THEOREM XXV. 429. If from a point without a circle a secant and tangent be drawn, the rectangle of the whole secant and the part outside the circle is equal to the square of the tangent. Hypothesis. P, a point with out a circle; PT, a tangent touch- Conclusion. PA.PB ± PT2. P Then T 1. Because the angle ABT stands upon the arc TA, Angle ABT angle arc AT. = 2. Because PT is a tangent, and AT a chord, = B (§ 235) (§ 234) 3. Therefore angle ABT angle PTA, and the triangles PAT and PTB have the angles at P identical and two other angles equal. Therefore the third angles PTB and PAT are also equal, and the triangles are similar (§ 390). 4. Comparing homologous sides, we have 5. Therefore PA: PT:: PT : PB. Rect. PA.PB = PT (§ 420). Q.E.D. 430. Corollary. Because T the rectangles formed by all secants from P are equal to the square of the same tangent PT, they are all equal to each other. P THEOREM XXVI. 431. When the bisector of an angle of a triangle meets the base, the rectangle of the two sides is equal to the rectangle of the segments of the base plus the square of the bisector. Hypothesis. ABC, any triangle; CD, the bisector of the angle at C, cutting the base at D. Conclusion. Rect. CA. CB rect. AD.DB + CD2. Proof. Circumscribe a circle ACBE around the given triangle, and continue the bisector till it meets the circle in E. Join BE. Then 1. In the triangles CAD and CEB we have Angle ACD = angle BCE (hyp.). = D Angle CAD angle BEC (on same arc BC). Whence Rect. CA. CB = rect. CE. CD, =rect. (CD+DE) CD, = CD2 + rect. CD.DE, (§ 287) = CD2+rect. AD.DB (§ 428). Q.E.D. THEOREM XXVII. 432. In a right-angled triangle the area of any polygon upon the hypothenuse is equal to the sum of the areas of the similar and similarly described polygons upon the two other sides. put Hypothesis. ABC, a triangle, right-angled at A. We also Proof. From A drop AD 1 BC. Then A D 1. Because ABC is right-angled at A, and AD is a perpendicular upon the hypothenuse, DC CA: CA: BC. DB: BA :: BA : BC. 2. Because a', a", and a are the areas of similar polygons described upon CA, BA, and BC, a'a: DC: BC. a": a: BD: BC. 3. Therefore, taking the sum of the ratios (§ 366), a' + a": a :: BD + DC: BC. (§ 427) 4. Because BD + DC = BC, the second ratio is unity; therefore the first also is unity, and a' a" a. Q.E.D. Scholium. This result includes the Pythagorean proposition (§ 308), as a special case in which the polygons are squares. CHAPTER IV. PROBLEMS IN PROPORTION. PROBLEM I. 433. To divide a straight line similarly to a given divided straight line. Given. A line, AB; another line, CD, divided at the points M and N. Required. To divide AB similarly to CD. Analysis. By § 388 two sides of a triangle are similarly divided by any lines parallel to the base. Therefore, if we put together the lines AB and CD in such a way as to form two sides of a triangle, all lines parallel to the third side will A B. M N C D N M M N B divide these two sides similarly. Hence Construction. 1. Form a triangle ABD, such that AD = = given line CD. BD any convenient length. 2. Through the points M and N draw MM' and NN' parallel to BD, meeting AB in M' and N'. Then ($388) AM': M'N' : N'B :: AM : MN: ND. Therefore the line AB is divided at the points M' and N' similarly to CD. Q.E.F. PROBLEM II. 434. To divide a straight line internally into segments which shall be to each other as two given straight lines. Given. Two straight lines, q. p and q; a third straight line, AB. Required. To divide AB into segments having the same ratio as p to q. Construction. 1. From one end of the line AB draw an indefinite straight line AD. 2. From this line cut off AC=p and CD = q. 3. Join DB. 4. From C draw a line parallel to DB, and let E be the point in which it cuts AB. The line AB will be cut internally at E into the segments AE and EB, having to each other the same ratio as the lines p and q. Proof. As in Problem I. PROBLEM III. 435. To divide a straight line externally so that the segments shall be to each other as two given straight lines. Given. A straight line, AB; two other lines, p, q. Required. To cut AB externally so that the segments shall have to each other the ratio p: q. Construction. 1. From either end of the line AB, as A, 4. From C draw a line parallel to DB, and let E be the point in which it cuts AB produced. B C The line AB will be divided externally at E, so that Proof. As in Problem I. 436. Corollary 1. If p = q, the point D would fall upon A, and the line DB would coincide with AB. The line drawn through C parallel to DB would then be parallel to AB, so there would be no point of intersection E. Therefore there would be no external point for which the ratio of the segments would be unity. 437. Cor. 2. If we combine Problems II. and III. on the same straight line, using the same ratio, we shall divide the line harmonically, and the ends of the line, together with the points of division, will form four harmonic points. PROBLEM IV. 438. To find a fourth proportional to three given straight lines. Given. Three straight lines, b D a E Analysis. To solve the problem it is only necessary to form two similar triangles, one of which shall have a and c for two of its sides, while the other shall have b as the side homologous to a. B The side homologous to c will then be x. |