XXIX. An acute-angled triangle has three acute angles.

XXX. A square has all its sides equal, and all its angles right angles.

XXXI. An oblong has all its angles right angles, but not all its sides equal.

XXXII. A rhombus has its sides equal, but its angles not right angles.

XXXIII.—A rhomboid has its opposite sides equal, but all its sides not equal, nor its angles right angles.

XXXIV. All other four-sided figures besides these are called trapeziums.

XXXV. Parallel straight lines are such as are in the same plane, and which, being produced ever so far both ways, do not meet.

POSTULATES.

I. Let it be granted that a straight line may be drawn from any one point to any other point.

II. That a terminated straight line may be produced to any length in a straight line.

III. That a circle may be described from any centre, at any distance from that centre.

AXIOMS.

I. Things equal to the same are equal to one another. II. If equals be added to equals, the wholes are equal.

III. If equals be taken from equals, the remainder are equal.

IV. If equals be added to unequals, the wholes are unequal.

V. If equals be taken from unequals, the remainders are unequal

VI. Doubles of the same are equal to one another. VII. Halves of the same are equal to one another. VIII. Magnitudes which coincide with one another, or exactly fill the same space, are equal to one another. IX. The whole is greater than its part.

X. Two straight lines cannot enclose a space. XI. All right angles are equal to one another. XII. "If a straight line meets two straight lines, so as to make the two interior angles on the same side of it taken together less than two right angles, these straight lines being continually produced, shall at length meet upon that side on which are the angles which are less than two right angles."

PROPOSITION I.

To describe an aquilateral triangle on a given finite straight line.

Let AB be the given straight line; it is required t describe an equilateral triangle on it. From the centre A, at the distance AB, describe the circle BCD, and from the centre B, at the distance BA, describe the circle ACE; and from C, where the circles cut one another, join CA, CB; ABC is an equilateral triangle.

Because A is the centre of BCD, AC is equal to AB; and because B is the centre of ACE, BC is equal to BA, but it has been proved that CA is equal to AB; therefore CA, CB, are each of them equal to AB, but things which are equal to the same are equal to one another; therefore CA is equal to CB; wherefore CA, AB, BC, are equal to one another; and the triangle ABC is equilateral, and is described on the given straight line AB.

II.-From a given point to draw a straight line equal to a given straight line.

Let A be the given point, and BC the given straight line; it is required to draw from the point A a straight line equal to BC.

Join AB, and on it describe the equilateral triangle DAB, and produce DA, DB, to E and F; from the centre B, at the distance BC, describe the circle CGH, and from the centre D, at the distance DG, describe the circle GKL, AL, to BC.

K

H

D

Because B is the centre of CGH, BC is equal to BG; and because D is the centre of GKL, DL is equal to DG; and DA, DB, parts of them, are equal; therefore the remainder AL, is equal to the remainder 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 AL is equal to BC. Wherefore from the given point A a straight line AL has been drawn equal to the given straight line BC.

B

G

F

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

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

E

From A draw AD equal to C; and from the centre A, and at the distance AD, describe the circle DEF; and because A is the centre of DEF, AE shall be equal to AD; but C is likewise equal to AD; whence AE and C are each of them equal to AD; wherefore the straight line AE is equal to C, and from AB, the greater of two straight lines, a part AE has been cut off equal to C the less.

IV.-If two triangles have two sides of the one equal to

two sides of the other, each to each; and 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 sides AB, AC, equal to DE, DF, each to each, viz., AB to DE, and AC to DF; and the angle BAC equal to the B 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., ABC to DEF, and 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 C shall coincide with F, because AC is equal to DF: but the point B coincides 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. Therefore BC shall coincide with 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., ABC to DEF, and ACB to DFE.

V.-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 AB is qual to AC, and let AB, AC, be produced to D and E the angle ABC shall be equal to ACB, and the angle CBD to 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

F

D

B

G

E

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 to the base GB, and the triangle AFC to the triangle AGB; and the remaining angles of the one are equal to the remaining angles of the other, each to each, to which the equal sides are opposite; viz., ACF to ABG, and AFC to 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 to the remainder CG; and FC was proved to be equal to GB; therefore the two BF, FC, are equal to the two CG, GB, each to each, and the angle BFC is equal to CGB, and BC is common to the two triangles BFC, CGB; wherefore the triangles are equal and their remaining angles each to each, to which the equal sides are opposite; therefore FBC is equal to GCB, and BCF to CBG: and, since it has been demonstrated that the whole angle ABG is equal to the whole ACF; the parts of which, 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 also has 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.

COROLLARY.-Hence every equilateral triangle is also equiangular.

VI.-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 ABC; the side AB is also equal to the side AC.

For if AB be not equal to AC, one of them is greater: let AB be the greater and from it cut off