Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

Then because AD is equal to AC, (constr.)

therefore the angle ACD is equal to the angle ADC; (1. 5.)
but the angle BCD is greater than the angle ACD; (ax. 9.)
therefore also the angle BCD is greater than the angle ADC.
And because in the triangle DBC,

the angle BCD is greater than the angle BDC,

and that the greater angle is subtended by the greater side; (1. 19.)
therefore the side DB is greater than the side BC;
but DB is equal to BA and AC,

therefore the sides BA and AC are greater than BC.
In the same manner it may be demonstrated,
that the sides AB, BC are greater than CA;
also that BC, CA are greater than AB.
Therefore any two sides, &c. Q. E. D.

PROPOSITION XXI. THEOREM.

If from the ends of a side 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.

Let ABC be a triangle, and from the points B, C, the ends of the side BC, let the two straight lines BD, CD be drawn to a point D within the triangle.

Then BD and DC shall be less than BA and AC the other two sides of the triangle,

but shall contain an angle BDC greater than the angle BAC.

[merged small][graphic][subsumed][merged small][merged small]

Produce BD to meet the side AC in E.

Because two sides of a triangle are greater than the third side, (1. 20.) therefore the two sides BA, AE of the triangle ABE are greater than BE;

to each of these unequals add EC;

therefore the sides BA, AC are greater than BE, EC. (ax. 4.) Again, because the two sides CE, ED of the triangle CED are greater than DC; (1. 20.)

add DB to each of these unequals ;

therefore the sides CE, EB are greater than CD, DB. (ax. 4.) But it has been shewn that BA, AC are greater than BE, EC; much more then are BA, AC greater than BD, DC. Again, because the exterior angle of a triangle is greater than the interior and opposite angle; (I. 16.)

therefore the exterior angle BDC of the triangle CDE is greater than the interior and opposite angle CED;

for the same reason, the exterior angle CED of the triangle ABE is greater than the interior and opposite angle BAC;

and it has been demonstrated,

that the angle BDC is greater than the angle CEB; much more therefore is the angle BDC greater than the angle BAC. Therefore, if from the ends of the side, &c. Q. E. D.

H

PROPOSITION XXII. PROBLEM.

To make a triangle of which the sides shall be equal to three given straight lines, but any two whatever of these must be greater than the third. Let A, B, C be the three given straight lines,

of which any two whatever are greater than the third, (1. 20.) namely, A and B greater than C;

A and C greater than B;

and B and C greater than A.

It is required to make a triangle of which the sides shall be equal to A, B, C, each to each.

[graphic][merged small][subsumed][merged small][merged small][subsumed][merged small][merged small]

Take a straight line DE terminated at the point D, but unlimited towards E,

make DF equal to A, FG equal to B, and GH equal to C; (1. 3.) from the center F, at the distance FD, describe the circle DKL; (post. 3.)

from the center G, at the distance GH, describe the circle HLK; from K where the circles cut each other draw KF, KG to the points F, G;

then the triangle KFG shall have its sides equal to the three straight lines A, B, C.

Because the point Fis the center of the circle DKL,
therefore FD is equal to FK; (def. 15.)

but FD is equal to the straight line A;
therefore FK is equal to A.

Again, because G is the center of the circle HKL;
therefore GH is equal to GK, (def. 15.)
but GH is equal to C;

therefore also GK is equal to C'; (ax. 1.)
and FG is equal to B;

therefore the three straight lines KF, FG, GK, are respectively equal to the three, A, B, C:

and therefore the triangle KFG has its three sides KF, FG, GK, equal to the three given straight lines A, B, C.

Q. E. F.

PROPOSITION XXIII. PROBLEM.

At a given point in a given straight line, to make a rectilineal angle equal to a given rectilineal angle.

Let AB be the given straight line, and A the given point in it, and DCE the given rectilineal angle.

It is required, at the given point A in the given straight line AB, to make an angle that shall be equal to the given rectilineal angle DCE.

[blocks in formation]

In CD, CE, take any points D, E, and join DE;

on AB, make the triangle AFG, the sides of which shall be equal to the three straight lines CD, DE, EC, so that AF be equal to CD, AG to CE, and FG to DE. (1. 22.)

Then the angle FAG shall be equal to the angle DCE.
Because FA, AG are equal to DC, CE, each to each,

and the base FG is equal to the base DE;

therefore the angle FAG is equal to the angle DCE. (1. 8.) Wherefore, at the given point A in the given straight line AB, the angle FAG is made equal to the given rectilineal angle DCE.

PROPOSITION XXIV. THEOREM.

Q.E.F.

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

Let ABC, DEF be two triangles, which have the two sides AB, AC, equal to the two DE, DF, each to each, namely, AB equal to DE, and AC to DF; but the angle BAC greater than the angle EDF. Then the base BC shall be greater than the base EF.

