to the same BG, are equal to one another (7). There- (7) Ax. 1.
fore from the given point A a right line AL has been
drawn, equal to the given right line BC.

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

PROP. III. PROB.

From the greater of two given right lines (AB and CF) Fig. 14. to cut off a part equal to the less.

From either extremity A of the greater line draw AD, equal to CF the less of the given right lines (1). From the centre A, with the radius AD, describe a circle which shall cut off AE equal to AD (2), and therefore also equal to the given right line CF (3).

PROP. IV. THEOREM.

See N.

(1) Prop. 2.

(2) Def. 15.

(3) Constr.
& Ax. 1.

If two triangles (EDF, ABC) have two sides of the Fig. 15. one respectively equal to two sides of the other (ED and See N DF to AB and BC) and the angles contained by those sides also equal to one another (D to B); their bases (EF and AC) are equal, and the angles at the bases, which are opposite to the equal sides, are equal (E to A and F to C); and also the triangles themselves.

For if the triangle EDF be so applied to the triangle ABC, that the point D may be on B, and the side DE on BA, and that DF and BC may lie at the same side, the point E must coincide with A, because the sides DE and BA are equal: and because the angles D and B are equal, the side DF must fall on BC; and because the side DF is equal to BC, the point F must coincide with C.

But as the points E and F coincide with the points A and C, the right lines EF and AC must coincide (1), (1) Ax. 10. and therefore the bases EF and AC are equal (2).

(2) Ax. 8.

d

fi

(3) Ax. 8.

Fig. 16.

a

And as the sides of the angles E and F coincide with the sides of the angles A and C, the angles themselves must coincide, and are therefore equal (3).

And as the right lines, which bound the triangle EDF, coincide with the right lines which bound the triangle ABC, the triangles themselves must coincide, and are therefore equal (3).

PROP. V. THEOR.

In an isosceles triangle (BAC) the angles at the base (ABC and ACB) are equal to one another; and if the equal sides be produced, the angles below the base (FBC and GCB) shall also be equal.

Take any point F in the side produced, cut off AG (1) Prop. 3. equal to AF (1), and join FC, GB.

(2) Constr.
& Hypoth.

(3) Prop. 4.

(4) Ax, 3.

In the triangles FAC and GAB, the sides FA and AC are equal to the sides GA and AB (2), and the angle A is common to both, therefore the angle ACF is equal to ABG, and the angle AFC to AGB; and the side FC is equal to the side GB (3). Therefore in the triangles BFC, CGB, the angle BFC is equal to the angle CGB, and the side FC is equal to BG, and taking away the equals AB and AC from the equals AF and AG, the side BF is equal to the side CG, therefore the angle FBC is equal to the angle GCB (3); but these are the angles below the base BC.

And in the same triangles, the angle FCB is is equal to the angle GBC (3), and taking away these from the equals FCA and GBA, the remaining angles ACB and ABC are equal (4), but these are the angles at the base BC of the given triangle..

Cor. Hence every equilateral triangle is also equiangular; for whatever side is taken for the base, the angles adjacent to it are equal, since they are opposite to equal sides.

PROP. VI. THEOR.

If two angles (B and C) of a triangle (BAC) be Fig. 17.1 equal, the sides opposite to them (AC and AB) are also equal.

If the sides be not equal, let one of them AB be greater than the other, and from it cut off DB equal to AC (1), and join CD.

(1) Prop. 3.

Then in the triangles DBC, and ACB, the sides DB, BC, are equal to the sides AC, CB; and the angles DBC and ACB are also equal (2); therefore the tri- (2) Hypoth. angles themselves, DBC and ACB, are equal (3), a part (3) Prop. 4. equal to the whole, which is absurd: therefore neither of the sides AB or AC is greater than the other; they are therefore equal to one another.

Cor. Hence every equiangular triangle is also equilateral; for whatever side is taken for the base, the angles adjacent to it are equal, and therefore the sides which subtend them.

PROP. VII. THEOR.

On the same right line (AB) and on the same side of Fig. 18, 19, it, there cannot be constructed two triangles (ACB, and 20, ADB), whose conterminous sides (AC and AD, BC and BD) are equal.

