Sidebilder
PDF
ePub
[ocr errors]

rior angle ACD is greater than either of the interior opposite angles CBA, BAC.

Join B and E the middle point of the line AC, and produce BE to F,

А. and make EF equal to BE; join also FC, and produce AC to G.

Because AE is equal to EC, and BE to EF; AE, EB are equal to CE, EF, each to each; and the angle

E AEB is equal (8. 1.) to the angle CEF, because they are vertical angles; therefore the base AB is equal (1. 1.) to the base CF, and the trian

B

D gle AEB to the triangle CEF, and the remaining angles to the remaining angles each to each, to which the equal sides are opposite ; wherefore the angle BAE is equal to the angle

Ġ ECF; but the angle ECD is greater than the angle ECF; therefore the angle ECD, that is ACD, is greater than BAE: In the same manner, if the side BC be bisected, it may be demonstrated that the angle BCG, that is (8. 1.) the angle ACD, is greater than the angle ABC.

[blocks in formation]

Any two angles of a triangle are together less than two right angles. Let ABC be any triangle; any two of its angles together are less than two right angles.

Produce BC to D; and because ACD is the exterior angle of the triangle ABC, ACD is greater (9. 1.) than the interior and opposite angle

А ABC; to each of these add the angle ACB; therefore the angles ACD, ACB are greater than the angles ABC, ACB ;' but ACD, ACB are together equal (6.1.) to two right angles: therefore the angles ABC, BCA are less than two right angles. In like manner, it may be demonstrated, that BAC, ACB, as also CAB, ABC, are less than two right angles.

B

D

PROP. XI. THEOR.

The greater side of every triangle has the greater angle opposite to it. Let ABC be a triangle of which the side AC is greater than the side

A

AB; the angle ABC is also greater than the angle BCA.

Make AD equal to AB, and join BD: and because ADB is the exterior angle of the triangle BDC, it is greater (9. 1.)

D than the interior and opposite angle DCB; but ADB is equal (3. 1.) to ABD, because the side AB is equal to the side.

B AD; therefore the angle ABD is likewise greater than the angle ACB; wherefore much more is the angle ABC greater than ACB.

PROP. XII. THEOR.

The greater angle of every triangle is subtended by the greater side, or has

the greater side opposite to it. Let ABC be a triangle, of which the angle ABC is greater than the angle BCA ; the side AČ is likewise greater than the side AB.

For, if it be not greater, AC must either be equal to AB, or less than it; it is not A equal, because then the angle ABC would be equal (3. 1.) to the angle ACB; but it is not; therefore AC is not equal to AB; neither is it less; because then the angle ABC would be less (11. 1.) than

d the angle ACB; but it is not; therefore B the side AC is not less than AB; and it has been shewn that it is not equal to AB; therefore AC is greater than AB.

PROP. XIII. THEOR.

Any two sides of a triangle are together greater than the third side. Let ABC be a triangle; any two sides of it together are greater than the third side, viz. the sides BA, AC greater than the side BC; and AB, BC greater than AC; and BC, CA greater than AB. Produce BA to the point D, and

D make AD equal to AC; and join DC.

Because DA is equal to AC, the angle ADC is likewise equal (3. 1.) to AČD; but the angle BCD is greater than the angle ACD; therefore the angle BCD is greater than the angle ADC; and because the angle BCD of the triangle DCB is greater than its

B

С angle BDC, and that the greater (12. 1.) side is opposite to the greater angle: therefore the side DB is greater than the side BC; but DB is equal to BA and AC together; therefore BA and AC together are greater than

BC. In the same manner it may be demonstrated, that the sides AB, BC are greater than CA, and BC, CĂ greater than AB.

SCHOLIUM.
This

may be demonstrated without producing any of the sides : thus, the line BC, for example, is the shortest distance from B to C; therefore BC is less than BA+AC, or BA+AC>BC.

[blocks in formation]
[ocr errors]

If from the ends of one side of a triangle, there be drawn two straight lines to

a point within the triangle, these two lines shall be less than the other two sides of the triangle, but shall contain a greater angle.

Let the two straight lines BD, CD be drawn from B, C, the ends of the
side BC of the triangle ABC, to the point D within it; BD and DC are
less than the other two sides BA, AC of the triangle, but contain an angle
BDC greater than the angle BAC.
Produce BD to E; and because two sides

A
of a triangle (13. 1.) are greater than the
third side, the two sides BA, AE of the trian-
gle ABE are greater than BE. To each

D
of these add EC ; therefore the sides BA,
AC are greater than BE, EC: Again, be-
cause the two sides CE, ED, of the triangle
CED are greater than CD, if DB be added
to each, the sides CE, EB, will be greater
than CD, DB ; 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 (9. 1.) is greater than the interior and opposite angle, the exterior angle BDC of the triangle CDE is greater than CED; for the same reason, the exterior angle CEB of the triangle ABE is greater than BAC ; and it has been demonstrated that the angle BDC is greater than the angle CEB; much more then is the angle BDC greater than the anglo BAC.

