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. spectively equal to the three sides of the 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. Take any point D in AB. join DE. join AF. 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. Bisect – ACB by CD. 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. given straight line, from a given point in pt. in it; it is required to draw from C a Take any point D in AC, make CE = CD. Prop. 3. Then FC I AB. PROP. XI. PROB. 12. 1 E. given straight line of an unlimited length, 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. |