On AB describe the square AGHB; through D draw DKL parallel to AG: then KH is the square on BC. Then it may be seen that EL, DO, BF are equal figures; but the difference of the squares on AB, BC is the figure made up of AO and EL, that is, it is equivalent to the figure made up of AO and BF, that is, to AF, which is the rectangle contained by the sum and difference of AB, BC. COR. If a straight line is bisected and divided in any point, the rectangle contained by the segments is equal to the difference of the squares on half the line and the line between the points of section. Proof. For let AB be bisected in C, and divided internally or externally in P. Then AP is the sum of AC and CP, and PB is their difference, since BC= AC. Therefore the rectangle contained by AP, PB is equal to the difference of the squares of AC, and CP. Remark. The student will begin here to suspect, what he will afterwards find to be true, that there is an intimate relation between geometry and algebra. Algebraical or analytical geometry as it is called, investigates this relation and applies it to the establishment of theorems in geometry, and will occupy him at a later stage of his mathematical studies. We shall at present use the expression AB2, which is read 'AB squared,' only as an abbreviation for "the square on AB," and AB × AC or AB.AC, as an abbreviation for "the rectangle contained by AB and AC." These three theorems may be used to demonstrate other properties of divided lines. For example, THEOREM 28. If a straight line be divided into two equal and also into two unequal segments, the squares of the two unequal segments are together double of the square of half the line bisected, and the square on the line between the points of section. Let AB be bisected in C, and divided internally or externally in D. A टं D B B Ď Then the squares on AD, DB will be double of the squares on AC, CD. Proof. For AD2 = AC2 + CD2 + 2AC × CD by Th. 25; and DB3 = CB2 + CD3 − 2BC × CD by Th. 26; therefore, adding, and remembering that AC-BC, and that therefore the rectangle AC× CD = the rectangle BC × CD, we get that AD2 + DB2 = 2AC2 + 2 CD2. THEOREM 29. In any right-angled triangle the square on the hypothenuse is equivalent to the sum of the squares on the sides which contain the right angle. [Воок Therefore the triangle DAC is on the same base DA, and between the same parallels DA, EC with the square DABE. Therefore the triangle DAC is half the square DABE (Th. 23, Cor. 3). Similarly the triangle BAH is half the rectangle AJ. But the triangles DAC, BAH are equal (Th. 16); for the sides DA, AC are respectively equal to BA, AH, and the contained angle DAC the contained angle BAH, each of them being a right angle together with BAC. = Therefore the rectangle AJ= the square DABE. Similarly it may be shewn that the rectangle CJ = the square BCGF, and therefore, since AJ and CJ make up the whole square AHIC, the square AHIC is equivalent to the sum of the squares ABDE and BCGF, that is, AC2 = AB2 + BC2. COR. I. The square of a side subtending an obtuse or acute angle is not equal to the sum of the squares of the side containing that angle. For if BD is drawn at right angles to BC and equal to BA, and DC joined, then AC is greater or less than DC, according as the angle CBA is obtuse or acute by Th. 17. Therefore AC is greater or less than DC, that is, than DB2+BC2 or than AB2 + BC2. COR. 2. Hence it follows that the converse theorem holds, viz. that if the square on one side of a triangle is equal to the squares on the other two sides, the triangle is right-angled. COR. 3. It follows that in a triangle right-angled at B, AB AC - BC2 and BC2 = AC2 – AB2. Def. 39. The projection of one line on another line is the portion of the latter intercepted between perpendiculars let fall on it from the extremities of the former. Thus the projections of AB, CD on EF are the lines ab, cd respectively. It is clear that the line EF must be supposed indefinitely long. There could be no projection of AB on the terminated line GF. THEOREM 30. In any triangle the square on a side opposite an acute angle is less than the squares on the sides containing that angle by twice the rectangle contained by either of those sides and the projection on it of the other. Let ABC be a triangle, B an acute angle, BD the projection of AB on BC, then will AC = AB + BC2 - 2 CB × BD. Proof. For AC2 = AD2 + DC by Theorem 29, B D C but AD2 = AB3 – BD3, by the same Theorem, and DC' = BC2 + BD3 − 2 CB × BD (by Theorem 26). Therefore AC2 = AB2 + BC2 - 2 CB × BD. W. G. F THEOREM 31. In an obtuse-angled triangle the square on the side subtending the obtuse angle is greater than the squares on the sides containing that angle by twice the rectangle contained by either of these sides and the projection on it of the other side. Let ABC be the triangle, ABC being the obtuse angle, BD the projection of AB on BC, BC being produced backward. Then will AC2 = AB2 + BC2 + 2CB . BD, for but and therefore ACAD + DC, by Theorem 29, DCCB+BD2 + 2CB. BD, by Th. 25, EXERCISES ON EQUIVALENT FIGURES. Ex. 1. If through any point in the diagonal of a parallelogram lines be drawn parallel to the sides, the two parallelograms so formed through which the diagonal does not pass are equivalent to one another. |