE, EBD, have been proved to be equal to the same three angles. And things that are equal to the same thing are equal (4x. 1) to one another. Therefore the two angles CBE, EBD, are equal to the two angles DBA, ABC. But the two angles CBE, EBD, are two right angles. Therefore the two angles DBA, ABC, are together equal (4x. 1) to two right angles. Wherefore, the angles which one straight line, &c. Q. E. D.

PROP. XIV. (THEOREM).-If, at a point (B) in a straight line (AB), two other straight lines (BC, BD), upon the opposite sides of it (A B), make the two adjacent angles (ABC, ABD) together equal to two right angles, these two straight lines (BD, CB) are in one and the same straight line.

For, if BD be not in the same straight line with CB, let BE be in the same straight line with it. Because the straight line AB makes with the straight line CBE, upon one side of it, the two angles ABC, ABE, these two angles are together equal (I. 13) to two right angles. But the two angles ABC, ABD, are likewise together equal (Hyp.) to two right angles. Therefore the twoangles CBA, ABE, are equal (4x. 1) to the two angles CBA, ABD. From each of these equals, take away the common angle ABC. Therefore the remaining angle ABE is equal (4x. 3) to the remaining angle ABD, the less to the greater, which is impossible. Wherefore BE is not in the same straight line with BC. In like manner it may be shown that no other straight line but BD can be in the same straight line with BC. Therefore BD is in the same straight line with CB. Wherefore, if at a point, &c. Q. E. D.

C

B

D

PROP. XV. (THEOREM.)-If two straight lines (AB, CD) cut one another (in E), the vertical or opposite angles (AEC, DE B, and CEB, AED) are equal.

Because the straight line A E makes with CD the two angles CEA, AED, these angles are together equal (I. 13) to two right angles. Again, because the straight line D E makes with AB the

B

two angles AED, DEB, these angles are together equal (I. 13) to two right angles. But the two angles CEA, AED, have been proved to be equal to two right angles. There

fore the two angles CEA, AED, are equal c (Ax. 1) to the two angles AED, DEB. From these equals take away the common angle A AED. Therefore the remaining angle CEA

is equal (Ax, 3) to the remaining angle DEB.

B

In the same manner it can be demonstrated that the angle CEB is equal to the angle AED. Therefore, if two straight lines, &c. Q. E.D.

PROP. XVI. (THEOREM.)-If one side (BC) of a triangle (ABC) be produced (to D), the exterior angle (ACD) is greater than either of the interior opposite angles (CBA, BAC).

Bisect (I. 10) AC in E, join BE, and produce it to F. Make EF equal (I. 3) to BE. Join FC.

A

F

E

Because AE is equal (Const.) to EC, and BE (Const.) to EF. Therefore, in the triangles AEB, CEF, the two sides AE, EB, are equal to the two sides CE, EF, each to each. But the angle AEB is equal (I. 15) to the angle CEF, because they are vertical angles. Therefore the base AB is equal (I. 4) to the base CF, the triangle AEB to the triangle CEF, and the remaining angles B of the one to the remaining angles of the other, each to each, viz., those to which the equal sides are opposite. Wherefore the angle BAE is

equal to the angle ECF. But the angle ECD is greater (Ax. 9) than the angle ECF. Therefore the angle ACD is greater than the angle BAE. In the same manner, if the side BC be bisected, and AC be produced to G, it may be demonstrated that the angle BCG is greater than the angle ABC. But the angle ACD is equal (I. 15) to the angle BCG. Therefore the angle ACD is greater than the angle ABC. Therefore, if one side, &c. Q. E. D.

PROP. XVII.

(THEOREM.)--Any two angles of a triangle (ABC) are together less than two right angles.

A

Produce BC to D. Because ACD is the exterior angle of the triangle ABC, the angle ACD is greater (I. 16) than the interior and opposite angle ABC. To each of these unequals add the angle ACB. Therefore the two angles ACD, ACB, are greater (Ax. 4) than the two angles ABC, ACB. But the two angles ACD, ACB, are together equal (I. 13) to two right angles. Therefore the two angles ABC, BCA, are together less than two right angles. In like manner, it may be demonstrated that the two angles BAC, ACB, as also the two angles CAB, ABC, are together less than two right angles. Therefore any two angles, &c. Q. E. D.

B

C

D

PROP. XVIII. (THEOREM.)--The greater side (AC) of every triangle (ABC) is opposite to the greater angle (ABC).

From AC cut off (I. 3) AD equal to AB. `Join BD.