Obs. Propositions 27, 28, 29 being cumbrous in form and of little value as geometrical results are now very generally omitted. DEFINITION. A straight line is said to be divided in extreme and mean ratio, when the whole is to the greater segment as the greater segment is to the less. [Book VI. Def. 5.] PROPOSITION 30. PROBLEM. To divide a given straight line in extreme and mean ratio. A Let AB be the given st. line. It is required to divide AB in extreme and mean ratio. Divide AB in C so that the rect. AB, BC may be equal to the sq. on AC. Then because the rect. AB, BC the sq. on AC, II. 11. ... AB AC :: AC: BC. VI. 17. Q.E.F. EXERCISES. 1. ABCDE is a regular pentagon; if the lines BE and AD intersect in O, shew that each of them is divided in extreme and mean ratio. 2. If the radius of a circle is cut in extreme and mean ratio, the greater segment is equal to the side of a regular decagon inscribed in the circle. In a right-angled triangle, any rectilineal figure described on the hypotenuse is equal to the sum of the two similar and similarly described figures on the sides containing the right angle. P R Let ABC be a right-angled triangle of which BC is the hypotenuse; and let P, Q, R be similar and similarly described figures on BC, CA, AB respectively. Then shall the fig. P be equal to the sum of the figs. Q and R. ... CB BA :: BA: BD; VI. 8. CB: BD: the fig. P: the fig. R; VI. 20, Cor. inversely, BD : BC :: the fig. R: the fig. P. In like manner, DC : BC :: the fig. Q : the fig. P; V. 2. .. the sum of BD, DC: BC :: the sum of figs. R, Q : fig. P; but BC the sum of BD, DC; .. the fig. Pthe sum of the figs. R and Q. v. 15. Q.E.D. NOTE. This proposition is a generalization of Book I., Prop. 47. It will be a useful exercise for the student to deduce the general theorem (VI. 31) from the particular case (I. 47) with the aid of VI. 20, Cor. 2. EXERCISES. 1. In a right-angled triangle if a perpendicular is drawn from the right angle to the opposite side, the segments of the hypotenuse are in the duplicate ratio of the sides containing the right angle. 2. If, in Proposition 31, the figure on the hypotenuse is equal to the given triangle, the figures on the other two sides are respectively equal to the parts into which the triangle is divided by the perpendicular from the right angle to the hypotenuse. 3. AX and BY are medians of the triangle ABC which meet in G: if XY is joined, compare the areas of the triangles AGB, XGY. 4. Shew that similar triangles are to one another in the duplicate ratio of (i) corresponding medians, (ii) the radii of their inscribed circles, (iii) the radii of their circumscribed circles. 5. DEF is the pedal triangle of the triangle ABC; prove that the triangle ABC is to the triangle DBF in the duplicate ratio of AB to BD. Hence shew that the fig. AFDC: the ▲ BFD :: AD2: BD2. Shew that 6. The base BC of a triangle ABC is produced to a point D such that BD DC in the duplicate ratio of BA : AC. AD is a mean proportional between BD and DC. 7. Bisect a triangle by a line drawn parallel to one of its sides. 8. Shew how to draw a line parallel to the base of a triangle so as to form with the other two sides produced a triangle double of the given triangle. 9. If through any point within a triangle lines are drawn from the angles to cut the opposite sides, the segments of any one side will have to each other the ratio compounded of the ratios of the segments of the other sides. 10. Draw a straight line parallel to the base of an isosceles triangle so as to cut off a triangle which has to the whole triangle the ratio of the base to a side. 11. Through a given point, between two straight lines containing a given angle, draw a line which shall cut off a triangle equal to a given rectilineal figure. Obs. The 32nd Proposition as given by Euclid is defective, and as it is never applied, we have omitted it. In equal circles, angles, whether at the centres or the circumferences, have the same ratio as the arcs on which they stand: so also have the sectors. Let ABC and DEF be equal circles, and let BGC, EHF be angles at the centres, and BAC and EDF angles at the Oces. Then shall (i) Then the BGC, CGK, KGL are all equal, for they stand on the equal arcs BC, CK, KL: III. 27. .. the BGL is the same multiple of the BGC that the arc BL is of the arc BC. Similarly, the EHR is the same multiple of the EHF that the arc ER is of the arc EF. And if the arc BL= the arc ER, the BGL the EHR; = and if the arc BL is greater than the arc ER, the BGL is less than the EHR. III. 27. Now since there are four magnitudes, namely the 45 BGC, EHF and the arcs BC, EF; and of the antecedents any equimultiples have been taken, namely the BGL and the arc BL; and of the consequents any equimultiples have been taken, namely the EHR and the arc ER: L and since it has been proved that the BGL is greater than, equal to, or less than the EHR, according as BL is greater than, equal to, or less than ER; .. the four original magnitudes are proportionals; v. Def. 5. that is, the BGC: the EHF :: the arc BC: the arc EF. (iii) Join BC, CK; and in the arcs BC, CK take any points X, O. .. the remaining arc BAC = the remaining arc CAK : = III. 27. .. the segment BXC is similar to the segment COK; and these segments stand on equal chords BC, CK; III. Def. 10. the segment COK. III. 24. |