the proposition; since we are at liberty to suppose any line to be drawn, or any figure to be constructed, the existence of which does not involve an absurdity. The first proof of this proposition is that of Euclid, who, to avoid the possibility of the taking for granted that which may imply a contradiction, never supposes a thing to be done, the manner of doing which has not been previously explained.

The equality of the s on the oppo. side of the base will follow from Prop. 12.

PROP. VI. THEOR.

6. 1 Eu.

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 have ABC = ACB; then AB AC.

one of them is > the other;

let AB > AC;

from AB cut off DB = AC;
join DC.

Prop. 3.

{the greater, which is

AB is not AC,

i. e. AB AC.

Wherefore, if two angles, &c.

COR.-Hence every equiangular triangle is

equilateral.

PROP. VII.

THEOR.

If two triangles have three sides of the one respectively equal to the three sides of the other, each to each, the triangles are equal, and the angles are equal which are opposite to the equal sides.

In ▲s CBA, CDA, if AB, BC, CA = AD, DC, CA, respectively, ea. to ea. then CBA

CDA, ▲ BAC = L

C

D

A

D

A

Conceive the As to be so applied to each other that the side CA may be common to both, and the equal sides BC of the one, and DC of the other, to terminate at the same pt. C, and to be on oppo. sides of CA. Join BD.

ABAD,
ABD = ADB;
CD,

Prop. 5.

Hyp.

and

≤ CBD = ▲ CDB,

▲ CBA / CDA.

AB, BC = AD, DC, ea. to ea. Hyp.

and incl. CBA incl. ▲ CDA;
▲ CBA = A CDA,

Prop. 5.

Ax. 2, 3.

To bisect a given rectilineal angle, that is, to divide it into two equal angles.

Let BAC be the given rectilin. ; it is required to bisect it.

Take any point D in AB.

From AC cut off AE = AD,
join DE.

On DE descr. equilat. DEF,

join AF.

Prop. 4.

Prop. 3.

Prop. 1.

PROP. IX. PROB. 10. 1 Eu.

To bisect a given finite straight line, that is, to divide it into two equal parts.

Constr. Prop. 4.

A D B

ACCB,

CD common to ◇s ACD, BCD,

LACD BCD;

=

base AD base DB.

Therefore the str. line AB is divided into two parts in the point D.

PROP. X. PROB. 11. 1 Eu. To draw a straight line at right angles to a given straight line, from a given point in the same.

Let AB be the given line, and C a given

pt. in it; it is required to draw from C a

str. line AB.

Take any point D in AC, make CE = CD. Prop. 3.
On DE descr. equilat. DFE; join FC.
Then FCL AB.

F

Prop. 1.

FC is common to As DCF, ECF,

|FD, DC= FE, CE, ea. to ea.

Constr.

▲ DCF = ≤ ECF, which are adj. ▲ s, Prop. 7.

and FC LAB.

Therefore there has been drawn, &c.

PROP. XI. PROB.

12. 1 E.

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

Let AB be the given line, and Ca pt. without it; it is required to draw CH | AB. Take any pt. D upon the other side of AB. From cent. C, dist. CD, descr. meeting AB in F and G; bisect FG in

CH.

Then CHL AB.

Def. 10.

EGDF Post. 3. H; join Prop. 9.