coincide with the base EF, because the point B coinciding with E, and C with F, if the base BC does not coincide with the base EF, two straight lines would inclose a space, which is impossible. (10 Ar.) Therefore the base BC shall coincide with the base EF, and be equal to it. Wherefore the whole triangle ABC shall coincide with the whole triangle DEF, and be equal to it; and the other angles of the one shall coincide with the remaining angles of the other, and be equal to them, viz. the angle ABC to the angle DEF, and the angle ACB to DFE. Therefore, if two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise the angles contained by those sides equal to one another, their bases shall likewise be equal, and the triangles be equal, and their other angles to which the equal sides are opposite shall be equal, each to each. Which was to be demonstrated.

Proposition. V. Theorem.

The angles at the base of an Isosceles triangle are equal to one another; and if the equal sides be produced, the angles upon the other side of the base shall be equal.

Let ABC be an Isosceles triangle, of which the side AB is equal to AC, and let the straight lines AB, AC be produced to D and E, the angle ABC shall be equal to the angle ACB, and the angle CBD to the angle BCE.

In BD take any point F, and from AE the greater, cut off AG equal (3.1.) to AF, the less, and join FC, GB.

Because AF is equal to AG, and AB to AC, the two sides FA, AC are equal to the two GA, AB, each to each; and they contain the angle FAG common to the two triangles AFC, AGB; therefore the base FC is equal (4. 1.) to the base GB, and the triangle AFC to the triangle AGB, and the remaining angles of the one are equal (4. 1.) to the remaining angles of the other, each to each, to which the equal sides are opposite; viz. the angle ACF to the angle ABG, and the angle AFC to the angle AGB.

B

And because the whole AF, is equal to the whole AG of which the parts, AB, AC, are equal; the remainder BF shall be equal (3 Ax.) to the remainder CG; and FC, was proved to be equal to GB; therefore the two sides BF, FC are equal to the two CG, GB, each to each; and the angle BFC is equal to the angle CGB, and the base BC is common to the two triangles BFC, CGB; wherefore the triangles are equal, (4. 1.) and their remaining angles, each to each, to which the equal sides are opposite; therefore the angle FBC is equal to the angle GCB, and the angle BCF to the angle CBG.

And since it has been demonstrated, that the whole angle ABG is equal to the whole ACF, the parts of which, the angles CBG, BCF are also equal; the remaining angle ABC is therefore equal to the remaining angle ACB, which are the angles at the base of the triangle ABC.

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

COROLLARY. Hence every equilateral triangle is also equiangular.

Proposition VI. Theorem.

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

:

D

A

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 (3. 1.) off DB equal 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; and the angle DBC is equal to the angle ACB; therefore the base DC is equal to the base AB, and the triangle DBC, is equal to the triangle (4.1.) 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.

B

COR. Hence every equiangular triangle is also equal equilatera.

Proposition VII. Theorem.

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.

D

If it be possible, 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 the extremity A of the base equal to one another, and likewise their sides 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 (5. 1.) to the angle ADC: But the angle ACD is greater than the angle BCD; therefore the angle ADC is greater also 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 to the angle BCD; but it has been demonstrated to be greater than it; which is impossible.

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 (3. 1.) to one another; but the 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 BDC is equal (3.1.) to 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. Q. E. D.

Proposition VIII. Theorem.

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 contained by the two sides equal to them, of the other. Let ABC, DEF be two triangles ha

ving 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.

B

E

For, if the triangle ABC be applied to DEF, so that the point B be on E, and the straight line BC upon EF; the point C shall also coin. cide 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 impossible; 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, to it. Therefore (8. Ax.) if two triangles, &c. Q. E. D.

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Proposition IX. Problem.

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 3. 1.) off AE equal to AD; join DE, and apon it describe (1. 1.) an equilateral triangle DEF; then join AF; the straight line AF bisects the angle BAC.

F

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; and the base DF is equal to the base EF; therefore the angle DAF is equal (8. 1.) to the angle EAF; wherefore the given rectilineal angle BAC is bisected by the straight line AF. Which was to be done.

Proposition X. Problem.

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.

Because AC is equal to CB, and CD coinmon to the two triangles ACD, BCD; the two sides AC, CD are equal to BC, CD, each to each; and the angle ACD is 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 equal parts in the point D. Which was to be done.

Proposition XI. Problem.

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; it is required to draw a straight line from the point C at right angles to AB.

Take any point D in AC, and (3. 1.) make CE equal to CD, and upon DE describe (1. 1.) he 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 cqual to CE, and FC common to the two triangles DCF,ECF; the 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 anglo 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 (10 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 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 line, the angle CBE is equal (10 Def.) 1.) to the angle 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 is impossible therefore two straight lines cannot have a common segment.

Proposition XII. Problem.

D

A

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 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 Č.

H

E

Take any point D upon the other side of AB, and from the centre C, at the distance CD, describe (3 Post.) the circle EGF meeting AB in F and G; and bisect (10. 1.) 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.

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 (15 Def.) to the base CG; therefore the angle CHF is equal (8. 1.) to the angle 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.

Proposition XIII. Theorem.

The 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 it, the