1 Join A, B; bisect (E. 10. 1.) AB in C; from C draw (E. 11. 1.) CD 1 to AB, meeting XY in D. The point D is equidistant from A, B. For, join A, D and B, D. Then, since (constr.) AC=BC, and CD is common to the two ACD, BCD, and that (constr. and E. 10. def. 1.) ∠ACD = ∠ BCD, .. (E. 4. 1.) AD=BD; i.e. Dis 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 XY, 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 4 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 ACP, BCP, and that (hyp.) PA=PB, .. (E. 8. 1.) the ∠ ACP=∠ BCP, and ... (E. 10. def. 1.) the ∠ ACP is a right; but (hyp.) the ∠ACD is a right; the ∠ ACP is equal to the ∠ ACD, the less to the greater, which is impossible; ... the point Pis not equidistant from A and B. 1 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; and join A, B; and let CD be drawn bisecting AB at right; 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; ... (Cor. 2.) D is equidistant from A and B; and... a circle described from D, as a centre, at the distance DA, which (constr. E. 15. 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 given A, 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. 11. 1.) DG 1 to AB, and EG 1 to AC, and let them meet in G: Join G, F. Then, (constr. and S. 3. 1. Cor. 2...) BG = AG, and AG=CG; CG=BG. Again, since (hyp.) BF=CF (constr.) and F G is common to the two BFG, CFG, and that BG=CG, ... the ∠ BFG = ∠ CFG (E. 8. 1.); i.e. (E. 10. def. 1.) GF is 1 to BC: And there cannot (E. 10. def. 1.) be drawn from F more than one straight line 1 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 line. Let A, B, C. be three given 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. L Join A, B, and B, C, and C, A; bisect (E. 10. 1.) AB in D, and AC in E; draw (E. 11. 1.) from D and E, DG to AB, and EG 1 to AC, and let them meet in G. Then, (S. 3. 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. 11. THEOREM. PROP. VI. There cannot be drawn more than two equal straight lines, to another straight line, from a given point without it. Let A be a given point, without the given 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; ... (E. 5. 1.) ∠ACB = ∠AGC: Also ∠ACB=∠ ABC; .. ∠AGC=∠ABC; i. e. the exterior is equal to the interior opposite ∠, when the side |