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PROPOSITION IV. 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 one another, they shall also have their bases or third sides equal ; and the two triangles shall be equal, and their other angles shall be equal, each to each, namely those to which the equal sides are opposite.

Let ABC, DEF be two as which have the two sides AB, AC equal to the two sides DE, DF, each to each, namely AB to DE, and AC to DF, and the included Z BAC equal to the included LEDF.

Then shall the base BC be equal to the base EF, and A ABC to A DEF, and the other 2 s, to which the equal sides are opposite, shall be equal, each to each, namely

ABC to 2 DEF, and L ACB to _DFE.

[blocks in formation]

Dem. For if A ABC be applied to A DEF,
So that the pt. A may be on D, and the st. line AB on
DE;
.: AB = DE,

(hyp.) .. the pt. B shall coincide with the pt. E; and AB coinciding with DE, : _ BAC= _EDF, (hyp:)

.. the st. line AC shall fall on DF;

also .. AC = DF,
.. the pt. C shall coincide with F;

but the pt. B coincides with E;
.. the base BC shall coincide with the base EF;

for, the pt. B coinciding with E, and C with F, if the base BC do not coincide with the base EF, the two st. lines BC and EF would enclose a space, which is impossible.

(ax. 10) .. the base BC does coincide with the base EF, and is equal to it.

(ax. 8) Wherefore the whole A ABC coincides with the whole A DEF, and is equal to it;

(ax. 8) and the other Zs of the one coincide with the other Ls

of the other, and are equal to them, namely L ABC to _DEF, and LACB to _DFE. Therefore, if two triangles, &c.

Q. E. D.

[The hypothesis in this, the first theorem, is that two As have two sides and the included < of the one equal to two sides and the included L of the other, each to each; the conclusion is that their bases shall be equal, &c. The eighth axiom contains the principle of superposition, by which we conceive one figure to be placed on another, and prove that their boundaries exactly coincide, and we thence infer that the figures are in all respects equal. It may be regarded as the definition of geometrical equality, and it is first employed in this proposition.]

1. A straight line bisects the vertical angle of an isosceles triangle ; show that it also bisects the base.

2. ABDE, BFGC are squares on two sides of the triangle ABC, and AF, CD are joined; show that AF, CD are equal.

3. Two st. lines bisect one another at right angles ; show that any point in either of them is equidistant from the extremities of the other.

PROPOSITION V. THEOREM.

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

Let ABC be an isosceles A in which AB=AC, and let the equal sides AB, AC be produced to D and E.

Then shall Z ABC = LACB,

and < DBC= _ ECB.

[blocks in formation]

Constr. In BD take any pt. F; from AE the greater, cut off AG = AF the less, (i. 3)

join FC, GB. Dem. : AF= AG, (constr.) and AB = AC; (hyp.)

the two sides FA, AC = the two GA, AB, each to each, and they contain the _FAG common to the two As AFC, AGB;

.. base FC = base GB,
and A AFC = A AGB;

(i. 4) also the rem. Ls of the one = the rem. Ls of the other, each to each, to which the equal sides are opposite;

(i. 4) namely LACF = L ABG, and Z AFC= LAGB.

And :: the whole AF = the whole AG,
of which the part AB = the part AC, (hyp.)

.:. the remainder BF = the remainder CG; (ax. 3)

and it was proved that FC=GB; hence : the two sides BF, FC = the two CG, GB, each to each, and it was proved that BFC= _ CGB,

.. As BFC, CGB are equal; (i. 4) and their other <s, each to each, to which the = sides are opp.; namely _ FBC= _ GCB, and _BCF=_ CBG.

(i. 4) And since it has been demonstrated that

LABG=LACF, the parts of which, Ls CBG, BCF are also =

.. rem. _ABC= rem. LACB, (ax. 3)

which are <s at the base of A ABC; and it has been proved that FBC= _ GCB, which are 4s on the other side of the base. Therefore, the angles at the base, &c.

Q. E. D. Cor. Hence every equilateral triangle is also equiangular.

[Hypothesis, an isosceles A; conclusion that Zs at the base are equal, and if the equal sides be produced, the Zs on the other side of the base shall be equal.

Observe, the construction is simple enough, and the demonstration falls naturally into three parts : (1) As ABG,ACF are proved equal by prop. 4, (2) As BCF,BCG are also proved equal by prop. 4, (3) Zs ABC,ACB are proved equal by ax. 3.

A corollary (cor.) is an inference which at once follows from the demonstration of a proposition : in fact, it is a simple rider' (see preface). The student should not leave it without seeing clearly why it so follows.]

1. The middle point of the base of an isosceles triangle is joined to the vertex : show that the triangles so formed are equal in all respects.

2. The straight line which bisects the base of an isosceles triangle at right angles shall pass through the vertex.

PROPOSITION VI.

THEOREM.

If two angles of a triangle be equal to one another, the sides also which subtend, or are opposite to, the equal angles, shall be equal to one another.

Let ABC be A in which _ ABC= LACB.
Then shall AB = AC.

A

D

B В

Constr. For, if AB be not equal to AC, one of them is

greater than the other. Let AB be greater than AC; and at the pt. B, from BA cut off BD = CA the less,

(i. 3) and join DC. Dem. Then in As DBC, ABC, :: DB=AC and BC is

common, The two sides DB, BC = the two sides AC, CB, each to

each;

and _DBC= LACB; (hyp.) .. base DC= base AB, and A DBC=A ACB, (i. 4)

the less = the greater, which is absurd. (ax. 9) .. AB is not unequal to AC, that is, AB = AC. Wherefore, if two angles, &c.

Q. E. D. Cor. Hence every equiangular triangle is also equilateral.

[Hypothesis, two angles of a triangle are equal. Conclusion, the opposite sides are also equal. This is the converse of the first part of prop. 5. One prop. is said to be the converse of another when the conclusion of each is

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