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PROPOSITION XV.

The greater angle of every triangle has the greater side opposite to it.

B

Let ABC be a ▲ having the ABC > the ▲ ACB,

then shall the side AC be> the side AB.

For if not, AC must either be

=

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AC is not AB, for then would ▲ ABC be = ▲ ACB, but it is not.

(I. 2)

And AC is not < AB, for then would ▲ ABC be < LACB, but it is not.

(I. 14)

.. AC is > AB.

PROPOSITION XVI.

Any two sides of a triangle are together greater than the third.

Let ABC be a ▲.

Then shall any two of its sides, as AB, AC, be

together > BC.

B

Produce BA to K, making AK= AC, and join KC.

Then. AK is = AC; .. LACK is = LAKC; (1. 2)

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.. BA, AC are together > BC.

COR. Let two points A, B be joined by a path which is as short as any which can be drawn between them.

every point in this path shall lie in the straight line AB.

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Then

Then to AE, BF may be applied the portions of the path AQ, BQ, meeting in some point Z, and thus forming a path AZB between A and B shorter than the former, which is impossible.

Hence it follows that

the shortest distance between two points is a straight line.

PROPOSITION XVII.

If from the ends of one of the sides of a triangle there be drawn two straight lines to a point within the triangle: these shall be less than the other two sides of the triangle but shall contain a greater angle.

D

From the ends B, C of the side BC of the ▲ ABC let the two straight lines BD, CD be drawn to a point D within the a;

then shall BD, DC together be less than BA and AC,

but the

BDC shall be greater than ▲ BAC.

Produce BD to meet the side AC in E.

Then BA and AE are together greater than BE;

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.. BA and AC are together greater than BE and EC. Again, DE and EC are together greater than DC;

add to each BD;

(I. 16) .. BE and EC are together greater than BD and DC. Much more then are BA and AC greater than BD and DC.

Also, since ED of ▲ CED has been produced to B; .. L BDC is greater than ▲ DEC; - (I. 13) and. AE of ▲ ABE has been produced to C;

.. ▲ BEC is greater than ▲ BAE.

Much more then is ▲ BDC greater than ▲ BAE,
i.e. ▲ BAC.

PROPOSITION XVIII.

If two triangles have two sides of the one equal to two sides of the other; but the angle contained by the two sides of one of them greater than the angle contained by the two sides of the other; then shall the base of that which has the greater angle be greater than the base of the other.

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In the two As ABC, DEF,

let AB, AC be respectively equal to DE, DF,
but the BAC be > the ▲ EDF,

then shall BC be > EF.

Place the ▲ ABC so that AB shall fall on DE, the point C falling on Q on the same side of DE as F.

Ist.

Let EQ cut DF: join FQ.

Then. DF is = DQ, . L DFQ is = LDQF; (1.2)

and is .. >

EQF.

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Then EQ, QD are together > EF, FD, (1. 17) and QD is FD;

=

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If two triangles have two sides of the one equal to two sides of the other, but the base of one greater than the base of the other; then the angle contained by the sides of that triangle which has the greater base shall be greater than the angle contained by the sides equal to them of the other.

D

B

E

Let ABC, DEF be two as having the sides BA, AC respectively equal to the sides ED, DF;

but the base BC > than the base EF.

Then shall the BAC be > the EDF

For if the

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BAC were < the ¿ EDF,

then would the base BC be< the base EF, (1. 18) but it is not;

and if the BAC were the ▲ EDF,

then would the base BC be = the base EF, (1.4) but it is not;

... the BAC must be the EDF

L

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