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PROPOSITION XXII.

PROBLEM. To describe a triangle of which the sides shall be equal to three given straight lines respectively; provided always, thal any two of these be together greater than the third.

Let A, B, C be the three given straight lines; of which any two are together greater than the third; (for if they are not, it has been *I. 20. Cor. shown* that they cannot form a triangle). It is required to make

+ I. 3.

*INTERC.13.

Cor. 5.

+Hyp.

K

F M

NG

E

ABC

a triangle, of which the sides shall be equal to A, B, C, respectively. In a straight line DE, maket FG equal to any one of the straight lines as A; FH equal to another as B; and GI equal to INTERC.13. C. About the centre F, with the radius FH, describe a circle; Cor. 3. and let it cut FE in the* point N. About the centre G, with the radius GI, describe another circle; and because FH and GI are together+ greater than FG, this last circle will cut GD in a point (as M) nearer to D than is the point N. For if M and N should coincide, FH and GI must be together equal to FG; and if M should lie on the other side of N, they must be together less than FG; neither of which is possible, for they are‡ greater. Also because FN (which is equal to FH) is* less than the sum of FG and GI, the point N will fall on the side of I which is towards D ; and because GM (which is equal to GI) is† less than the sum of GF and FH) the point M will fall on the side of H which is towards E. Wherefore the circumference of each of the circles will pass through that of the other; and in passing from the outside of it to the inside, and from the inside to the outside again, 1. 7. Cor. will necessarily cut it in two points. Let one of these be K. Join KF, KG. The triangle FKG has its sides equal to the three straight lines A, B, C.

+ Hyp.

Hyp.

+ Hyp.

*INTERC.13. Cor. 4.

+ Constr.

Cor. 1.

Because the point F is the centre of the circle HKL, FK is* equal to FH. But FH is† equal to the straight line B; thereINTERC. 1. fore FK ist equal to B. Again, because G is the centre of the circle IKL, GK is equal to GI. But GI is equal to the straight line C; therefore GK is equal to C. Also FG is equal to A; therefore the three straight lines FG, FK, GK (which are the sides of the triangle FKG) are respectively equal to the three straight lines A, B, C. Which was to be done.

• 1.8.

And by parity of reasoning, the like may be done in every other instance.

COR. 1. In this manner may be described a triangle of which the sides shall be equal to those of a given triangle respectively. And if it is required that one particular side of the triangle shall be on an assigned straight line DE, and terminated at an assigned point F in it, and that another particular side shall likewise be terminated at F; this also may be done.

For it may be effected by making FG equal to the side which is to be on DE, and FH equal to the other side which is to be terminated at F, and GI equal to the remaining side.

COR. 2. By the help of this, on an assigned straight line and at an assigned point in it, may be made an angle equal to a given angle.

For if AB be the assigned straight

line, A the point in it, and C the given

G

angle; in the two straight lines between
which is the angle C, take any points D,
E, and join DE; and (by Prop. XXII
above) describe the triangle AFG, in which the side AF shall be
equal to CD, and AG to CE, and FG to DE.
are equal to CD, CE respectively, and the
equal to the third side DE, the angle FAG
the angle DCE.

Because AF, AG third side FG is will be* equal to

COR. 3. Hence also on an assigned straight line and at an assigned point in it, may be described a rectilinear figure having its sides and angles respectively equal to those of a given rectilinear figure of any number of sides; and the two rectilinear figures shall be equal.

For the given rectilinear figure may be divided into triangles, by straight lines drawn from some of its angular points to others. And triangles having their sides equal to the sides of these respectively, may (by Cor. 1 above) be constructed in succession, and in the corresponding positions. And because in the corresponding triangles the angles will bet equal respectively, viz. those to which equal sides are opposite; by adding equal angles to equal, the INTERC. 1. angles of one of the rectilinear figures will be equal to the angles of the other respectively. And because the several triangles are equal respectively, the sums, which are the rectilinear figures, are equal.

+1.8.

Cor. 5.

See Note.

*1.22.Cor.2. +1.3.

11.12.Cor.2.

*L.13.Cor.3.

+1.9.

PROPOSITION XXIII.

PROBLEM.-To draw a straight line perpendicular to an assigned straight line of unlimited length, from an assigned point without it.

Let AB be the assigned straight line, which may be prolonged to any length that is desired, both ways; and let C be the point without it. It is required to draw from the point C, a straight line which shall be perpendicular to AB.

