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Then shall the base BC 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. the angle ABC to the angle DEF, and the angle ACB to the angle DFE.

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For, if the triangle ABC be applied to the triangle DEF, 60 that the point A may be on D, and the straight line AB on DE; then the point B shall coincide with the point E,

because AB is equal to DE;

and AB coinciding with DE,

the straight line AC shall fall on DF,
because the angle BAC is equal to the angle EDF;
therefore also the point C shall coincide with the point F,

because AC is equal to DF;
but the point B was shewn to coincide with the point E;
wherefore the base B C shall coincide with the base EF;

because the point B coinciding with E, and C with F, if the base BC do not coincide with the base EF, the two straight lines BC and EF would enclose a space, which is impossible. (ax. 10.) Therefore the base BC does coincide with EF, and is equal to it;

and the whole triangle ABC coincides with the whole triangle DEF, and is equal to it; also the remaining angles of one triangle coincide with the remaining angles of the other, and are equal to them,

viz. the angle ABC to the angle DEF,

and the angle ACB to DFE. Therefore, if two triangles have two sides of the one equal to two sides, &c. Which was to be demonstrated.

PROPOSITION V. THEOREM. The angles at the base of an isosceles triangle are equal to each nther; and if the equal sides be produced, the angles on the other side of the base shall be equal. l.et ABC be an isosceles triangle of which the side AB is equal to AC,

and let the equal sides AB, AC be produced to D and E.
Then the angle ABC shall be equal to the angle ACB,
and the angle DBC to the angle ECB.

In B D take any point F;
from AE the greater, cut off AG equal to AF the less, (I. 3.)

and join FC, GB. Because AF is equal to AG, (constr.) and AB to AC; (hyp.) the 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;

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therefore the base FC is equal to the base GB, (I. 4.)

and the triangle AFC is equal to the triangle ÀGB, also 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. the angle ACF to the angle ABG,

and the angle AFC to the angle AGB.
And because the whole AF is equal to the whole AG,

of which the parts AB, AC, are equal ; therefore the remainder is equal to the remainder CG; (ax. 3.)

and FC has been proved to be equal to GB; hence, because the two sides BF, FC are equal to the two CG, GB,

each to each; and the angle BFC has been proved to be ual to the angle CGB, also the base BC is common to the two triangles BFC, CGB;

wherefore these triangles are equal, (I. 4.) and their remaining angles, each to each, to which the equal sides are opposite; therefore the angle FBC is equal to the angle GCB,

and the angle BCF to the angle CBG. And, since it has been demonstrated,

that the whole angle ABG is equal to the whole ACF,

the parts of which, the angles CBG, BCF are also equal; therefore the remaining angle A B C'is equal to the remaining angle ACB,

which are the angles at the base of the triangle ÅBČ;
and it has also 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. Q.E.D.
COR. Hence an equilateral triangle is also equiangular.

PROPOSITION VI. THEOREM,
If tro angles of a triangle be equal to each other; 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 ACB.

Then the side AB shall be equal to the side AC.

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For, if AB be not equal to AC, one of them is greater than the other.

If possible, let AB be greater than AC; and from BA cut off BD equal to CA the less, (1. 3.) and join DC.

Then, in the triangles DBC, ABC, because DB is equal to AC, and BC is common to both triangles, the two sides DB, B C are equal to the two sides AC, CB, each to each ;

and the angle DBC is equal to the angle ACB; (hyp:)
therefore the base DC is equal to the base AB, (1. 4.)
and the triangle DBC is equal to the triangle ABC,

the less equal to the greater, which is absurd. (ax. 9.) Therefore AB is not unequal to A C, that is, AB is equal to AC

Wherefore, if two angles, &c. Q.E.D.
COR. Hence an equiangular triangle is also equilateral.

PROPOSITION VII. THEOREM. Upon the same base, and on the same side of it, there cannot be tro triangles that have their sides which are terminated in one extremity of the base, equal to one another, and likewise those which are terminated in the other extremity.

If it be possible, on the same base AB, and upon the same side of it, let there be two triangles ACB, ADB, which have their sides CA, DA, terminated in the extremity A of the base, equal to one another, and likewise their sides CB, DB, that are terminated in B.

C D

B

Join CD. First. When the vertex of each of the triangles is without the other triangle.