PROP. XV. THEOR.

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 the one greater than the angle contained by the two sides 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, viz. AB equal to DE, and AC to DF; but the angle BAC greater than the angle EDF; the base BC is also greater than the base EF.

Of the two sides DE, DF, let DE be the side which is not greater than the other, and at the point D, in the straight line DE, make the angle

EDG equal to the angle BAC: and make DG equal to AC or DF, and join EG, GF Because AB is equal

А

D to DE, and AC to DG, the two sides BA, AC are equal to the two ED, DG, each to each, and the angle BAC is equal to the angle EDG, therefore the base BC is equal (1. 1.) to the base EG; and because DG is equal to DF, the angle DFG is equal B

C E (3.1.) to the angle DGF; but the angle DGF is greater than the angle EGF; therefore the angle DFG is greater than EGF; and much more is the angle EFG greater than the angle EGF; and because the angle EFG of the triangle EFG is greater than its angle EGF, and because the greater (12. 1.) side is opposite to the greater angle, the side EG is greater than the side EF; but EG is equal to BC; and therefore also BC is greater than EF.

[graphic]

If two triangles have two sides of the one equal to two sides of the other, each

to each, but the base of the one greater than the base of the other; the angle contained by the sides of that which has the greater base, shall be greater than the angle contained by the sides 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, viz. AB equal to DE, and AC to DF: but let the base CB be greater than the base EF, the angle BAC is likewise greater than the angle EDF. For, if it be not greater, it must

D either be equal to it, or less;

A

but the angle BAC is not equal to the angle EDF, because then the base BC would be equal (1. 1.) to EF;

but it is not ; therefore the angle BAC is not equal to the angle EDF; neither is it less; because then the base BC would be less (15. 1.) than the base EF; but it is not; therefore the angle

B

C BAC is not less than the angle EDF: and it was shewn that it is not equal to it: therefore the angle BAC is greater than the angle EDF.

[ocr errors]

1

PROP. XVII. THEOR.

A perpendicular is the shortest line that can be drawn from a point, situated with

out a straight line, to that line : any two oblique lines drawn from the same point on different sides of the perpendicular, cutting off equal distances on the other line, will be equal; and any two other oblique lines, cutting off unequal distances, the one which lies farther from the perpendicular will be the longer.

If AB, AC, AD, &c. be lines drawn from the given point A, to the indefinite straight line DE, of which AB is perpendicular ; then shall the perpendicular AB be less than AC, and AC less than AD, and so on.

For, the angle ABC being a right one, the angle ACB is acute, (Th. 10.) or less than the angle ABC. But the less angle of a triangle is subtended by the less side (Th. 12.); therefore, the side AB is less than the side AC.

Again, if BC=BE; then the two oblique lines AC, AE, are equal. For the side AB is common to the two triangles ABC, ABE, and the contained D

C B angles ABC and ABE equal; the two triangles must be equal (Th. 1.); hence AE, AC are equal.

Finally, the angle ACB being acute, as before, the adjacent angle ACD will be obtuse ; since (Th. 6.) these two angles are together equal to two right angles ; and the angle ADC is acute, because the angle ABD is right; consequently, the angle ACD is greater than the angle ADC; and, since the greater side is opposite to the greater angle (Th. 12.); therefore, the side AD is greater than the side AC.

Cor. 1. The perpendicular measures the true distance of a point from a line, because it is shorter than any

other distance. Cor. 2. Hence, also, every point in a perpendicular at the middle point of a given straight line, is equally distant from the extremities of that line.

Cor. 3. From the same point, three equal straight lines cannot be drawn to the same straight line; for if there could, we should have two equal oblique lines on the same side of the perpendicular, which is impossible.

PROP. XVIII. THEOR.

When the hypotenuse and one side of a right angled triangle, are respectively

equal to the hypotenuse and one side of another ; the two right angled triangles are equal.

Suppose the hypotenuse AC=DF, and the side AB=DE; the right angled triangle ABC will be equal to the right angled triangle DEF.

Their equality would be manifest, if the third sides BC and EF were equal. If possible, suppose that those sides are not equal, and that BC is the greater. Take BH=EF; and join AH. The triangle ABH=DEF;

« ForrigeFortsett »