the base of the triangle ABC: And it has also been proved, that the angle FBC is equal to the angle GCB, which are the angles upon the other side of the base. Therefore, the angles at the base, &c. Q. E. D.

COR. Hence, every equilateral triangle is also equiangular.

PROP. VI. THEOR.

If two angles of a triangle be equal to one another, the sides which subtend, or are opposite to them, are also equal to one another.

Let ABC be a triangle having the angle ABC equal to the angle ACB; the side AB is also equal to the side AC.

A

For, if AB be not equal to AC, one of them must be greater than the other: Let AB be the greater, and from it cut (I. 3) off DB equal to AC the less, and join DC; therefore, because in the triangles DBC, ACB, DB is equal to AC, and BC common to both, the two sides DB, BC are equal to the two AC, CB, each to each; but the angle DBC is also equal to the angle ACB; therefore the base DC is equal to the base AB, and the area of the triangle DBC is B equal to that of the triangle (I. 4) ACB, the less to the greater; 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. COR. Hence, every equiangular triangle is also equilateral.

PROP. VII. THEOR.

C

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 equal to one another.*

Let there be two triangles ACB, ADB, upon the same base AB, and upon the same side of it, which have their sides CA, DA terminated in A equal to one another; then their sides CB, DB, terminated in B, cannot be equal to one

another.

B

Join CD, and if possible let CB be equal to DB; 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 (I. 5) to the angle ADC: But the angle ACD is greater than the angle BCD (Ax. 9); A therefore also, the angle ADC is greater than BCD; much more then is the angle BDC greater than the angle BCD. Again, because CB is equal to DB, the angle BDC is equal (I. 5) to the angle BCD; but it has been demonstrated to be greater than it; which is impossible.

See Notes

F

But if one of the vertices, as D, be within the other triangle ACB; produce AC, AD to E, F; therefore, because AC is equal 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 than the angle BCD (Ax. 9); wherefore the angle FDC is likewise greater than BCD; much more than is the angle BDC greater

B

than the angle BCD. Again, because CB is equal to DB, the angle BDC is equal (1. 5) 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, 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 equal to one another. 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 is equal to the angle contained by the two sides of the other.*

B

A

C E

D

Let ABC, DEF be two triangles having the two sides AB, AC, equal to the two sides DE, DF, each to each, viz., AB to DE, and AC 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 must also coincide with the point F, because BC is equal to EF; therefore BC coinciding with EF, BA and AC must 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 (I. 7); therefore, if the base BC coincide 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 (Ax. 8) to it. Therefore, if two triangles, &c. Q. E. D.

See Note to Prop. IV.

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 D in AB, and from AC cut (I. 3) off AE equal to AD; join DE, and upon it describe (I. 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 is equal (I. 8) to the angle EAF; wherefore the given rectilineal angle BAC is bisected by the straight B line AF. Which was to be done.

PROP. X. PROB.

DA

F

E

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 (I. 1) upon it an equilateral triangle ABC, and bisect (I. 9) the angle ACB by the straight line CD. AB is cut into two equal parts in the point D.

C

Because 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 (I. 4) DB, and the straight line AB is divided into two equal parts in the point D. Which was to be done.

PROP. XI. PROB.

D

B

To draw a straight line at right angles to a given straight

line, from a given point in that line.

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 angles to AB.

Take any point D in AC, and (I. 3) and 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.

Because DC is equal to CE, and FC

make CE equal to CD,

F

A

is common to the two triangles DCF, A D

C

E B

ECF, the two sides DC, 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 (I. 8) 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 (Def. 7) 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 done.

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 perpendicular

to AB from the point C.

G;

A

H

E

B

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 bisect (I. 10) FG in H, and join CF, CH, CG; the straight line CH, drawn from the given point C, is perpendicular to the given straight line AB.

D

Because FH is equal to HG, and HC is 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 (Def. 11) to the base CG; therefore the angle CHF is equal (I. 8) 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 line 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 CBA be equal to ABD, each of them is a right angle (Def. 7); but, if not, from the point B draw BE at

right angles (I. 11) to

ABE together; add the D

angle EBD to each of these equals, and the two angles CBE,

EBD will be equal (Ax. 2) to the three angles 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, A BC 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 thing are equal (Ax. 1) 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, the angles which one straight line, &c. Q. E. D.

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 are 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 together equal to two right angles. BD is in the same straight line with CB.

B

A

E

For, if BD be not in the same straight line with CB, let BE be in the same straight line with it; therefore, because the straight line AB makes angles with the straight line c CBE, upon one side of it, the angles ABC, ABE are together equal (I. 13) 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 (Ax. 3) 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.

If two straight lines cut one another, the vertical or opposite 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 (I. 13) to two right angles; and the angles AED, DEB, which the straight line DE makes with the straight line AB are also together equal (I. 13)