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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 FĀG 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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F

E

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.'

Demonstration. Then, in the triangles DBC, ABC, because DB is equal to AC, and BC is common to both triangles,

А.

B

1. The two sides DB, BC are equal to the two sides AC, CB,

each to each ; and the angle DBC is equal to the angle ACB (hyp.), therefore

2. The base DC is equal to the base AB (I. 4), and the triangle

DBC is equal to the triangle ABC, the less equal to the greater, which is absurd (Ax. 9). Therefore

3. AB is not unequal to AC; that is, AB is equal to AC.

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

PROPOSITION 7.-Theorem. Upon the same base, and on the same side of it, there cannot be two 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.

First. Let the vertex of each of the triangles be without the other. Construction. Join CD.

Demonstration. Because AC is equal to AD in the triangle ACD; therefore,

B

1. The angle ACD is equal to the angle ADC (I. 5). But the angle ACD is greater than the angle BCD (Ax. 9); therefore also

2. The angle ADC is greater than BCD; much more therefore

3. The angle BDC is greater than the angle BCD. Again, because the side BC is equal to BD in the triangle BCD (hyp.), therefore

4. The angle BDC is equal to the angle BCD (I. 5). But the angle BDC was proved greater than the angle BCD, he 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 AČB.

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

Demonstration. Then, because AC is equal to AD in the triangle ACD, the angles upon the other side of the base CD, namely,

1. The angles ECD, FDC are equal to one another (I. 5); but the angle ECD greater than the angle BCD (Ax. 9); therefore also

2. The angle FDC is greater than the angle BCD; much more then

3. The angle BDC is greater than the angle BCD. Again, because BC is equal to BD in the triangle BCD, therefore

4. The angle BDC is equal to the angle BCD (I. 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 8.-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 ĀC to DF; and also the base BC equal to the base EF.

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

Demonstration. For, if the triangle ABC be applied to DEF, so that the point B be on E, and the straight line BC on EF; then, because BC is equal to EF (hyp.), therefore

B

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