Because AC is equal to AD in the triangle ACD, therefore the angle ADC is equal to the angle ACD; (1. 5.) but the angle ACD is greater than the angle BCD; (ax. 9.)

therefore also the angle ADC is greater than BCD;

much more therefore is the angle BDC greater than BCD. Again, because the side B C is equal to BD in the triangle BCD, (hyp.)

therefore the angle BDC is equal to the angle BCD; (1. 5.)

but the angle BDC was proved greater than the angle BCD, hence the angle BDC is both equal to, and greater than the angle BCD;

which is impossible. Secondly. Let the vertex D of the triangle ADB fall within the triangle ACB.

E с

F

B

Produce AC to E, and AD to F, and join CD.

Then because AC is equal to AD in the triangle ACD, therefore the angles ECD, FDC upon the other side of the base CD, are equal to one another; (1. 5.)

but the angle ECD is greater than the angle BCD; (ax. 9.) therefore also the angle FDC is greater than the angle BCD; much more then is the angle BDČ greater than the angle BCD.

Again, because BC is equal to BD in the triangle BCD,
therefore the angle BDC is equal to the angle BCD, (1.

5.) but the angle BDC has been proved greater than BCD, wherefore the angle BDC is both equal to, and greater than the

angle BCD; which is impossible. Thirdly. The case in which the vertex of one triangle is upon a side of the other, needs no demonstration. Therefore, upon the same base and on the same side of it, &c. Q.E.D.

PROPOSITION VIII. THEOREM. If two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise their bases equal; the angle which is contained by the two sides of the one shall be equal to the angle contained by the two sides equal to them, of the other.

Let ABC, DEF be two triangles, having the two sides AB, AC, equal to the two sides DE, DF, each to each, viz. AB to DE, and À C to DF, and also the base B C equal to the base EF.

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Then the arigle BAC shall be equal to the angle EDF.

For, if the triangle ABC be applied to DEF, 80 that the point B be on E, and the straight line BC on EF;

then because BC is equal to EF, (hyp.) therefore the point C shall coincide with the point F.

wherefore BC coinciding with EF,

BA and AC shall coincide with ED, DF; for, if the base B C coincide with the base EF, but the sides BA, AC, do not coincide with the sides ED, DF, but have a different situation as EG, GF:

then, upon the same base, and upon the same side of it, there can he two triangles which have their sides which are terminated in one extremity of the base, equal to one another, and likewise those sides, which are terminated in the other extremity; but this is impossible. (1.7.)

Therefore, if the base BC coincide with the base EF, the sides BA, AC cannot but coincide with the sides ED, DF; wherefore likewise the angle BAC coincides with the angle EDF, and is equal to it. (ax. 8.)

Therefore if two triangles have two sides, &c. Q.E.D.

PROPOSITION IX. PROBLEM. To bisect a given rectilineal angle, that is, to divide it into two equal angles.

Let BAC be the given rectilineal angle.

is required to bisect it.

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In AB take any point D;
from A C cut off AE equal to AD, (1. 3.) and join DE;

on the side of DE remote from A,
describe the equilateral triangle DEF (1. 1.), and join AF.
Then the straight line AF shall bisect the angle BAC.

Because AD is equal to AE, (constr.) and AF is common to the two triangles DAF, EAF; the two sides DA, AF, are equal to the two sides EA, AF, each to each;

and the base DF is equal to the base EF: (constr.) therefore the angle DAF is equal to the angle EAF. (1. 8.) Wherefore the angle BAC is bisected by the straight line AF. Q.E.F.

PROPOSITION X. PROBLEM. To bisect a given finite straight line, that is, to divide it into two equal parts.

Let AB be the given straight line.

It is required to divide AB into two equal parts. Upon A B describe the equilateral triangle ABC; (I. 1.)

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and hisect the angle ACB by the straight line CD meeting A B in the point D. (1. 9.) Then AB shall be cut into two equal parts in the point D.

Because AC is equal to CB, (constr.) and CD is common to the two triangles ACD, BCD; the two sides AC, CD are equal to the two BC, CD, each to each ;

and the angle ACD is equal to BCD; (constr.)

therefore the base AD is equal to the base BD. (1. 4.) Wherefore the straight line AB is divided into two equal parts in the

point D.

Q.E.F.

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