If it be possible, let the two triangles be constructed: Fig. 18. and first let the vertex of each of the triangles be without the other triangle, and draw CD.

(2) Prop. 5.

(3) Ax. 9.

Because the sides AD and AC of the triangle CAD are equal (1), the angles ACD and ADC are equal (1) Hypoth. (2), but ACD is greater than BCD (3), therefore ADC is greater than BCD; but the angle BDC is greater than ADC (3), and therefore BDC is greater than BCD but in the triangle CBD, the sides BC and BD are equal (4), therefore the angles BDC and (4) Ilypoth BCD are equal (5), but the angle BDC has been (5) prop. 6. proved to be greater than BCD, which is absurd. Therefore the triangles constructed upon the same

Fig. 19.

(6) Hypoth. (7) Prop. 5. (8) Ax. 9.

Fig. 20.

right line cannot have their conterminous sides equal, when the vertex of each of the triangles is without the other.

Secondly. Let the vertex D of one triangle be within the other: produce the sides AC and AD, and join CD. Because the sides AC and AD of the triangle CAD * are equal (6), the angles ECD and FDC are equal (7); but the angle BDC is greater than FDC (8), and therefore greater than ECD; but ECD is greater than BCD (8), and therefore BDC is greater than BCD: but in the triangle CBD, the sides BC and BD are equal (6), therefore the angles BDC and BCD are equal (7); but the angle BDC has been proved to be greater than BCD, which is absurd. Therefore triangles constructed upon the same right line cannot have their conterminous sides equal, if the vertex of one of them be within the other.

Thirdly. Let the vertex D of one triangle be on the side of the other AC; and it is evident that the sides AC and AD are not equal. Therefore in no case can two triangles, whose conterminous sides are equal, be constructed at the same side of a given line.

PROP. VIII. THEOR.

If two triangles (ABC and EFD) have two sides of the one respectively equal to two sides of the other (AB to EF and BC to FD), and also have the base (AC) equal to the base (ED), then the angles (B and F) contained by the equal sides are equal.

For if the equal bases AC and ED be so applied to one another, that the equal sides AB and EF, CB and DF may be conterminous, the vertex B must fall upon F(1), and the equal sides AB and EF, CB and DF must coincide (2): therefore the angles B and F must coincide, and are therefore equal (3).

Schol. It is evident that the remaining angles A and E, C and D, opposite to the equal sides, are equal and also that the triangles themselves are equal.

PROP. IX. PROB.

To bisect a given rectilineal angle (BAC).

Take any point D in the side AB, and from AC cut

Fig 22.

off AE equal to AD (1); draw DE, and upon it describe (1) Prop. 3. an equilateral triangle DFE (2), at the side remote from (2) Prop. 1. A. The right line, joining the points A and F, bisects

the given angle BAC.

Because the sides AD and AE are equal (3), and the (3) Constr. side AF is common to the triangles FAD and FAE, and the base FD is also equal to FE (3); the angles

DAF and EAF are equal (4), and therefore the right (4) Prop. 8. line AF bisects the given angle.

Cor. By this proposition an angle can also be divided into 4 parts, 8, 16, &c. &c. by bisecting again each part.

PROP. X. PROB.

To bisect a given finite line (AB).

Fig. 23.

Upon the given line AB describe an equilateral triangle ACB (1); bisect the angle ACB by the right (1) Prop. 1. line CD (2): this line bisects the given line in the point (2) Prop. 9.

D.

Because the sides AC and CB are equal (3), and CD (3) Constr. is common to the triangles ACD and BCD, and the

angles ACD and BCD are also equal (3); therefore the

bases AD and DB are equal (4), and the right line AB (4) Prop. 4. is bisected in the point D.

PROP. XI. PROB.

From a given point (C), in a given right line (AB), to Fig. 24. draw a perpendicular to the given line.

In the given line take any point D, and make CE equal to CD (1); upon DE describe an equilateral tri- (1) Prop. 3. angle DFE (2), draw FC, and it is perpendicular to the (2) Prop. 1. given line.

Because the sides DF and DC are equal to the sides