From the point C draw the straight line CIL parallel to AD, meeting AB in I, and DĚ in L. Through G draw GM parallel to AB, meeting AC or CF in M; and through E draw EN parallel to BC, meeting CL in N. L D Then, since 2ACB= a right angle = 2 BCF, ACF is a straight line (30. Cor. 1.), and it is parallel to BG, because BCFG is a parallelogram. Also BAMG is a parallelogram; therefore, since BCFG, BAMG, are parallelograms upon the same base BG, and between the same parallels AF, BG, they are equal to each other, that is, the square described on BC = the parallelogram BAMG. Again AB=BE, and BG=BC; also 2 ABG= -ABC + a right angle = 4CBE; therefore the two parallelograms BAMG, BCNE have two adjacent sides of the one equal to two adjacent sides of the other, each to each, and likewise the angles between those sides equal; and the parallelograms are equal, (40. Cor. 1.); that is, the parallelogram BAMG= the parallelogram BCNE; the square described on BC = the parallelogram BCNE= the parallelogram BILE, which is on the same base BE, and between the same parallels BE, CL. Similarly it may be shewn, by joining DN, and drawing through H a line parallel to AB, that the square on AC= the parallelogram AILD; the square on BC + the square on AC = the parallelogram BILE + the parallelogram AILD = the square ABED= square on AB. N. B. The square described upon a line is generally called, for shortness, the square of that line. Thus the square described upon the line AB is called the square of AB' square of B COR. The converse of this proposition is also true, viz. that if the square described upon one of the sides of a triangle be equal to the sum of the squares described upon the other two sides of it, the angle between these two sides is a right angle.' Let ABC be a triangle, such that square of AC + square of BC = square of AB; from C draw CD at right angles to BC, making CD= AC; and join BD. Then, since CD= AC, the square of CD-square of AC, (42, Cor. 1) and square of CD+ square of BC = square of AC + square of BC; but square of CD+ square of BC A BD, because BCD is a right angle; and square of AC + square of BC= square of AB by the supposition; .. square of BD-square of AB, and .. BD = AB (42, Cor. 3). Hence the two triangles ABC, BCD have all the sides of the one equal to the sides of the other, each to each, and .. the two triangles are equal, and their angles equal, each to each, to which the equal sides are opposite, .- ACB=- BCD= a right angle. 44. PROP. XXII. If a straight line be divided into any two parts, the square of the whole line is equal to the sum of the squares of the two parts together with twice the rectangle* contained by the parts. [DEF. A rectangle is said to be contained by any two of its adjacent sidest.] * A rectangle has been already defined (13) as a plane surface in the form of a parallelogram with all its angles right angles. Care must be taken not to confound it with right angle'. + Since the opposite sides of every parallelogram and therefore of a rectangle, are equal to one another, and likewise the opposite angles, Let AB be a straight line divided into any two parts in C; upon AB describe the C B square ADEB (42); join BD; through the point C draw CGF parallel to AD or BF, and TI! meeting BDin G; and through K G draw HGK parallel to AB. Then, since BCGK is a parallelogram by construction, its opposite sides are equal to each other, and likewise its opposite angles (40), that is, BC= D E KG, BK = CG, - CBK = 2 CGK, and _BCG=<BKG. But since BE=ED, .. - EBD=2 EDB (26); and since KG is parallel to ED, 2 KGB = - EDB (34), :. - KBG, which is the same as 2 EBD,= 2 KGB, and · . BK= KG; but BC= KG, and BK = CG, .. BCGK is equilateral. Again, since BC is parallel to KG, - CBK + 2 BKG = two right angles (34); but - CBK is a right angle, .. BKG is also a right angle; and the opposite angles are equal, .. BCGK has all its angles right angles. And it has been proved to have all its sides equal. It is therefore a square ; and it is the square of BC. Similarly it may be shewn, that HGFD is a square; and it is the square of HG, or AC, since ACGH is a parallelogram of which the side AC=HG. Also, since - BCG is a right angle, .. LACG is a right angle, and :: the parallelogram ACGH is a rectangle, and it is contained by'* AC, CG, or by AC, CB, since CB=CG. And, similarly, EFGK is a rectangle, contained by FG, GK, or by HG, GC, or by AC, CB. And these make up the whole area ABED, which is the square of AB;.. the square of AB= the square of AC + the square BC + twice the rectangle AC, CBt. (40), and since each of the angles of a rectangle is always a right angle, two adjacent sides alone will obviously serve to fix any rectangle; and hence it is, that the rectangle is said to be "contained' by those sides, because nothing more is needed to determine the rectangle. * The expression contained by' is mostly, omitted ; and the rectangle contained by any two lines, as AC, and CB, is simply called the rectangle AC, CB'. + By the help of this Proposition a very elegant proof of the important Theorem in (43) may be given as follows : 45. Prop. XXIII. If a straight line be divided into any two parts, the squares of the whole line and one of the parts are together equal to twice the rectangle contained by the whole and that part, together with the square of the other part. With the same construction as in (44), and in the same manner it may be shewn, that BCGK, and HGFD, are the squares of BC, and AC; and that ACGH is a rectangle, and = EFGK, which is also a rectangle. Add to each of these equals the square BCGK, and then the rectangle ABKH = the rectangle EBCF; .. ABKH+ EBCF = twice the rectangle ABKH = twice the rectangle contained by AB, BC. Now add to these equals the square HDFG, which is the square of AC; then ABKH + EBCF + square of AC = twice the rectangle AB, BC + square of AC. But the former of these equals make up the square ABED + the square BCGK; .. the squares of AB and BC are together equal to twice the rectangle AB, BC, together with the square of AC. 46. Prop. XXIV. In any obluse-angled triangle if a perpendicular be drawn from tħe vertex of either of the acute angles upon the opposite side produced, the square of the side subtending the obtuse angle is greater than the sum of the squares of the sides forming the obtuse angle by twice the rectangle contained by the side which is produced and the part produced, viz. the part intercepted between the perpendicular and the vertex of the obtuse angle. Let - ACB be an obtuse angle of the triangle ABC; and from A draw AD perpendicular to BC produced; Let ABC be a right-angled triangle; C the right angle; produce CA to D, F making ÅD=CB. Upon CD describe the square DCEF. Take EG=CB, and FH =CB; and join BG, GH, HA. It may H then easily be shewn, that ABGH is a square, and that the four triangles ABC, B BGE, GHF, HAD are equal to one an. other in all respects, so that the sum of them is equal to twice the rectangle AC, CB, or AC, AD, since AD=CB. Hence D the square of CD=the square of AB + twice the rectangle AC, AD. But by (44) the square of CD=the square of AC +square of AD+twice the rectangle AC, AD. Therefore the square of AB=square of AC + square of AD= square of AC + square of CB. the square of AB shall be greater than the sum of the squares of AC, and BC, by twice the rectangle BC, CD. For the square of BD= the square of BC + the square of CD + twice the rectangle BC, CD (44). Add to these equals the square of AD; then the B c D square of BD + square of AD = square of BC + square of CD + square of AD + twice the rectangle BC, CD. But (43) square BD + square of AD= square of AB; and square of CD + square of AD = square of AC; .. square of AB= square of BC + square of AC + twice the rectangle BC, CD; that is, square of AB is greater than the sum of the squares of AC, and BC, by twice the rectangle BC, CD. 47. PROP. XXV. In every triangle the square of the side opposite to any acute angle is less than the sum of the squares of the sides forming that angle by twice the rectangle contained by either of these sides and that part of it which is intercepted between a perpendicular let fall upon it from the vertex of the opposite angle and the acute angle. Let ABC be an acute angle of the triangle ABC; from A draw AD perpendicular to BC meeting it in the point D. Then the square of AC shall be less than the squares of AB and BC by twice the rectangle BC, BD. For by (45) the square of BC +the square of BD= twice the rectangle BC, BD+the square of CD. Add to each of these equals the square of AD; then the square of BC + the square of BD + the square of AD= twice the rectangle BC, B BD + the square of CD+ the square of AD. But, since - ADB is a right angle, the squares of BD, and AD, are equal to the square of AB (43); also the squares of CD and AD are equal to the square of AC; .. the square of BC + the square of AB = twice the rectangle BC, BD+the square of AC; that is, the square of AC is less than the sum of the squares of AB, and BC by twice the rectangle BC, BD, |