§ 32. (Bookwork, EUCLID, II. 1-4.) DEFINITION.-If a point be taken in a straight line, its distances from the extremities of the line are called segments of the line, and the point is called a Point of Internal Section. 1. In a right-angled triangle, the square on the altitude from the right angle is equal to the rectangle contained by the segments into which the altitude divides the hypotenuse; and the square on either of the sides which contain the right angle is equal to the rectangle contained by the hypotenuse and its adjacent segment. (A Standard Theorem.) Let ABC be a triangle, right-angled at A, and let AN be the altitude from A, it is required to prove that But BC2 BN2+NC2+2BN NC; [Euc. II. 4. 2. Prove the above theorem by means of the geometrical construction employed by Euclid in proving Euc. I. 47. K 3. In any triangle in which the altitude from the vertex falls within the base, the square on the base, together with twice the square on the altitude, is equal to the squares on the sides, together with twice the rectangle contained by the segments of the base. Use Euc. I. 47 and II. 4, BC2+2AN2=AB2+AC2+2BN.NC. B N 4. If a straight line be divided into three segments, the square on the whole line shall be equal to the squares on the segments, together with twice the rectangles contained by each pair of segments. D B Consider AB as divided in C, and CB as divided in D, and apply Euc. II. 4 to each. Use also Euc. II. 1. 5. In the ▲ ABC, A is a right angle, and D, E are points in BC, such that BE=AB, DC=AC; show that DE2=2BD EC. Obtain two expressions for BC2 by means of Euc. I. 47, II. 4, and by the method of the previous Ex. 6. ABCD is a square, and through P a point in AB, PQR is drawn AB, meeting AC in Q, and DC in R; show that AP PB =PQQR=AQʻQC. B D R Apply Euc. II. 4 to AB and AC, etc. 7. Obtain a second proof of Euc. II. 4 from the theorem :-If, in the sides of a given square, four points be taken at equal distances from the angular points towards the same parts, the figure contained by the straight lines which join these points shall be a square. (§ 12, Ex. 18.) Use Euc. I. 4, I. 13, and I. 32, to show that EFGH is a square. Use § 18, Ex. 5, to show that each of the triangles= AE EB, and... AB2 EF2+2AE EB. Apply Euc. I. 47. H B D 8. If, in a right-angled triangle, one of the sides which contain the right angle be equal to twice the other, the altitude from the right angle shall divide the hypotenuse into segments, one of which is four times the other. Use Ex. 1 (2), and § 31, Ex. 3 B N 9. ABC is a triangle, right-angled at A, and on AB and AC squares ABFG, ACKH are described; show that 10. ABC is an isosceles triangle, having AB-AC, and BN is the altitude from B; show that (1) BN2=2AN NC+NC2. (2) BC2=2AC NC. (1) Use Euc. I. 47 and II. 4. (2) Deduce this from (1) by Euc. II. 3. 11. If, in a triangle, the square on the altitude be equal to the rectangle contained by the segments into which the altitude divides the base internally, the angle from which the altitude is drawn shall be a right angle. See Ex. 1 (1), to which this is a converse. 12. O is the centre of a square ABCD, the diagonal AOC is produced to E, so that CE-AB; show that AE2=20E2. Draw EFL AB, and prove EF=OE, etc. 13. ABC is a triangle, right-angled at A, D is the mid-point of BC, and from D, DE is drawn 1 BC, meeting AB produced if necessary in E; show that BC2=2BA BE. Join EC, and use Euc. I. 47 and II. 4, etc. 14. AB is divided into five equal parts at C, D, E, F; show that AB2 AE2+AF2. 15. Prove Ex. 4 by means of a geometrical construction similar to that used in Euc. II. 4. 16. A, B, C, D are four points in order in a straight line, and BC=CD; show that AB2+2AC CD=AC2+CD2. 17. A, B, C, D, E are five points in order in a straight line, and BC=CD; show that AC2+BE DE CE2+AB'AD. § 33. (Bookwork, EUCLID, II. 1-6.) 1. The rectangle contained by the sum and the difference of two straight lines is equal to the difference between their squares. (A Standard Theorem.) Let AB, CD be the two given straight lines, and let AB> CD, it is required to prove that (AB+CD). (AB-CD)-AB2-CD2. Produce AB to E, so that BE=AB. From BE cut off BF CD. AFFE+BF2=AB2. ... AF.FE=AB2- BF2, or (AB+CD). (AB-CD)-AB2-CD2. 2. Deduce the same theorem from Euc. II. 6. B c [Euc. II. 5. Produce AB to F, so that BF CD. From BA cut off BG=CD. Use Euc. II. 6. 3. Prove the same theorem by means of a geometrical construction. K Let AB, AC be the given straight lines. Describe squares on them as in the figure, and produce ED to K, so that AC. Produce GF and draw KLÖGF. DK Prove EL EH+FB, etc. Observe that this is a general theorem, of which both Euc. II. 5 and II. 6 are particular cases. |