AD2=AC2+CD2+2(CD. CE), Prop. XIX. .; BC2 DB2+CD2+2(CD. DF); and, by addition, EF.CD CD.EC+CD. CD+CD.DF; ... AD2+BC2=AC2+DB2+2(EF. CD); 17. The figure has been already constructed. Prove AN = to AL, and NB=BM. DG2=AG2+AD2+2(AG. AL), Prop. XIX. The student who has attended carefully to the previous demonstration will find no difficulty in establishing the truth of the theorem. BOOK III. DEFINITIONS. 1. An Arc is any portion of the circumference of a circle. 2. A Chord is a line joining the extremities of an arc. 3. A Tangent is a line without a circle, which touches it in one point only. 4. A Line is inscribed in a circle when its extremities terminate in the circumference. 5. A Secant is a line which cuts the circumference in two points. 6. A Segment of a circle is that part of it bounded by a chord and the arc subtending it. 7. A Sector of a circle is that part of it bounded by two radii and the intercepted arc. 8. An Angle is inscribed within a circle when the vertex and the extremities of the sides are in the circumference. 9. A Polygon is inscribed within a circle when all its vertices are in the circumference. 10. The Zone of a circle is a part bounded by two parallel lines and the intercepted arcs. PROPOSITION I.-THEOREM. Every diameter bisects a circle and its circumference. Let ACBD be a circle; then will the diameter AB bi sect it and its circumference. A Conceive the part ADB turned over and applied to the part ACB: it must exactly coincide with it; for if it does not, there must be points in either portion of the circumference unequally distant from the centre; which is contrary to the definition of a circle. Therefore the two parts exactly coincide and are equal (Ax. 10). PROPOSITION II-THEOREM. D B A line perpendicular to a radius at its extremity is tangent to the circumference. Let the line CD be perpendicular to the radius AB at its extremity; then will CD touch the circumference at one point, B, only. From any point, as E, in the line CD, draw the line AE to the centre A. Since AB is perpendicular to CD, it is shorter than any oblique line, AE (Prop. XXXI., Bk. I.). Hence the point E is without the circle; and in like manner it can be proved that any C ર E ब A other point in the line CD is without the circle. Hence CD touches the circumference in one point, B, only. PROPOSITION III.-THEOREM. When a line is tangent to the circumference of a circle, a radius drawn to the point of contact is perpendicular to the tangent. Let the line AB be tangent to the circumference of a circle at the point D; then will the radius CD be perpendicular to the tangent AB. For the line AB being wholly without the circumfer ence, except at the point D, it follows that any line, as PROPOSITION IV.-THEOREM. If a line drawn through the centre of a circle bisect a chord, it will be perpendicular to the chord; or, if it be perpendicular to the chord, it will bisect both the chord and the arc of the chord. A B E D Let the line CD, drawn from the centre of the circle, bisect the chord AB in E; then will CD be perpendicular to AB. Draw the radii CA and CB. In the two triangles CAE and CBE the sides AC and CB are equal, being radii of the same circle; AE and BE are also equal (hyp.), and CE is common to both. Hence the two triangles have three sides of the one equal to three sides of the other, each to each, and are equal in all their parts (Prop. VII., Bk. I.). Therefore the angle AEC is equal to the angle BEC; and, since they are equal, each must be a right angle, and the line CD must be perpendicular to AB (def. 9). Again, if CD be perpendicular to AB, then will the chord AB be bisected at the point E, and the arc ADB be bisected at the point D. For, in the two right-angled triangles AEC and BCE, the hypothenuse AC is equal to the hypothenuse BC (def. 19), and the side CE is common to both; hence the two triangles are equal in all their parts (Prop. XXXII., Bk. I.); the side AE is equal to the side BE, and the angle ACE is equal to the angle BCE. Then apply the sector ACD to the sector DCB, so that DC will be common; and since the angle ACD is equal to the angle DCB, the line AC will coincide with CB; and since the lines AC and CB are equal, the point A must fall upon B. The sectors must, therefore, coincide; for if they did not, some point of the arc AD would fall within or without the circumference, which is contrary to the definition of a circle. Hence the arc AD is equal to the arc DB. PROPOSITION V.-PROBLEM. To describe a circle which shall pass through three given points not in the same straight line. Let A, B, and C be three points not in the same straight line; it is required to describe a circle passing through these three points. A. ៩. F D Join the points A and B, and B and C; bisect AB at the point D, and BC at the point E (Prop. IX., Bk. I.). From D erect the perpendicular DF, and from E erect the perpendicular EG (Prop. X., Bk. I.); the point of intersection, H, is the centre of the circle whose circumference will pass through A, B, and C. The perpendiculars DF and EG must intersect; for if they do not, they are parallel, which is impossible, because, if DF and EG were parallel, since the angles HDB and HEB are right angles, the lines AB and BC must be the secant line, and form one and the same straight line, which is contrary to hypothesis. In the two triangles ADH and BDH, AD and BD are equal (const.). DH is common, and the angles ADH and BDH are equal, being right angles; hence the two triangles have two sides and the included angle of the one equal to two sides and the included angle of the other, each to each, and are there D |