BF BE.ED+ EF2; therefore, AE.EC+ EF2 BE.ED Book III. + EF2; and taking EF2 from both, there remains the rectangle AE.EC-BE.ED. Lastly, let neither of the straight lines AC, BD, pass through the centre F. Through the point E, the intersection of AC. BD, draw the diamater HFEG. Then, by the last case, AE, EC= GE.EH, and BE. EDGE.EH; therefore, AE.EC-BE.ED. Wherefore, if two straight lines &c. Q,ED. H c Cor. 1.3.3. Cor. If, from any point in the circumference of a circle, a perpendicular be drawn to the diameter, the square of this perpendicular, is equivalent to the rectangle under the segments into which it divides the diameter. PROP. VII. THEOREM. If from any point D, without a circle, two straight lines be drawn, one of which DA cuts the circle, and the other DB touches it; the rectangle AD.DC contained by the whole line which cuts the circle, and the part of it without the circle, is equal to the square of DB, the line which Book III. d 1. 2. Again, suppose that DCA does = F DA E Cor. 1. If, from any point without a circle, there be drawn two straight lines cutting it, the rectangles contained by the whole lines, and the parts of them without the circle, are equivalent. Cor. 2. If from any point without a circle, there be drawn two tangents to it, these tangents are equal. Cor. 3. If from a point without a circle, there be drawn two straight lines, one of which cuts the circle, and the other meets it; if the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, be equivalent to the square of the line which meets it, the line which meets also tauches the circle. ELEMENTS OF GEOMETRY. BOOK IV. DEFINITIONS. I. SIMILAR rectilineal figures are those which have their Book IV. several angles equal, each to each, and the sides about the equal angles proportionals. II. Two sides of one figure are said to be reciprocally proportional to two sides of another, when one of the sides of the first is to one of the sides of the second, as the remaining side of the second is to the remaining side of the first. III. The altitude of any figure is the straight line drawn from its vertex perpendicular to its base. N Book IV. PROP. I. THEOREM. Triangles of the same altitude are to one another as their bases. Let the triangles ABC, ACD, have the same altitude, viz. the perpendicular drawn from the point A to BD. Then the triangle ABC is to the triangle ACD, as the base BC is to the base CD. For BC being divided into as many equal parts as there are units in m, BD will contain as many such equal parts as there are units in n; the division being made, and straight lines drawn from A to the several points of section, it is manifest that the whole triangle ABD will be divided into m+n, equal a Cor.1.16.1. triangles. Now, the ▲ ABC will contain as many of these equal triangles as there are units in m, and ACD as many of them as there are units in n. Hence, ▲ ABC: ▲ ACD :: m b 111. Alg.: n; but BC: CD by supposition as mn, and therefore A ABC: A ACD :: BC: CD. Case 2. When the bases are incommensurable, From the demonstration of Case I., it is obvious, that if BC be greater than of CD, but less than + of CD, then the n m n m+1 n ABC will be greater than of ACD, but less than m+1 of ACD, however great the magnitudes m and n may n c 124. Alg. be; and hence, that the ▲ ABC: a ACD :: BC : CD.— Wherefore triangles, &c. Q. E. D. Cor. 1. Parallelograms of the same altitude are to one another as their bases. Cor 2. Triangles and parallelograms on the same base are to one another as their altitudes. Book IV. PROP. II. THEOREM. If a straight line DE be drawn parallel to one of the sides BC, of a triangle ABC, it will cut the other sides, or the other sides produced proportionally; and if the sides, or the sides produced, be cut proportionally, the straight line which joins the points of section, will be parallel to the remaining side of the triangle. Join BE and CD; then the triangle BDE is equivalent to the triangle CDE, and therefore the triangle BDE: triangle ADE:: triangle CDE: triangle ADE*. But the triangle BDE: triangle ADE :: BD: DA; and the triangle CDE : triangle ADE ::CE: EA. Hence BD: DA :: ČE : EAC a Cor. 1. 15. 1. b 1. 4. c 111. Alg. Again, if BD : DA :: CE: EA, then is DE parallel to BC. The same construction being made, we have, by supposition, BD: DA :: CE : EA, but BD: DA :: triangle BDE: triangle ADE, and CE: EA :: triangle CDE : triangle ADE, therefore the triangle BDE: triangle ADE : : triangle CDE : triangle ADE, that is, the triangles BDE and CDE have the same ratio to ADE, and, therefore, BDE and CDE are equivalent; and they are upon the same base DE, hence they are between the same parallels, there- d Cor. 3. It is hardly necessary to prove, that equal magnitudes have to the same the same ratio; or that magnitudes which have to the same the same ratio, are equal to one another. If a and b be two equal quantities, and c a third quantity, then a b = and hence a:c::b: c. Con and hence a=b. 15. 1. |