II. 4. If a straight line be divided into any two parts, the square on the whole line = the squares on the parts, and twice the rectangle contained by the parts. II. 5. If a straight line be divided into two equal and also into two unequal parts, the rectangle contained by the unequal parts, together with the square on the line between the points of section, is equal to the square on half the line. Rect. AD, DB and sq. on CD=sq. on CB. II. 6. If a straight line be bisected, and produced to any point, the rectangle contained by the whole line thus produced, and the part of it produced, together with the square on half the line bisected, is equal to the square on the straight line which is made up of the half and the part produced. Rect. AD, DB and sq. on CB=sq. on CD. II. 10. If a straight line be bisected and produced to any point, the square on the whole line thus produced, and that on the part produced, are double of the squares on half the line bisected, and on the line made up of the half and the part produced. Sqq. on AD, DB = twice sqq. on AC, CD. LECTURE XXI. PROPOSITIONS 11-14.-ANALYSIS OF BOOK II. THE second book concludes with two problems and two theorems depending on the principles laid down in the previous ten propositions. We will first consider the theorems. Props. XII. and XIII. which complete the subject begun in I. 47, show the relation of the square subtending (1) an obtuse, (2) an acute angle, to the squares on the sides containing either of these angles.. B In the enunciation of these theorems the term projection is made use of. The projection of one line upon another is that part of the second (produced if necessary) intercepted between perpendiculars let fall from the extremities of the first. If, however, the lines meet at an angle, the projection is the line intercepted between that angle and the perpendicular let fall from the other extremity. Prop. XII. shows that in obtuseangled triangles the square on the D B C side subtending the obtuse angle is greater than the squares on the sides containing it by twice the rectangle contained by one of these sides and the projection of the other upon it. Let ABC be a triangle having an obtuse angle ACB, from A let fall a perpendicular upon BC produced to D. We are required to show that sq. on AB: =sqq. BC, CA, and twice the rectangle BC, CD. What do we know about the squares on these lines? (1) Sq. on AB = sqq. AD, DB (I. 47). (2) sq. on DB =sqq. on BC, CD and twice rectangle Then sq. on AB = sqq. on AD, CD, BC, and twice the rect. BC, CD. But sq. on AC = sqq. on AD, DC. Therefore the sq. on AB = sqq. on AC, CB and twice the rect. BC, CD.-Q.E.D. Prop. XIII. shows that in any triangle the square on the side subtending an acute angle is less than the squares on the sides containing it by twice the rectangle contained by either of those sides and the projection of the third side upon it. Let ABC be any triangle of which ABC is an acute angle; let fall a perpendicular from A upon BC (produced if necessary). We are required to show that the square on AC with twice rect. CB, BD = the sqq. on CB, BA. The perpendicular may fall upon BC produced, as in the figure of II. 12, or it may fall within the figure, or lastly, if the triangle is right-angled, it may coincide with a side of the triangle. 44 4 B D C B In the last case BC is itself the projection of AB upon BC, therefore the rectangle CB, BD is represented by the square on BC. Hence the demonstration of this case is obvious, for Sq. on AB sqq. on AC and BC. therefore the sq. on AC is less then sqq. on AB, BC by twice the sq. on BC (that is rect. CBBC). Let us examine the first case. Sq. on AC sqq. on AD, DC (I. 47). Twice rect. CB, BD and sq. on CD = sqq. on BC and BD (II. 7). Then the sq. on AC and twice the rect. CB, BD = sqq. on AD, BD, and BC. |