of the circle CGH, BC is equale to BG; and because D is the

centre of the circle GKL, DL is equal to DG; and DA, DB, parts of them, are equalf; therefore the remainder AL is equal f 19. Def. to the remainders BG: but it has been shown, that BC is g 3. Ax. equal to BG; wherefore AL and BC are each of them equal to BG; therefore the straight line AL is equal to BCh. h1. Ax. Wherefore, from the given point A a straight line AL has been drawn equal to the given straight line BC. Which was

to be done.

PROP. III. PROB.

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

he centre of the circle DEF, AE is equal to AD; but he straight line C is likewise equal to AD; whence AE and Care each of them equal to AD; wherefore the straight ine AE is equal to C, and from AB, the greater of two $ 1. Ax.

straight lines, a part AE has been cut off, equal to C the less.

Which was to be done.

Book I.

Ν.

PROP. IV. THEOREM.

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, or third sides, shall be 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 each, viz.
AB to DE, and AC
to DF; and the angle
BAC equal to the

D

angle EDF, the base
BC shall be equal to
the base EF; and the
triangle ABC to the

the other angles to

which the equal sides are opposite, shall be equal, each to each, viz. the angle ABC to the angle DEF, and the angle AСВ to DFE.

For, if the triangle ABC be applied to the triangle 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 AC is equal to DF: but the point B coincides with the point E; wherefore the base BC shall co

a cor. def. 3. incide with the base EFa, and shall be equal to it. Therefore also the whole triangle ABC shall coincide with the whole triangle DEF, and be equal to it: and the remaining ancles of the one shall coincide with the remaining angles of the o and be equal to them, viz. the angle ABC to the angle I and the angle ACB to the angle DFE. Therefore, if tw angles have two sides of the one equal to two sides o other, each to each, and have likewise the angles contain those sides equal to one another; their bases shall be equal

the triangles shall be equal, and their other angles, to which Book I. 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 N. equal to one another; and, if the equal sides be-produced, the angles upon the other side of the base shall also 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 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

A

A

they contain the angle FAG
common to the two triangles,
AFC, AGB; therefore the
base FC is equal to the base
GB, and the triangle AFC to
the triangle AGB; and the re-
maining angles of the one are
equal to the remaining an-
gles of the other, each to each,
to which the equal sides are op-
posite, viz. the angle ACF to
the angle ABG, and the angle D/
AFC to the angle AGB: and

F

B

C

G

a 3. 1.

b 4. 1.

because the whole AF is equal to the whole AG, and the part AB to the part AC; the remainder BF shall be equale c 3. Ах. 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; but the angle BFC is equal to the angle CGB; wherefore the triangles BFC, CGB are equalb, and their remaining angles are equal, 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. Now, since it has been demonstrated, that the whole angle ABG is equal to

Book 1. the whole ACF, and the part CBG to the part BCF, 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 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 equi angular.

PROP. VI. THEOR.

J

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

a 3. 1.

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

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 DB

equal to AC, the less, and join DC; there

64.1.

fore, 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 triangle DBC is equal to the
triangleb ACB, the less to the greater,
which is absurd. Therefore AB is not
* unequal to AC, that is, it is equal to it. B
Wherefore, if two angles, &c. Q. E. D.

A

D

C

COR. Hence every equiangular triangle is also equilateral.

PROP. VII. THEOR.

Book I.

UPON the same base, and on the same side of it, See N.

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.

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 A, equal to one another, and likewise their sides

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

CAD

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 equala to the angle ADC: but the angle ACD is greater than

the angle BCD; therefore the

A

B

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 tri

angle ACB; produce AC, AD to

E

E, F; therefore, because AC is e

F

qual to AD, in the triangle ACD, the angles ECD, FDC, upon the other side of the base CD, are equala to one another; but the angle ECD is greater than the angle BCD; wherefore the angle FDC

C

D

is likewise greater than BCD; much more then is the angle BDC greater than the angle BCD. A

gain, because CB is equal to DB,

B

the angle BDC is equal to the angle BCD; but BDC has

a 5. 1.