PROPOSITION V. THEOREM. The area of any parallelogram is equal to the product of its base by its altitude. : For, the parallelogram ABCD is equivalent FD E C to the rectangle ABEF, which has the same base AB, and the same altitude BE (Prop. I. Cor.): but this rectangle is measured by AB XBE (Prop. IV. Sch.); therefore, AB x BE A B is equal to the area of the parallelogram ABCD. Cor. Parallelograms of the same base are to each other as their altitudes ; and parallelograms of the same altitude are to each other as their bases : for, let B be the common base, and C and D the altitudes of two parallelograms: then, BxC : BxD::C: D, (Book II. Prop. VII.) And if A and B be the bases, and C the common altitude, we shall have AC: BXC ::A: B. And parallelograms, generally, are to each other as the products of their bases and altitudes. PROPOSITION VI. THEOREM The area of a triangle is equal to the product of its base by half its altitude. For, the triangle ABC is half of the parallelogram ABCE, which has the same base BC, and the same altitude AD (Prop. II.); but the area of the parallelogram is equal to B D BC AD (Prop. V.); hence that of the triangle must be BC x AD, or BC RAD. Cor. Two triangles of the same altitude are to each other as their bases, and two triangles of the same base are to each other as their altitudes. And triangles generally, are to each other, as the products of their bases and altitudes. PROPOSITION VII. THEOREM. The area of a trapezoid is equal to its altitude multiplied by the half sum' of its parallel bases. DE B Let ABCD be a trapezoid, EF its a ti с K *tude, AB and CD its parallel bases; then will its area be equal to EF x }(AB+CD). Through I, the middle point of the side BC, draw KL parallel to the opposite side AD; and produce DC till it meets KL. T L In the triangles IBL, ICK, we have the side IB=IC, by construction; the angle LIB=CIK; and since CK and BL are parallel, the angle IBL=ICK (Book I. Prop. XX. Cor. 2.); hence the triangles are equal (Book I. Prop. VI.); therefore, the trapezoid ABCD is equivalent to the parallelogram ADKL, and is measured by EFX AL. But we have AL=DK; and since the triangles IBL and KCI are equal, the side BL=CK: hence, AB+CD=AL+ DK=2AL; hence AL is the half sum of the bases AB, CD; hence the area of the trapezoid ABCD, is equal to the altitude EF multiplied by the half sum of the bases AB, CD, a result AB+CD which is expressed thus : ABCD=EFx 2 Scholium. If through I, the middle point of BC, the line IH be drawn parallel to the base.AB, the point H will also be the middle of AD. For, since the figure AHIL is a parallelogram, as also DHIK, their opposite sides being parallel, we have AH=IL, and DH=IK; but since the triangles BIL, CIK, are equal, we already have IL IK ; therefore, AHODH. It may be observed, that the line HI=AL is equal to AB+CD ; hence the area of the trapezoid may also be ex2 pressed by EFX HI: it is therefore equal to the altitude of the trapezoid multiplied by the line which connects the middle points of its inclined sides Х PROPOSITION VIII. THEOREM. HD If a line is divided into two parts, the square described on the whole line is equivalent to the sum of the squares described on AC? or (AB+BC)?=AB + BC2+2AB BC. F G The square ACDE is made up of four parts; the first ABIF is the square described on AB, since we made AF=AB: the second IDGH is A вс the square described on IG, or BC; for since we have AC= AE and AB=AF, the difference, AC—AB must be equal to the difference AE—AF, which gives BC=EF ; but IG is equal to BC, and DG to EF, since the lines are parallel ; therefore IGDH is equal to a square described on BĈ. And those two squares being taken away from the whole square, there remains the two rectangles BCGI, EFIH, each of which is measured by AB X BC: hence the large square is equivalent to the two small squares, together with the two rectangles. Cor. If the line AC were divided into two equal parts, the two rectangles EI, IC, would become sqnares, and the square described on the whole line would be equivalent to four times the square described on half the line. Scholium. This property is equivalent to the property demonstrated in algebra, in obtaining the square of a binominal ; which is expressed thus : (a+b)=a?+ 2ab +6. PROPOSITION IX, THEOREM. The square described on the difference of two lines, is equivalent to the sum of the squares described on the lines, minus turce the rectangle contained by the lines. Let AB ard BC be two lines, AC their difference; then is AC, or (AP-BC)=ABP + BC2_2AB x BC. Describe the square ABIF; take AE L G =AC; draw CG parallel to to BI, HK parallel to AB, and complete the square K H EFLK. E D The two rectangles CBIG, GLKD, are each measured by AB x BC ; take them away from the whole figure А. ABILKEA, which is equivalent to AB? + BC?, and there will evidently remain the square ACDE; hence the theorem is true. Scholium. This proposition is equivalent to the algebraical formula, (a−b)=a_2ab+b?. PROPOSITION X. THEOREM. of The rectangle contained by the sum and the difference lincs, is equivalent to the difference of the squares of those lines. F Let AB, BC, be two lines; then, will (AB+BC) x (AB-BC)=AB2-BC?. On AB and AC, describe the squares o I ABIF, ACDE; produce AB till the produced part BK is equal to BC ; and H E complete the rectangle AKLE. D The base AK of the rectangle EK, is the sum of the two lines AB, BC; its altitude AE is the difference of the same lines; therefore the rectangle с в K AKLE is equal to (AB+BC) x (ABBC). But this rectangle is composed of the two parts ABHE + BULK ; and the part BHLK is equal to the rectangle EDGF, because BH is equal to DE, and BK to EF; hence AKLE is equal to ABHE+EDGF. These two parts make up the square ABIF minus the square DHIG, which latter is equal to a square described on BC: hence wo have (AB+BC) (AB-BC)=ABS_BC?. Scholium. This proposition is equivalent to the algebraical formula, (a+b) x (a-6)=a?_6%. G* PROPOSITION XI. THEOREM. The square described on the hypothenuse of a right angled tri angle is equivalent to the sum of the squares described on the other two sides, E G Let the triangle ABC be right angled at A. Having described xquares on the three sides, let fall from A, on the hypothenuse, the perpendicular AD, which produce to E; and draw the H diagonals AF, CH. B C The angle ABF is made up of the angle ABC, together with tbe right angle CBF; the angle CBI is made up of the same angle ABC, logether with the right angle ABH; hence the angle ABF is equal to HBC. But we have AB=BII, being sides of the same square ; and BF=BC, for the same reason : therefore the triangles ABF, HBC, have two sides and the included angle in each equal; therefore they are themselves equal (Book 1. Prop. V.). The triangle ABF is half of the rectangle BE, because they have the samo base BF, and the same altitude BD (Prop. II. Cor. 1.), The triangle HBC is in like manner half of the square AH: for the angles BAC, BAL, being both right angles, AC and AL form one and the same straight line parallel to HB (Book I. Prop. III.); and consequently the triangle HIBC, and the square AH, which have the common base BH, have also the common altitude AB; hence the triangle is half of the square. The triangle ABF has already been proved equal to the triangle HBC; hence the rectangle BDEF, which is double of the triangle ABF, must be equivalent to the square AH, which is double of the triangle HBC. In the same manner be proved, that the rectangle CDEG is equivalent to the square AI. But the two rectangles BDEF, CHEG, taken together, make up the square BCGF; therefore the square BCGF, described on the hypothenuse, is equivalent to the sum of the squares ABHL, ACIK, described on the two other sides; in ether words, BC?=AB? + AC%. it may |