Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

Ex. 2. Shew that the sum of the straight lines, joining the angles of a triangle with a point within the triangle, is less than the perimeter of the triangle, and greater than half the perimeter.

PROPOSITION XXII. PROBLEM.

To make a triangle, of which the sides shall be equal to three given straight lines, any two of which are together greater than the third.

[blocks in formation]

Let A, B, C be the three given lines, any two of which

are together greater than the third.

It is required to make a ▲ having its sides respectively.

Take a st. line DE of unlimited length.

In DE make DF=A, FG=B, and GH=C.
With centre F and distance FD, describe
With centre G and distance GH, describe

Join FK and GK.

=

A, B, C

I. 3.

DKL.
HKL.

Def. 13.

Def. 13.

Then A KFG has its sides A, B, C respectively.

=

For FK-FD;

.. FK=A; and GK=GH; .. GK=C; and FG=B;

.. a A KFG has been described as reqd.

Q. E. F.

Ex. 1. Draw an isosceles triangle having each of the equal

sides double of the base.

[blocks in formation]

At a given point in a given straight line, to make an angle equal to a given angle.

[blocks in formation]

Let A be the given pt., BC the given line, DEF the given 4.

It is reqd. to make at pt. A an angle
= L DEF.
In ED, EF take any pts. D. F; and join DF.
In AB, produced if necessary, make AG=DE.
In AC, produced if necessary, make AH=EF.
In HC, produced if necessary, make HK=FD.

With centre A, and distance AG, describe ✪ GLM.
With eentre H, and distance HK, describe o LKM.

[blocks in formation]

and ·.· HL=HK, .. HL=FD.

Then in AS LAH, DEF,

· LA=DE, and AH-EF, and HL=FD;

.. ¿ LAH= L DEF.

.. an angle LAH has been made at pt. A as was reqd.

Ax. 1.

Ax. 1.

I. c.

Q. E. F.

NOTE.-We here give the proof of a theorem, necessary to the proof of Prop. XXIV. and applicable to several propositions in Book III.

PROPOSITION D. THEOREM.

Every straight line, drawn from the vertex of a triangle to the base, is less than the greater of the two sides, or than either, if they be equal.

B

In the ▲ ABC, let the side AC be not less than AB.

Take any pt. D in BC, and join AD.

Then must AD be less than AC.

For AC is not less than AB;

.. ABD is not less than ▲ ACD.

I. A. and 18.

I. 16.

But ADC is greater than ▲ ABD ;

.. ADC is greater than ▲ ACD ;

.. AC is greater than AD.

I. 19.

Q. E. D.

[blocks in formation]

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 one of them greater than the angle contained by the two sides equal to them of the other; the base of that which has the greater angle must be greater than the base of the other.

[blocks in formation]

Then must BC be greater than EF.

Of the two sides DE, DF let DE be not greater than DF.* At pt. D in st. line ED make ▲ EDG= 2 BAC,

and make DG=AC or DF, and join EG, GF.

I. 23.

Then. AB=DE, and AC=DG, and ▲ BAC=▲ EDG,

Again,

..BC=EG,

:: DG=DF,

:. ▲ DFG= L DGF ;

I. 4.

I. A.

[ocr errors]

:: ▲ EFG is greater than ▲ DGF; much more then ▲ EFG is greater than ▲ EGF;

.. EG is greater than EF.

But EG=BC;

.. BC is greater than EF.

I. 19.

Q. E. D.

proof.

*This line was added by Simson to obviate a defect in Euclid's Without this condition, three distinct cases must be discussed. With the condition, we can prove that F must lie below EG.

For since DF is not less than DE, and DG is drawn equal to DF, DG is not less than DE.

Hence by Prop. D, any line drawn from D to meet EG is less than DG, and therefore DF, being equal to DG, must extend beyond EG.

For another method of proving the Proposition, see p. 113.

[blocks in formation]

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 also, contained by the sides of that which has the greater base, must be greater than the angle contained by the sides equal to them of the other.

[blocks in formation]

In the As ABC, DEF,

let AB=DE and AC=DF,

and let BC be greater than EF.

Then must ▲ BAC be greater than ▲ EDF.

For BAC is greater than, equal to, or less than ▲ EDF. Now BAC cannot ▲ EDF,

=

for then, by 1. 4, BC would=EF; which is not the case. And BAC cannot be less than ▲ EDF,

for then, by 1. 24, BC would be less than EF; which is not the case;

.. BAC must be greater than ▲ EDF.

Q. E. D.

NOTE.-In Prop. xxvI. Euclid includes two cases, in which two triangles are equal in all respects; viz., when the following parts are equal in the two triangles :

1. Two angles and the side between them.

2. Two angles and the side opposite one of them.

Of these we have already proved the first case, in Prop. B, so that we have only the second case left, to form the subject of Prop. XXVI., which we shall prove by the method of superposition.

For Euclid's proof of Prop. XXVI., see rn. 114-115.

« ΠροηγούμενηΣυνέχεια »