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Let A be the given point, and BC the given straight line. It is required to draw, from the point A, a straight line equal to BC.

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Construction. From the point A to B draw the straight line AB (Post. 1); upon AB describe the equilateral triangle ABD (I. 1), and produce the straight lines DA, DB to E and F (Post. 2); from the centre B, at the distance BC, describe the circle CGH (Post. 3), cutting DF in the point G; and from the centre D, at the distance DG, describe the circle GKL, cutting AE in the point L.

Then the straight line AL shall be equal to BC.

Proof. Because the point B is the centre of the circle CGH, 1. BC is equal to BG (Def. 15) ;

and because D is the centre of the circle GKL,

2. DL is equal to DG,

and DA, DB, parts of them are equal (I. 1); therefore

3. The remainder AL, is equal to the remainder BG (Ax. 3). But it has been shown that BC is equal to BG; wherefore, AL and BC are each of them equal to BG; and things which are equal to the same thing are equal to one another (Ax. 1); therefore

4. The straight line AL is equal to BC.

Wherefore, from the given point A, a straight line AL has been drawn equal to the given straight line BC.

Which was to be done.

PROPOSITION 3.- Problem.

From the greater of two given straight lines to cut off a part equal to the less.

Let AB and C be the two given straight lines, of which AB is the greater. It is required to cut off from AB the greater, a part equal to C the less.

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Construction. From the point A draw the straight line AD equal to C (I. 2); and from the centre A, at the distance AD, describe the circle DEF (Post. 3), cutting AB in the point E.

Then AE shall be equal to C.

Proof. Because A is the centre of the circle DEF,

1. AE is equal to AD (Def. 15).

But the straight line C is equal to AD (constr.); whence AE and C are each of them equal to AD; wherefore

2. The straight line AE is equal to C (Ax. 1).

And therefore from AB, the greater of two straight lines, a part AE has been cut off equal to C the less. Q.E.F.*

PROPOSITION 4.-Theorem.

If two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise the angles contained by those sides equal to each other; (1) they shall have their bases, or third sides, equal; (2) and the two triangles shall be equal; (3) and their other angles shall be equal, each to each-viz., those to which the equal sides are opposite.

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 to DE, and AC to DF, and the included angle BAC equal to the included angle EDF.

Then (1) shall the base BC be equal to the base EF; and (2) the triangle ABC to the triangle DEF; and (3) 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 DFE.

* An abbreviation for quod erat faciendum—that is, "which was to be done."

Demonstration. For, if the triangle ABC be applied to the triangle DEF, so that the point A may be on D, and the straight line AB on DE; then

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1. The point B shall coincide with the point E, because AB is equal to DE; and AB coinciding with DE, 2. The straight line AC shall fall on DF,

because the angle BAC is equal to the angle EDF; therefore also 3. The point C shall coincide with the point F,

because AC is equal to DF. But the point B was shown to coincide with the point E; wherefore, the base BC 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 inclose a space which is impossible (Ax. 10);

therefore

4. The base BC does coincide with EF, and is equal to it,

and

5. 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.,

6. The angle ABC is equal to the angle DEF, and the angle ACB to the angle DFE.

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

PROPOSITION 5.-Theorem.

The angles at the base of an isosceles triangle are equal to each other; and if the equal sides be produced, the angles on the other side of the base shall be equal.

Let 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.

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Construction. In BD take any point F; and from AE the greater, cut off AG equal to AF the less (I. 3), and join FC, GB.

Demonstration. 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; therefore

1. The base FC is equal to the base GB (I. 4), and the triangle AFC to the triangle AGB;

and 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.

2. The angle ACF is equal 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

3. The remainder BF 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 equal to the angle CGB, also the base BC is common to the two triangles BFC, CGB; wherefore

4. The triangles BFC, CGB are equal (I. 4),

and their remaining angles are equal, each to each, to which the equal sides are opposite; therefore

5. 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

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equal to the whole ACF, the parts of which, the angles CBG, BCF, are also equal; therefore,

6. The remaining angle ABC is equal to the remaining angle ACB,

which are the angles at the base of the triangle ABC. 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.* Corollary. Hence an equilateral triangle is also equiangular.

PROPOSITION 6.-Theorem.

If two 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.

Construction. 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 (I. 3), and join DC.

*An abbreviation for quod erat demonstrandum—that is, “which was to be demonstrated or proved."

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