Book I.

IF

F two triangles have two fides of the one equal to two fides of the other, each to each; and have likewise the angles contained by those fides equal to one another; their bases, or third fides, fhall be equal; and the two triangles fhall be equal; and their other angles fhall be equal, each to each, viz. thofe to which the equal fides are oppofite.

A.

D

Let ABC, DEF be two triangles which have the two fides AB, AC equal to the two fides DE, DF, each to each, viz. AB to DE, and AC to DF; and the angle BAC equal to the angle EDF, the base BC fhall be equal to the base EF; and the triangle ABC to the triangle DEF; and

the other angles, to B

E

which the equal fides are oppofite, fhall be equal, each to each, viz. the angle ABC to the angle DEF, and the angle ACB to DFE.

For, if the triangle ABC be applied to the triangle DEF, fo that the point A may be on D, and the ftraight line AB upon DE; the point B fhall coincide with the point E, because AB is equal to DE; and AB coinciding with DE, AC fhall coincide with DF, because the angle BAC is equal to the angle EDF; wherefore also the point C fhall coincide with the point F, because AC is equal to DF: But the point B coincides with the point E; wherefore the base BC fhall coa cor.def.3.incide with the base EF a, and fhall be equal to it. Therefore also the whole triangle ABC fhall coincide with the whole triangle DEF, and be equal to it; and the remaining 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 the angle DFE.. Therefore, if two triangles have two fides of the one equal to two fides of the other, each to each, and have likewise the angles contained by thofe fides equal to one another, their bafes fhall be equal, and

the

the triangles fhall be equal, and their other angles, to which Boo kl. the equal fides are oppofite, fhall be equal, each to each. Which was to be demonftrated.

THE

HE angles at the base of an Ifofceles triangle are equal to one another; and, if the equal fides be produced, the angles upon the other fide of the base shall also be equal.

Let ABC be an ifofceles triangle, of which the fide AB is equal to AC, and let the straight lines AB, AC be produced to D and E, the angle ABC fhall 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 a to AF, the lefs, and join FC, GB.

A

Because AF is equal to AG, and AB to AC, the two fides 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 bafe FC is equal b to the bafe GB, and the triangle AFC to the triangle AGB; and the remaining angles of the one are equal b to the remaining angles of the other, each to each, to which the equal fides are oppofite, viz. the angle ACF to the angle ABG, and the angle AFC to the angle AGB: And

D

T

B

G

E

a 3. L

b 4. I...

because the whole AF is equal to the whole AG, and the part AB to the part AC; the remainder BF shall be equal to the remainder CG; and FC was proved to be equal c 3. Ax....... to GB, therefore the two fides BF, FC are equal to the two CG, GB, each to each; but the angle BFC is equal to the angle CGB; wherefore the triangles BFC, CGB are equal b, and their remaining angles are equal, to which the equal fides are oppofite; therefore the angle FBC is equal to the angle GCB, and the angle BCF to the angle CBG. Now, fince it has been demonftrated, that the whole angle ABG is equal to the

whole

Book I whole ACF, and the part CBG to the part BCF, the re maining 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 angle GCB, which are the angles upon the other fide of the bafe. Therefore the angles at the bafe, &c. Q. E. D.

COROLLARY. Hence every equilateral triangle is also equiangular.

IF

F two angles of a triangle be equal to one another, the fides which fubtend, or are oppofite to, those angles, fhall also be equal to one another.

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

D

For, if AB be not equal to AC, one of them is greater than 13. 1. the other: Let AB be the greater, and from it cut a off DB equal to AC, the lefs, and join DC; therefore, because in the triangles DBC, ACB, DB is equal to AC, and BC common to both, the two fides DB, BC are equal to the two AC, CB, each to each; but the angle DBC is alfo equal to the angle ACB; therefore the base DC is equal to the base AB, and the triangle DBC is equal to the 4.1. triangle b ACB, the less to the greater; which is abfurd. Therefore AB is not unequal to AC, that is, it is equal to it. Wherefore, if two angles, &c. QE. D. COR. Hence every equiangular triangle is alfo equilateral.

B

PROP.

13

U

PROP. VII.

THEOR.

Book I.

PON the fame bafe, and on the fame fide of it, See N, there cannot be two triangles that have their fides which are terminated in one extremity of the bafe equal to one another, and likewife those which are terminated in the other extremity, equal to one another.

If it be poffible, let there be two triangles ACB, ADB, upon the fame base AB, and upon the fame fide of it, which have their fides CA, DA, terminated in A equal to one another, and likewife their fides

CB, DB, terminated in B, equal to one another.

Join CD; then, in the cafe 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 a to the angle ADC: But the angle ACD is greater than the angle BCD; therefore the A angle ADC is greater also than

B

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 a to the angle BCD; but it has been demonftrated to be greater than it; which is impoffible.

E

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, DC upon he other fide of the bafe CD are equal a to one another, but the angle ECD is greater than the angle BCD; wherefore the angle FDC is likewife 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 a to the angle BCD; but BDC has

A

B

a 5.4.

been

Book I. been proved to be greater than the fame BCD; which is impoffible. The cafe in which the vertex of one triangle is upon a fide of the other, needs no demonstration.

Therefore, upon the fame bafe, and on the fame fide of it, there cannot be two triangles that have their fides 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.

F two triangles have two fides of the one equal to two fides of the other, each to each, and have likewise their bafes equal; the angle which is contained by the two fides of the one fhall be equal to the angle contained by the two fides of the other,

Let ABC, DEF be two triangles having the two fides AB, AC, equal to the two fides DE, DF, each to each, viz. AB to DE, and AC to DF; and also the bafe BC equal to the

bafe EF. The angle BAC is equal to the angle EDF.

For, if the triangle ABC be applied to the triangle DEF, fo that the point B be on E, and the ftraight line BC upon EF; the point C fhall alfo coincide with the point F, because BC is equal to EF: therefore BC coinciding with EF, BA and AC fhall coincide with ED and DF; for, if BA, and CA do not coincide with ED, and FD, but have a different fituation as EG and FG; then, upon the fame bafe EF, and upon the fame fide of it, there can be two triangles EDF, EGF, that have their fides which

are