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30. A parallelogram is a quadrilateral figure, Fig. 11. whose opposite sides are parallel.

31. A square is a quadrilateral figure, which has all its sides equal, and all its angles right angles.

32. Rectilineal figures, which have more than four sides, are called polygons.

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Postulates.

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

2. That a terminated right line may be produced any length in a right line.

3. That a circle may be described from any centre, See N. any

distance from that centre.

at

Arioms.

1. Things which are equal to the same are equal to one another.

2. If equals be added to equals, the wholes are equal.

3. If equals be taken from equals, the remainders are equal.

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

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

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

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 inclose a space. 11. All right habere are equal to one another.

12. If a right line meet two right lines, so as to make the two internal angles on the same side of it See N. taken together less than two right angles, these right lines being continually produced, shall at length meet upon that side on which are the angles, which are less than two right angles.

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Fig. 12,

To describe an equilateral triangle upon a given finite

right line (AB).

From the centre A, with the radius AB describe (1) Post. 3. the circle BCD, (1) and from the centre B, with

the radius BA describe the circle ACE. From the

point of intersection C, draw the right lines CA and (2) Post: 1. CB to the extremities of the given right line (2).

It is evident that ACB is a triangle constructed upon the given line; but it is also equilateral, for the

right line AC is equal to AB, as they are radii of the (3) Def. 13. same circle DCB (3); and the right line BC is equal

to BA, as they are radii of the same circle ACE.

Since then oth the lines A and BC are equal to (4) Ax. 1. the same AB, they must be equal to one another (4),

and therefore the triangle ACB is equilateral.

Schol. Draw the lines AG and GB and in the same manner it can be demonstrated that the triangle AGB is equilateral.

PROP. II, PROB.

Fig. 13.
See N.

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

a given finite right line (BC).

13

F

From the given point A draw a right line AB to (1) Post. l. either extremity B of the given line (1). Upon AB (2) Prop. 1. construct an equilateral triangle ADB (2). From (3) Post. 3. the centre B with the radius BC describe a circle

GCF (3), and produce the right line DB until it meets the circumference in G. From the centre D with the radius DG describe the circle GLO and produce the right line D A until it meets the circumference in L. The right line AL is equal to the given line BC.

For the right line DL is equal to DG, because they (4) Def. 15. are radii of the same circle GLO (4); and if the equals (3) Con

DA and DB (5) be taken away from both, the restruct.

mainders AL and BG are also equal (6); but BG is (6) AX. 3.

equal to BC as they are radii of the same circle GCF; therefore the right lines AL and BC, which are equal

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to the same BG, are equal to one another (7).(7) Ar. 1.
Therefore 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.

14

D

PROP. III. PROB.

From the greater of two right lines (AB and CF,) Fig. 14.

to cut off a part equal to the less.

See N.

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From either extremity A of the greater line draw
AD equal to FC the less of the given right lines (1). (!? Prop: 2.
From the centre A with the radius AD describe a circle (?) Def. 15.

(3) Constr. which shall cut off AE equal to AD (2), and therefore & Ax. . also equal to the given right line CF (3).

PROP. IV. THEOREM.

15

B В

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 Å and F to C); and also the triangles themselves.

For if the two triangles be so applied to one another 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) At. 8.

And as the sides of the angles E and F coincide with

the sides of the angles A and C, the angles themselves (3) Ax. 8.

must coincide, and are therefore equal (3).

And if the right lines which bound the triangle EDF coincide with the right lines which bound the

triangle ABC, the triangles themselves must coincide (4) Ax: 8. and are therefore equal (4).

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Figo 18.

A

In an isosceles triangle (B.1C) 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.

B

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

. In the triangles FAC and GAB, the sides FA and (2) ConstrAC are equal to the sides GA and AB (2), and the & hypoth.

angle A is common to both, therefore the angle ACF

is equal to ABG, and the angle AFC to AGB; and (3) Prop• 4 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, (4) Prop. 4. therefore the angle FBC is equal to the angle GCB

(4) ; but these are the angles below the base BC.

And in the same triangles, the angle FCB is equal to the angle GBC (4), and taking away these from the

equals FCA and GBA, the remaining angles ACB (5) Ax, 3.

and ABC are equal (5), 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.

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Il treo angles (B and C) of a triangle (BAC) be Fig. 17. equal, the sides opposite to them, (AC und AB) are

He also equal.

B

For if the sides be not equal, let one of them AB be greater than the other, and from it cut off DB equal (1) prop. 3. to AC(1), and join CD.

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 triangles themselves DBC and ACB, are equal, a part (2) hypoth. 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 F which subtend them.

PROP. VII. THEOR.

1

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On the same right line (AB) and onthe 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.

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Ifit 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., Because the sides AD and AC of the triangle CAD

(1) hypoth. are equal (1), the angles ACD and ADC are equal (2) prop. 5. (2), but ACD is greater than BCD (3), therefore (3) ax.o. 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 (4) hypoth:

(5) prop. 6. and BD are equal (4), therefore the angles BDC and BCD are equal (5) but the angle BDC has been 18 proved to be greater than BCD, which is absurd. Therefore the triangles constructed upon the same

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