In AB take any point, as D; and join CD. If CDB is a right angle, then there is drawn a straight line CD perpendicular to AB as required. But if CDB is not a right angle, then on the other side of AB make* an angle BDF equal to BDC, and maket DF equal to DC, and join CF. Because the equal angles CDB and FDB are either greater or less than right angles, and all the angles at the point D are‡ together equal to four right angles; DC and DF make an angle less than two right angles, either on one side or the other. Wherefore CF will* form a triangle with them, on that side on which is the angle CDF that is less than the sum of two right angles. Bisect+ CF, in G; and join DG. Because in the triangles CDG, FDG, DC is equal to DF, and CG to* FG, and DG is common; the triangles are† equal, and the angle CDG is equal to FDG, and DGC to DGF, and because they are adjacent angles they are right angles. And because the angle CDF has been shown to be bisected by the straight line DG, and was also* bisected by the straight line DB, DG and DB aret in 11.13.Cor.2. one straight line; wherefore EB shall pass through G, and cutt I. 12. CF. And because CGD and CGB are* together equal to two right angles, and CGD is a right angle, CGB is also a right angle, +1.Nom.37. and the straight line CG drawn from the point C ist perpendicular to AB. Which was to be done.

Constr.

Constr. + I. 8.

+I.Nom.37.

• Constr.

+ I. 9. Cor.

And by parity of reasoning, the like may be done in every other instance.

COR. 1. Of all the straight lines that can be drawn to a straight line from a point without it, the perpendicular is the shortest.

‡I.Nom.48.
*I. 19. Cor. and therefore* greater than the perpendicular.

For any other will be the hypotenuse of a right-angled triangle,

On Cor. 1 preceding, See Note.

1.23.

+Hyp.

I. 16.

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BOOK I.PROP, XXIV.

COR. 2. If two straight lines make an acute angle ACD; and from a point A in one of them, a straight line be drawn perpendicular to the other; the perpendicular shall fall on that side of C, on which is the acute angle.

B

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For the perpendicular cannot fall upon the point C; because ACD is† not a right angle but an acute angle. Neither can it fall on the side of C on which is the obtuse angle, as for instance upon the point B; for since ACD (which is the exterior angle of the triangle ACB) is less than a right angle, still more is ABC (which is less than ACD) less than a right angle. Therefore, because it neither falls upon C, nor on the side of it towards B, it falls on the side towards D; that is, on which is the acute angle.

COR. 3. If in any triangle a side AB be *1.17.Cor.3. taken which lies between* two acute angles, and from the opposite angular point a perpendicular CD be drawnt to the side; the perpendicular shall fall between the extremities of the side.

t 1.23.

D

For because BAC is an acute angle, CD (by Cor. 2 above) falls on the side of A which is towards B. And because ABC is an acute angle, CD falls on the side of B which is towards A. Therefore it falls between A and B.

See Note.

PROPOSITION XXIV.

THEOREM.-If two triangles have two sides of the one, equal to two sides of the other respectively, but the angle between the two sides of the one triangle is greater than the angle between the two sides which are equal to them in the other triangle; the remaining side of that which has the greater angle, shall be greater than the remaining side of the other.

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+ I. 19.

l. 16.

than the angle EDF. The remaining side BC is greater than the remaining side EF.

Of the two sides AB, AC, let AB be one that is not greater than the other; and at the point A in the straight line *1.22.Cor.2. AB, make* the angle BAI equal to the angle EDF. Because AB is not greater than AC, the angle ACB is not greater than ABC; for if it was greater, AB the side opposite to it would bet greater than AC the side opposite to the other. Again, the exterior angle AHC is greater than the interior opposite angle ABH or ABC; and because the angle ACB is not greater than ABC, the angle AHC is greater than ACB or ACH. Therefore in the triangle AHC, the side AC which is opposite to the angle AHC the greater, is* greater than AH which is opposite to the angle ACH the less. Since, therefore, AH is less than AC, maket AG equal to AC, and the point G will be on the different side of BC from that on which is the point A. Join BG, GC.

• I. 19.

+ I. 3.

Hyp. Constr. +Constr. II.4.

Because AB ist equal to DE, and AG to* AC or to DF, and the angle BAG ist equal to the angle EDF, BG is‡ equal to EF. And because AC is* equal to AG, the angle AGC ist equal to ACG. But the angle ACG is greater than BCG; therefore the angle AGC (which is equal to ACG) INTERC. 1. ist greater than BCG. Still more is the angle BGC (which is

*Constr. † I. 5.

Cor. 2.

* I. 19.

greater than AGC) greater than BCG. Wherefore in the triangle BGC, the side BC (which is opposite to the angle BGC the greater) is* greater than BG (which is opposite to the angle BCG the less). But BG has been shown to be +INTERC. 1. equal to EF; therefore BC (which is greater than BG) ist greater than EF.

Cor. 2.

And by parity of reasoning, the like may be proved in all other triangles under the same conditions. Wherefore, universally, if two triangles have two sides &c. Which was to be demonstrated.

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