from the vertex A, to the base BC, a straight line which shall exceed AC, as much as it is exceeded by AB. From AB cut off (E. 3. 1.) AD=AC; bisect (E. 10. 1.) DB in E; from the centre A, at the distance AE, describe (E. 3. Post.) the circle EF cutting BC in F; and join (E. 1. Post.) A, Then is AF the straight line which was to be drawn. = F: For, (E. 15. def. 1.) AF=AE; and (constr.) AD=AC; ... AF-AC AE- AD=DE. Also, AB-AE= BE; i. e. AB-AF= BE: and (constr.) BE=DE. ... AF-AC-AB-AF. PROP. III. 3. PROBLEM. In a straight line given in position, but indefinite in length, to find a point, which shall be equidistant from each of two given points, either on contrary sides, or both on the same side of the given line, and in the same plane with it; but not situated in a perpendicular to it. Let XY be a given straight line indefinite in length, and A, B, two given points without it; not-situated in a perpendicular to XY: It is required to find a point in XY that shall be equidistant from A and B. First, let A, B be both on the same side of XY: 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. ACD, For, join A, D and B, D. Then, since (constr.) AC BC, and CD is common to the two BCD, and that (constr. and E. 10. def. 1.) =2 BCD, .. (E. 4. 1.) AD=BD; i. e. D is equidistant from A and B. ACD 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 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 BCP, and .. (E. 10. def. 1.) the ACP L 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 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 |