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. Hyp. In ABD :: AB = AD, Prop. 5. Hyp. ≤ CBD = ▲ CDB, Prop. 5. and ▲ CBA CDA. Ax. 2, 3. AB, BC= AD, DC, ea. to ea. Hyp. and incl. CBA = incl. ▲ CDA; ▲ BAC = ▲ ACB Wherefore, if two triangles, &c. 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. ; it is re Prop. 1. Prop. 8. 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. C Constr. Constr. Prop. 4. A D B AC = CB, CD common to As ACD, BCD, CD common to ACD 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 Ca Take any point D in AC, make CE = CD. Prop. 3. F Prop. 1. ▲ 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 CHAB. Take any pt. Ꭰ upon the other side of AB. Def. 10. From cent. C, dist. CD, descr. EGDF Post. 3. meeting AB in F and G; bisect FG in H; join Prop. 9. CH. Const. Def. 15. Prop. 7. Def. 10. Join CF, CG; JFH, CF = HG, CG, ea. to ea. {and CH LAB. s, Therefore, the ICH has been drawn, &c. PROP. XII. THEOR. 13. 1 Eu. The angles which one straight line makes with another upon one side of it, are either two right angles, or are together equal to two right angles. Let AB make with DC, on the same side s DBA, ABC; then of it, DBA + L Prop. 10. But if ABC ‡≤ DBA, draw BE DC; Def. 10. Ax. 2. and :: s. Zs CBE, DBE are rt. ▲ CBE = ▲ ABC+ 2 ABE; add to these equals DBE; E={ ../CBE+DBE= Z ABC + ABE+ Again DBA = ▲ DBE +▲ ABE, :: ≤ DBA+≤ ABC={ ≤ ABC + ≤ ABE + Ax. 2. ={ LDBE: PROP. XIII. THEOR. 14. 1 Eu. If at a point in a straight line, two other straight lines, upon the opposite sides of it, make the adjacent angles together equal to two right angles, these two straight lines shall be in one and the same straight line. At the point B in AB let BC, BD, on the oppo. sides of AB, make the adj. s ABC + ABD 2 rt. Zs; then shall BD be in the same straight line with BC. = If BD be not in the same str. line with CB, let BE be in the same str. line with it; then : str. line AB makes with str. line CBE, on the same side of it, the s ABC, ABE: :: ≤ ABC + ≤ ABE = 2 rt. Ls; Prop. 12. |