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

the other. Fig. 19.

Secondly. Let the vertex Dof one triangle be within the other, produce the sides AC and A Dand join CD.

Because the sides AC and AD of the triangle CAD (6) hypotk. are equal (6), the angles ECD and FDCareequal (7);

(7) prop 5. but the angle BDC isgreater than FDC (8), and there-
(8) ax: 9. fore 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 trian -
gles constructed upon the same right line cannot have
their conterminous sides equal, if the vertex of one

of them is within the other. Fig. 20. Thirdly. Let the vertex D of one triangle be on

the side of the other AC, and itis 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,

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Fig. 21.
See N.

If two triangles, (ABC and EFD) have two sides of the one respectiv ly equal to two sides of the other, ( AB to EF and BC 10 FD), and also have the base (AC) equal to the base (ED) then the angles (B und F) contained by the equal sides are equal.

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For if the equal bases AC and ED be so applied to one another, that the equal sides A B and EF, CB and

DF, may be conterminous, the vertex B must fall (1) prop. 7. upon F (1), and the equal sides A B and EF, CB and (2) ar 10. DF must coincide (2), therefore the angles B and F

must coincide, and are therefore equal.

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.

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To bisect a given rectilineal angle (BAC.)

Fig. 22.

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Take any point D in the side AB and from AC cut off AE equal to AD (1); draw DE and upon it des- (1) prop. 3. cribe an equilateral triangle DFE (2), at the side re- (2) prop. 1. mote from 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 (3) constr. the 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 (4) prop. 8. the right 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.


To bisect a given finite right line (AB).

Fig. 23.


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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 (2) prop. 9 point D.

Because the sides AC and CB are equal (3) and (3) constr. CD common to the triangles ACD and BCD, and the angles ACD and BCD also equal (3), therefore the bases AD and DB are equal (4) and the right line (4) prop. 4. AB is bisected in the point D.



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

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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 (2) prop. 1. to the given line. Because the sides DF and DC are equal to the sides


prop. 8.

(3) constr. EF and EC (3), and CF is common to the triangles

DFC and EFC, therefore the angles opposite to the (4) Sch. equal sides DF and EF are equal (4), and therefore

FC is perpendicular to the given right line AB at the (5) Def. 11, point C (5).

Schol. In the same manner a perpendicular can be erected at one extremity of a given line, by first producing the line.

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Fig. 25.
See N.

To draw a perpendicular to a given indefinite right line (AB) from a point (C) given without it.

Take any point X on the other side of the given line, and from the centre C with the radius CX des.

cribe a circle, cutting the given line in E and F. Bi(1) prop. 10. sect EF in D (1), and draw from the given point to

the point of bisection the right line CD, it is perpendicular to the given line.

For draw CE and CF; and in the triangles EDC (2) Def. 15. and FDC, the sides EC and FC (2), and ED and (3) Constr. FD (3) are equal, and CD common; therefore the

angles EDC and FDC, opposite to the equal sides (4) Schol.

EC and FC, are equal (4), and therefore DC is per(5) * Def. 11. pendicular to the given line AB (5).

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Fig. 26.
See N.

When a right line (4B) standing upon another (DC) makes angles with it, they are either two right angles, or together equal to two right angles.



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If the right line AB is perpendicular to DC, the (1) Def. 11. angles ABC and ABD are right (1). If not, draw (2) Prop. 11. BE perpendicular to DC (2), and it is evident that the

angles CBA and ABD together are equal to the angles CBE and EBD, and therefore to two right angles.

Cor. 1. If several right lines stand on the same right line at the same point, and make angles with it, all the angles taken together, are equal to two right angles.

Cor 2. Two right lines intersecting one another make angles, which, taken together, are equal to four right angles.

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Cor. 3. If several right lines intersect one another in the same point, all the angles taken together are equal to four right angles.


If two right lines (CB and BD) meeting another Fig. 27. right line , AB) at the same point and at opposite

27 sides, make angles with it which are together cqual to two right angles, those right lines (CB and BD) form one continued right line.


For, if possible, let BE and not BD be the continuation of the right line CB, then the angles CBA and ABE are equal to two right angles (1), but CBA and (1) Prop. 13, ABD are also equal to two right angles (2), therefore (2) hypoth. CBA and ABD taken together are equal to CBA and ABE, take away from these equal quantities CBA which is common to both, and ABE shall be equal to ABD, a part to the whole, which is absurd. Therefore BE is not the continuation of CB; and in the same manner it can be proved that no other line, except BD, is the continuation of it ; therefore BD forms with BC one continued right line.


If two right lines (AB and CD) intersect one ano- Fig. 29. ther, the vertical angles are equal (CEA to BED, and CEB to AED).


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Because the right line CE stands upon the right line AB, the angle AEC, together with the angle CEB, is equal to two right angles (1); and because (1) Prop. 13 the right line BE stands upon the right line CD, the angle CEB, together with the angle BED, is equal to two right angles (1); therefore AEC and CEB (2) Ax 3. together are equal to CEB and BED, take away the common angle CEB, and the remaining angle AEC is equal to BED (2).


Fig. 29.

If one side (BC) of a triangle (BAC) be produced, the external angle (ACD) is greater than either of the internal opposite angles (A or B).

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(1) Prop. 10. For bisect the side AC in E (1), draw BE and (2) Prop. 3. produce it until EF be equal to BE (2) and join FC.

The triangles CEF and AEB have the sides CE (3) Constr. and EF equal to the sides AE and EB (3) and the (4) Prop. 15. angle CEF equal to AEB (4), therefore the angles (5) Prop. 4. ECF and A are equal (5), and therefore ACĎ is

greater than A. In like manner it can be shewn, that if AC be produced, the external angle BCG is greater than the angle B, and therefore that the angle ACD, which is equal to BCG (4), is greater

than the angle B, Fig. 30, Cor. 1. If from any point B, two right lines be

drawn to the same right line ED, one of them perpendicular to it, the other not, the perpendicular falls at the side of the acute angle.

For, if possible, let B A perpendicular to the line

ED fall at the side of the obtuse angle BCE, then the (1) Def.12. angle BAE is less than BCE (1), but BAE is, (2) Prop. 16. greater than the same angle BCE (2) which is ab

surd : BA, therefore, cannot fall at the side of the obtuse angle, and therefore falls at the side of the

acute angle. Fig. 33.

Cor. 2. Two perpendiculars cannot be drawn from the same point B, to the same right line ED.

For, if possible, let the lines BA and BC be both

perpendicular to ED, then the angle BAE is equal (1) Ar. 11. to the angle BCE (1), but it is also greater than it (?) Prop. 16, (2), which is absurd ; therefore the right lines BA and

BC cannot both be perpendicular to ED.

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Any two angles of a triangle (BAC) are together

less than two right angles,

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l'roduce any side BC, then the angle ACD is greater (1) Prop. 16. either the angle A or B (1), therefore ACB toge.

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