For, join A, D and B, D. Then, since (constr.) AC= BC, and CD is common to the two as ACD, BCD, and that (constr. and E. x. Def. 1.) ACD = ‹ BCD; .; (E. iv. 1.) AD = (E. IV. 1.) AD BD; i. e. D is equidistant from A and B. But, if the two given points, A and B, are on contrary sides of XY, let them be joined, as before, and let the straight line which joins them be bisected. Then, if the point of bisection be in XÃ, that, which was required, has been done, But, if that point be not in XY, draw from it, as before, a perpendicular to AB, and it may be shown, as in the first case, that the point, in which the perpendicular meets XY, is that which was required to be found. 4. COR. 1. By the help of this problem, it is manifest that a circle may be described, which shall have its centre in a given straight line, and which shall pass through two given points without that line. 5. COR. 2. It is evident from the demonstration, that any point in an indefinite straight line DZ, which bisects the given finite straight line AB, at right angles, is equidistant from the extremities A and B, of that given finite line: And, any point which is not in that indefinite line DZ, is not equidistant from the two extremities A and B of the given finite line. For, let P be any point, not in' DZ, which bisects AB at right angles in C; and, if it be possible, let P be equidistant from A and B: Join P, A and P, C and P, B; and since (hyp.) AC=CB, and CP is common to the two as ACP, BCP, and that the ACP = (hyp.) PA = PB; .... (E. vш. 1.) 4 BCP, and .. (E. x. Def. 1.), the ACP is a right angle; but (hyp.) the ACD is a right angle; therefore the ACP is equal to the ACD, the less to the greater, which is impossible; therefore the point P is not equidistant from A and B. 6. COR. 3. Hence, an indefinite number of circles may be described all of them passing through two given points: And if any number of circles pass, all of them, through the same two given points, their centres are all in the straight line that bisects at right angles the straight line joining the two given points. 7. COR. 4. Hence, also, a circle may be described which shall pass through two given points, and which shall have its semi-diameter equal to any given finite straight line, that exceeds the half of the straight line joining the two given points. For, let A, B be the two given points; andjoin A,B; and let CD be drawn bisecting AB at right angles; from A, as a centre, at a distance equal to the given finite straight line, describe a circle, and let it cut CD in D; therefore, (Cor. 2.) D is equidistant from A and B; and therefore a circle described from D, as a centre, at the distance DA, which (constr. E. xv. Def. 1.) is equal to the given semidiameter, will pass through B. PROP. IV. 8. THEOREM. If the three sides of a given triangle be bisected, the perpendiculars drawn to the sides, from the three several bisections, shall all meet in the same point: And that point is equidistant from the three angular points of the given triangle. Let ABC be a triangle, of which the three sides AB, AC, and CB are bisected in the points D, E and F, respectively: The perpendiculars drawn to the several sides from D, E, F, shall all meet in a point that is equidistant from A, B and C. = For, draw (E. xI. 1.) DG perpendicular to AB, and EG perpendicular to AC, and let them meet in G: Join G, F. Then, (constr. and Sect. iii. 1. Cor. 2.) .. BG=AG, and AG = CG; .. CG = BG. Again, since (hyp.) BF CF (constr.) and FG is common to the two As BFG, CFG, and that BG= CG; therefore the BFG = ▲ CFG (E. viii. 1); i. e. (E. x. Def. 1.) GF is perpendicular to BC: And there cannot (E. x. Def. 1.) be drawn from F more than one straight line perpendicular to BC. It is plain, therefore, that the perpendiculars drawn to the sides, from D, E and F, all meet in the same point G: And, since it has been shown that AG= BG=CG, the point G is equidistant from A, B and C. PROP. V. 9. PROBLEM. To find a point, in a given plane, which shall be equidistant from three given points in the plane, that are not all in the same straight linc. Let A, B, C, (See Fig. p. 6.) be three points, not all of them in the same straight line: It is required to find a point, that shall be equidistant from A, B and C. Join A, B, and B, C, and C, A; bisect (E. x. 1.) AB in D, and AC in E; draw (E. xI. 1.) from D and E, DG perpendicular to AB, and EG perpendicular to AC, and let them meet in G. Then, (Sect. iii. 1. Cor. 2.) the point G is equidistant from A, B, and C. 10. COR. By the help of this problem a circle may be described about a given triangle; or so as that its circumference shall pass through any three given points that are not in the same straight line. PROP. VI. 11. THEOREM. There cannot be drawn more than two equal straight lines, to another straight line, from a given point without it. Let A be a point, without the straight line BC: There cannot be drawn more than two equal straight lines, from A to BC. For, if it be possible, let AB=AG=AC; there fore (E. v. 1.) ≤ ACB=4 AGC: Also ACB= L'ABC; .. LAGCL ABC; i. e. the exterior is equal to the interior opposite angle, when the side BG, of the AGB, is produced: which (E. xvi. 1.) is absurd. 12. COR. A circle cannot cut a straight line in more points than two. PROP. VII. 13. THEOREM. The perpendicular let fall from the obtuse angle of an obtuse-angled triangle, or from any angle of an acute-angled triangle, upon the opposite side, falls within that side: But the perpendicular drawn to either of the sides containing the obtuse angle of an obtuse-angled triangle, from the angle opposite, falls without that side. Let ABC be an obtuse-angled triangle, obtuse angled at B, and let ABD be an acute-angled triangle: The perpendicular drawn from B to AC falls within AC; the perpendicular drawn from any other LA, of the ABC, to the opposite side BC, falls without BC; and the perpendicular from any A, of the ▲ ABD, to the opposite side BD, falls within BD. For, first, if it be possible, let AG, drawn (E. xII. 1.) from A perpendicular to BD, meet DB, produced, in G: Then, since (hyp.) the ABD is |