Because the point B is the centre of the circle CGH, BC is equal (15. Def.) to BG; and because D is the centre of the circle G KL, DL is equal to DG, and DA, DB, parts of them are equal (1. 1.): therefore the remainder AL is equal to the remainder (3 Ax.) BG: But it has been shown, that BC is equal to BG; wherefore AL and BC are each of them equal to BG; and things that are equal to the same are equal to one another; therefore the straight line A L is equal to B C. Wherefore from the given point A, a straight line AL has been drawn equal to the given straight line BC. Q. E. F.

In practice; from A as a centre, with a radius or semidiameter equal to the given line BC, describe an arc, or portion of the circumference of a circle in any convenient position; and draw a line from A to any point in the arc: the line so drawn will be equal to BC.

PROP. III. PROB.

FROM the greater of two given straight lines to cut off a part equal to the less.

Let A B and C be the two given straight lines, of which AB is the greater. It is required to cut off from AB, the greater, a part equal to c, the less.

From the point A draw (2. 1.) the straight line AD equal to C; and from

the centre A, and at the distance AD, describe (3. Post.) the circle DEF: then because A is the centre of the circle DEF, AE shall be equal to AD; but

D

C

B

E

F

the straight line c is likewise equal to AD; whence AE and c are each of them equal to AD: wherefore the straight line AE is equal (1. Ax.) to c, and from AB, the greater of two straight lines, a part AE has been cut off equal to c, the less. Q. E. F.

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; they shall likewise have their bases, or third sides, equal, and the two triangles shall be equal; and their other angles shall be equal, each to each, viz. those to which the equal sides are opposite.

Let ABC, DEF be two triangles, which have the two sides AB, AC equal to the two sides DE, DF, each to cach, viz.

A

D

AB to DE, and AC to DF, and the angle BAC equal to the angle EDF; the base BC shall be equal to the base EF; and the triangle ABC to the triangle DEF; and the other angles to which the equal sides are opposite shall be equal, each to each, viz. the angle

Да

B

ABC to the angle D EF, and the angle ACB to Dfe.

For if the triangle ABC be applied to DEF, so that the point A may be on D, and the straight line AB upon DE; the point B shall coincide with the point E, because AB is equal to DE; and AB coinciding with DE, AC shall coincide with DF, because the angle BAC is equal to the angle EDF; wherefore also the point c shall coincide with the point F, because the straight line AC is equal to DF: But the point B was proved to coincide with the point E; wherefore the base BC shall 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. Ax.) Therefore the base BC coincides with the base EF, and therefore is equal to it. Wherefore the whole triangle ABC coincides with the whole triangle DEF, and is equal to it; and the other angles of the one coincide with the remaining angles of the other, and are 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 shall be equal, and the other angles to which the equal sides are opposite shall be equal, each to each. Which was to be demonstrated.

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; also draw FC and GB.

F

B

C

G

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: And because the whole AF is equal to the whole AG, of which the parts AB, AC are equal; the remainder BF is 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 was proved to be equal to the angle CGB; wherefore the triangles BFC and CGB are equal (4.1.), and also their remaining angles, each to each, to which the equal sides are opposite; therefore the angle FBC is equal to the angle G C B, and the angle BCF to the angle CBG: And since it has been demonstrated, that the whole angle ABG is equal to the whole AC F, the parts of which, the angles CBG, BCF, are also equal; therefore the remaining angle ABC is 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 angle 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.

PROP. VI. THEOR.

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 shall be 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 off (3.1.) DB equal to AC, the less, and join D, C : 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

B

C

DBC is equal (4. 1.) to the triangle ACB, the less to the greater; which is absurd. Therefore A B 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.

UPON the same base, and on the same side of it, there cannot be two triangles having 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.

If it be possible, let there be two triangles ACB, ADB, upon the same base A B, 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.

C

D

Join C, D; then, in the case in which the vertex of each of the triangles is on the exterior of 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 A equal (5. 1.) to the angle BCD; but it has been demonstrated to be greater than it; which is impossible.

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 (5. 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 BD, the angle BDC is equal (5. 1.) to A

B

the angle BCD; but BDC has been proved to be greaterthan 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, &c. 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 shall be equal to the angle contained by the two sides equal to them of the other.

B

D

E

Let ABC, DEF be two triangles having the two sides AB, AC equal to the two sides DE, DF, each to each, viz. AB ot 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 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 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 having 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.

COR. Hence, and from prop. IV., it follows that all the angles of the one triangle are equal to all the angles of the other, each to each, as in the conclusion of that proposition. Also that the two triangles are equal to one another in every respect.

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