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NOTE 2. On the Conditions of Equality of two Triangles.

1

A Triangle is composed of six parts, three sides and three angles.

When the six parts of one triangle are equal to the six parts of another triangle, each to each, the Triangles are said to be equal in all respects.

There are four cases in which Euclid proves that two triangles are equal in all respects ; viz., when the following parts are equal in the two triangles. 1. Two sides and the angle between them.

I. 4. 2. Two angles and the side between them.

I. 26. 3. The three sides of each.

I. 8. 4. Two angles and the side opposite one of them. I. 26. The Propositions, in which these cases are proved, are the most important in our First Section.

The first case we have proved in Prop. iv.

Availing ourselves of the method of superposition, we can prove Cases 2 and 3 by a process more simple than that employed by Euclid, and with the further advantage of bringing them into closer connexion with Case 1. We shall therefore give three Propositions, which we designate A, B, and C, in the Place of Euclid's Props. V. VI. VII. VIII.

The displaced Propositions will be found on pp. 108-112.
Proposition A corresponds with Euclid I. 5.
B

I. 26, first part.
o

I. 8.

PROPOSITION A. THEOREM.

If two sides of a triangle be equal, the angles opposite those

sides vust also be equal.

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In the isosceles triangle ABC, let AC=AB. (Fig. 1.)

Then must ABC= L ACB. Imagine the a ABC to be taken up, turned round, and set down again in a reversed position as in Fig. 2, and designate the angular points A', B', C'.

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Then in As ABC, A'C'B',
:: AB=A'C', and AC=A'B', and <BAC= _ C' A'B',
:: LABC= L A'C'B'.

I. 4. But

L A'C'B'= L ACB ;
.: LABC= L ACB.

Ax. 1.
Q.E.D.

Cor. Hence every equilateral triangle is also equiangular.

Note. When one side of a triangle is distinguished from the other sides by being called the Base, the angular point opposite to that side is called the Vertex of the triangle.

PROPOSITION B.. THEOREM. If two triangles have two angles of the one equal to two angles of the other, each to each, and the sides adjacent to the equal angles in each also equal ; then must the triangles be equal in all respects.

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In As ABC, DEF, let / ABC= _DEF, and LACB= _ DFE, and BC=EF. Then must AB=DE, and AC=DF, and 2 BAC= _ EDF.

For if - DEF be applied to ABC, so that E coincides with B, and EF falls on BC; then ·:: EF=BC, .. F will coincide with C; and :: 2 DEF= L ABC, .. ED will fall on BA ;

.: D will fall on BA or BA produced.
Again, :: 2 DFE= ACB, .. FD will fall on CA ;

.: D will fall on CA or CA produced. :D must coincide with A, the only pt. common to BA and CA.

.: DE will coincide with and .. is equal to AB, and DF.

AC, and 4 EDF.

BAC, and A DEF.

Δ ABC; and :. the triangles are equal in all respects.

Q. E. D.

Cor. Hence, by a process like that in Prop. A, we can prove the following theorem :

If two angles of a triangle be equal, the sides which subtend them are also equal. (Eucl. I. 6.)

PROPOSITION C. THEOREM. If two triangles have the three sides of the one equal to the three sides of the other, each to each, the triangles must be equal in all respects.

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Let the three sides of the As ABC, DEF be equal, each to each, that is, AB=DE, AC=DF, and BC=EF.

Then must the triangles be equal in all respects. Imagine the - DEF to be turned over and applied to the - ABC, in such a way that EF coincides with BC, and the vertex D falls on the side of BC opposite to the side on which A falls ; and join AD.

; Case I. When AD passes through BC.

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that is,

Then in 1 ABD, : BD=BA,.: _BAD= . BDA, I. A. And in A ACD, :: CD=CA, .. CAD= 2 CDA, I. A. .: sum of 2s BAD, CAD=sum of _s BDA, CDA, Ax. 2.

< BAC= _ BDC. Hence we see, referring to the original triangles, that

BAC= _EDF. ..., by Prop. 4, the triangles are equal in all respects.

CASE II. When the line joining the vertices does not pass through BC.

B

D

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Then in A ABD, :: BD=BA, .. _BAD= LBDA, I. A. And in A ACD, ;: CD=CA, :: 2 CAD= 2 CDA, I. A. Hence since the whole angles BAD, BDA are equal.

and parts of these CAD, CDA are equal.

.: the remainders BAC, BDC are equal. Ax. 3. Then, as in Case I., the equality of the original triangles may be proved.

CASE III. When AC and CD are in the same straight line.

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Then in ABD, :: BD=BA,.. BAD= 2 BDA, I. A.

that is, .BAC= .BDC. Then, as in Case I., the equality of the original triangles may be proved.

Q. E. D.

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