[ocr errors][merged small][merged small][merged small][merged small][merged small][merged small]

Of the two sides DE, DF, let DE be not greater than DF, at the point D, in the line DE, and on the same side of it as DF, make the angle EDG equal to the angle BAC; (1. 23.) make DG equal to DF or AC, (1. 3.) and join EG, GF. Then, because DE is equal to AB, and DG to AC,

the two sides DE, DG are equal to the two AB, AC, each to each, and the angle EDG is equal to the angle BAC;

therefore the base EG is equal to the base BC. (1. 4.) And because DG is equal to DF in the triangle DFG, therefore the angle DFG is equal to the angle DGF; (1. 5.) but the angle DGF is greater than the angle EGF; (ax.9.) therefore the angle DFG is also greater than the angle EGF; much more therefore is the angle EFG greater than the angle EGF. And because in the triangle EFG, the angle EFG is greater than the angle EGF,

and that the greater angle is subtended by the greater side; (1. 19.) therefore the side EG is greater than the side EF;

but EG was proved equal to BC;

therefore BC is greater than EF. Wherefore, if two triangles, &c. Q. E. P

PROPOSITION XXV. THEOREM.

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

Let ABC, DEF be two triangles which have the two sides AB, AC, equal to the two sides DE, DF, each to each, namely, AB equal to DE, and AC to DF; but the base BC greater than the base EF. Then the angle BAC shall be greater than the angle EDF.

[blocks in formation]

For, if the angle BAC be not greater than the angle EDF,
it must either be equal to it, or less than it.

If the angle BAC were equal to the angle EDF,
then the base BG would be equal to the base EF'; (1. 4.)
but it is not equal, (hyp.)

therefore the angle BAC is not equal to the angle EDF.
Again, if the angle BAC were less than the angle EDF,
then the base BC would be less than the base EF; (1. 24.)
but it is not less, (hyp.)

therefore the angle BAC is not less than the angle EDF; and it has been shewn, that the angle BAC is not equal to the angle EDF; therefore the angle BAC is greater than the angle EDF. Wherefore, if two triangles, &c.

Q. E. D.

[blocks in formation]

If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, viz. either the sides adjacent to the equal angles in each, or the sides opposite to them; then shall the other sides be equal, each to each, and also the third angle of the one equal to the third angle of the other.

Let ABC, DEF be two triangles which have the angles ABC, BCA, equal to the angles DEF, EFD, each to each, namely, ABC to DEF, and BCA to EFD; also one side equal to one side.

First, let those sides be equal which are adjacent to the angles tha

are equal in the two triangles, namely, BC to EF

Then the other sides shall be equal, each to each, namely, AB ta DE, and AC to DF, and the third angle BAC to the third angle FDF.

[blocks in formation]

For, if AB be not equal to DE,
one of them must be greater than the other.
If possible, let AB be greater than DE,
make BG equal to ED, (1. 3.) and join GC..
Then in the two triangles GBC, DEF,

because GB is equal to DE, and BC to EF, (hyp.) the two sides GB, BC are equal to the two DE, EF, each to each;; and the angle GBC is equal to the angle DEF; therefore the base GC is equal to the base DF, (1. 4.) and the triangle GBC to the triangle DEF,

and the other angles to the other angles, each to each, to which the equal sides are opposite;

therefore the angle GCB is equal to the angle DFE;

but the angle ACB is, by the hypothesis, equal to the angle DFE; wherefore also the angle GCB is equal to the angle ACB; (ax. 1.) the less angle equal to the greater, which is impossible; therefore AB is not unequal to DE,

that is, AB is equal to DE.

Hence, in the triangles ABC, DEF;

because AB is equal to DE, and BC to EF, (hyp.) and the angle ABC is equal to the angle DEF; (hyp.) therefore the base AC is equal to the base DF, (1. 4.) and the third angle BAC to the third angle EDF. Secondly, let the sides which are opposite to one of the equal angle in each triangle be equal to one another, namely, AB equal to DL Then in this case likewise the other sides shall be equal, AC to DF, and BC to EF, and also the third angle BAC to the third angle EDF.

[blocks in formation]

For if BC be not equal to EF,

one of them must be greater than the other.
If possible, let BC be greater than EF;
make BH equal to EF, (I. 3.) and join AH.
Then in the two triangles ABH, DEF,
because AB is equal to DE, and BH to EF,
and the angle ABH to the angle DEF; (hyp.)
therefore the base AH is equal to the base DF, (1. 4.)
and the triangle ABH to the triangle DEF,

and the other angles to the other angles, each to each, to which the equal sides are opposite;

therefore the angle BHA is equal to the angle EFD;
but the angle EFD is equal to the angle BCA; (hyp.)
therefore the angle BHA is equal to the angle BCA, (ax. 1.)
that is, the exterior angle BHA of the triangle AHC, is
equal to its interior and opposite angle BCA;
which is impossible; (I. 16.)

wherefore BC is not unequal to EF,
that is, BC is equal to EF.
Hence, in the triangles ABC, DEF;

because AB is equal to DE, and BC to EF, (hyp.)

« ΠροηγούμενηΣυνέχεια »