4. The difference between the squares on two sides of a triangle is equal to twice the rectangle contained by the base and the segment of the base intercepted between its mid-point and the foot of the altitude. (A Standard Theorem.) A B D B Use § 20, Ex. 5 and $ 33, Ex. 1. Show that 2DN=BN - NC in figure 1, and that 2DN=BN+CN in figure 2, and hence that AB2 ~ AC=2BC.DN. 5. In the A ABC, D, E, F are the mid-points of BC, CA, AB, and AP, BQ, CR are the altitudes. If BC>CA>AB, show that BC.PD+AB.RF=CA EQ. As in the previous Ex., show CA2-AB2=2BC.PD, etc. 6. In the A ABC, AB=AC, and PQ || BC, meeting AB, AC in P, Q; show that BQ2-CQ2=BC.PQ. Draw PM, QNIBC. point in AB, or BA, produced. B M N 7. In the A ABC, AM bisects < A, CM I AM, and meets AB in D, and BN I CD; show that BC2 - BD2=4CM.MN. Compare Ex. 4. B 8. If a square and a rectangle have the same perimeter, show that the square has the greater area. In the figure of Ex. 3, show that AD and EL have the same perimeter, but that AD=EL+AF. 9. In a right-angled triangle, the square on one of the sides which contain the right angle is equal to the rectangle contained by the sum and difference of the hypotenuse and the other side. Use Ex. 1. 10. A, B, C, D are four points in order in a straight line, and AB=BC; show that AD2-CD2=2AC:BD. Use Ex. 1. 11. A, B, C, D are four points in order in a straight line, and AB=BD; show that AC2-CD2=2AD-BC. 12. A, B, C are three points in a straight line, such that AB=BC, and P is any point in the line; show that AP2-CP2=4 AB.BP. 13. ABC is an isosceles triangle, and D is the mid-point of the base BC; BC is produced to E, so that DE=AB; show that AD2=BECE. 14. ABC is a triangle, right-angled at A, E is the mid-point of AC, and EM is drawn IBC; show that BM2-CM2=AB2. * 15. A, B, C, D, E, F are six points in order in the same straight line, and AB=BC=CD=DE; show that AF2-EF2=2 (BF2-DF2). $ 34, (Bookwork, EUCLID, II. 1-8.) DEFINITION.-If a point be tuken in a straight line produced, its distances from the extremities of the line are called Segments of the line, and the point is called a Point of External Section. A straight line is equal to the sum of the segments into which it is divided by an internal point, and to the difference between the segments into which it is divided by an external point. 1. If, from the vertex of an isosceles triangle, a straight lino be drawn to meet the base, or base produced, the difference of the squares on this line, and on one of the equal sides of the triangle, is equal to the rectangle contained by the segments of the base. (A Standard Theorem.) In the A ABC, let AB=AC, and P be any point in BC, or BC produced, it is required to prove that AB2 ~ AP2= BP.PC. Bisect BC in D, and join AD. AD BC, =(BD+DP) (BD-DP); In figure 1, when P is in BC, this result becomes AB2-AP2=BP.PC; but in figure 2, where P is in BC produced, it is AP2- AB2=BP.CP. Note. It is to be observed that these two results are algebraically identical, if we assume that CP=-PC, and therefore BP.CP=-BP.PC; or, in other words, that the line PC changes its sign when P passes to the other side of C, so that the line is measured in the opposite direction. A similar use of the sign - to indicate the opposite direction is made in Trigonometry. 2. If a straight line be divided externally, the square on the straight line is equal to the sum of the squares on the segments diminished by twice the rectangle contained by the segments. (A Standard Theorem.) Use Euc. II. 7 to show that A B + P AB2= AP2+BP2-2AP.BP. The above theorem may be regarded as an alternative method of stating Euc. II. 7. It shows the relation which that proposition bears to Euc. II. 4. Note.--Here, as in Ex. 1, we observe that when the point of section is external instead of internal, the same theorem holds good if we consider that one of the segments, and therefore the rectangle contained by the segments has changed sign, and must be subtracted instead of added. 3. In any triangle, in which the altitude from the vertex falls beyond the base, the square on the base, together with twice the square on the altitude, is equal to the sum of the squares on the sides diminished by twice the rectangle contained by the segments of the base. Use Euc. I. 47 and II. 7 (or Ex. 1) to show that BC2+2AN=AB2+ACP-2BN.CN. Compare § 32, Ex. 3, and observe that the results are algebraically identical if we assume that CN=-NC. 4. If, in the produced 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. Hence obtain a second proof of Euc. II. 7. a B Compare $ 32, Ex. 7. |