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 Zs 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 A ABC have . ABC = L ACB; then AB = AC.

i 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 re-

spectively 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 As CBA, CDA, if AB, BC, CA = AD, DC, CA, respectively, ea. to ea. then A CBA = A CDA, 2CBA = _CDA, Z BAC = L CAD, L ACB = L ACD.

PROP. VIII. PROB. 9. 1 Eu. To bisect a given rectilineal angle, that is, to

divide it into two equal angles.
Let BAC be the given rectilin. Z; it is re-
quired to bisect it.

Take any point D in AB.
From AC cut off AE = AD, Prop. S.

join DE.
On DE descr. equilat. A DEF, Prop. 1.

join AF.
Then AF bisects < BAC.

Constr.

Constr.

(AD= AE,

AF common to As DAF, EAF,

(DF= EF; Prop. %. . _ DAF = LEAF.

Wherefore the rectilin. BAC is bisected by AF.

PROP. IX. PROB. 10. 1 Eu. To bisect a given finite straight line, that is, . to divide it into two equal parts.

To divide AB into two = parts.
Prop. 1. Upon AB descr. equilat. A ABC.
Prop. 8.

Bisect – ACB by CD.
Then AB is bisected at D.

A D B

AC = CB, Constr.

:: 3 CD common to As ACD, BCD, Constr.

12 ACD= LBCD; Prop. 4.

in 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 I AB.

Take any point D in AC, make CE = CD. Prop. 3.
On DE descr. equilat. ADFE; join FC. Prop. 1.

Then FC I AB.

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 I AB.

Take any pt. D upon the other side of AB.

From cent. C, dist. CD, descr. O EGDF Post. 3. meeting AB in F and G; bisect FG in H; join Prop. 9. CH.

Then CH | AB.