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

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

6. Things which are double of the same, equals, are equal to one another.

or of 7. Things which are halves of the same, or of equals, are equal to one another.

8. Magnitudes which coincide with one another are equal to one another.

9. The whole is greater than its part.

10. Two right lines cannot enclose a space.

11. All right angles are equal to one another.

12. If two right lines, meeting a right line, make the internal angles on the same side less than two right angles; these two right lines, being produced, will meet one another on that side at which the angles are less than two right angles.

On a given finite right line (AB) to describe an equilateral triangle.

From the centre A, and with the interval AB, describe the circle BCD, (by Post. 3). From the centre B and with the interval BA, describe the circle ACE. From the intersection C, draw the right lines CA, and CB, to the extremities of the given right line AB. (by Post. 1).

It is manifest that ABC is a triangle, constructed upon the given right line AB; but it is equilateral; for AC is equal to AB, because they are the radii of the same circle, DCB (by Def. 15). But BC is equal to BA, because they are the radii of the same circle ACE. Therefore, since both AC and BC are equal to the same, AB, they are equal to one another (by Ax. 1); and therefore the triangle ACB is equilateral.

SCHOL-Draw AF and FB, and it can be similarly demonstrated that the triangle AFB is equilateral.

From a given point (A) to draw a right line equal to a given finite right line (BC).

From the given point A draw a right line AB to either extremity B of the given right line (by Post. 1); upon AB construct an equilateral triangle ADB (by Prop. 1). From the centre B, with the interval BC, describe a circle GCF (by Post. 3), producing DB, till it meets its circum

F

D

A

ference in G. From D as a centre, with the interval DG, describe a circle GLO. The circumference of it

meets DA produced in L. AL is equal to the given right line BC. For DL is equal to DG, because they are the radii of the same circle GLO (by Def. 15); take away both DA and DB, which are also equal (by Const.) and the remainder AL will be equal to the remainder BG (by Ax. 3); but BG is equal to BC, because they are the radii of the same circle GCF: therefore both of them, AL and BC, are equal to the same BG and therefore are equal to each other (by Ax. 1). Therefore from the point A a right line AL is drawn equal to the given right line BC.

SCHOL. The position of the right line varies, according to the different extremity of the given line, to which the right line from the given point is drawn, and also according to the different side of that line on which the triangle is constructed.

From the greater of two given right lines (AB and CF) to cut off a part equal to the less.

From either extremity A of the greater right line, draw AD, equal to the less CF of the given lines (by Prop. 2). From the centre A, with the interval AD, describe a circle. It cuts off AE equal to AD (by Def. 15), and therefore equal to the given line CF (by Ax. 1).

PROPOSITION IV. THEOREM.

-F

If two triangles (EDF and ABC) have two sides of one equal to two sides of the other (ED and DF to AB and BC); and the angles contained by these sides also equal (D to B); the bases (EF and AC) will be equal; and the angles at the bases, opposite to equal sides, will be equal (E to A and F to C) and also the triangles themselves are equal.

E

D

FA

For if the triangles be so applied to each other that the point D may fall on B, and the side DE on BA, and that the sides DF and BC may be at the same side; then, because the sides DE and BA are equal, the point E must fall on A; and, because the angles D and B are equal, the side DF must fall on BC; and, because the side DF is equal to the side BC, the point F must fall on C. But, as the points E and F agree with the points A and C, the right lines FE and AC will agree (by Ax. 10); and therefore the bases EF and AC are equal (by Ax. 8). And, as the legs of the angles E and F agree with the legs of the angles A and C, the angles will agree, and are therefore equal (by Ax. 8). And, as the right lines which contain the triangle EDF agree with the right lines which contain the triangle ABC, the triangles themselves agree, and are therefore equal (by Ax. 8).

In any isosceles triangle (BAC) the angles at the base (ABC and ACB) are equal: and if the equal sides be produced, the angles below the base (FBC and GCB) will also be equal.

B

A

C

In either of the produced legs, assume any point F, and make AG equal to AF (by Prop. 3); and draw CF and BG. In the triangles FAC and GAB, the sides FA and AC are equal to the sides GA and AB (by Const. and Hypoth.), and the angle A is common, therefore the angle ACF is equal to ABG; and the angle AFC to AGB; and the side FC is equal to the side GB (by Prop. 4). Therefore in the triangles BFC and CGB, the angle BFC is equal to the angle CGB; and the side CF to the side BG; and taking away the

F

equals AB and AC from the equals AF and AG, the side BF is also equal to the side CG; therefore the angle FBC is equal to the angle GCB (by Prop. 4); but these are the angles below the base BC. And in the same triangles, the angle FCB is equal to the angle GBC (by Prop. 4); and taking away these from the equals FCA and GBA, the remaining angles ACB and ABC will be equal (by Ax. 3); but these are the angles at the base BC of the given triangle.

COR.-Hence, every equilateral triangle is also equiangular; for whatsoever side is assumed for the base, the adjacent angles will be equal, because the equal sides are opposite.

If two angles (B and C) of a triangle (BAC) be equal, the sides opposite to them (AC and AB) are equal.

If not, let one of them BA be made greater than the other, cut off a right line BD equal D to AC (by Prop. 3), and draw CD. Since, in the triangles DBC, ACB, the sides DB, BC, are equal to the sides AC, CB, and the angles DBC and ACB, which are contained by the B equal sides, are also equal (by Hypoth.), the triangles themselves, DBC and ACB, will be equal (by Prop. 4)—a part to the whole, which is absurd: therefore neither of the sides BA nor AC is greater than the other, therefore they are equal.

Hence every equiangular triangle is also equilateral; for whatsoever side is assumed for a base, the angles adjacent to it will be equal, and therefore the sides opposite to them are equal.

PROPOSITION VII. THEOREM.

Upon the same right line (AB) and on the same side of it, there cannot be constructed two triangles (ACB, ADB) whose conterminous sides (AC and AD, BC and BD) can be equal.