PROP.XV.—Theorem.
PROP.XV.—Theorem.
1. Two triangles equal in area(ACB,DCE), which have one angle(C)in oneequal to one angle(C)in the other, have the sides about these angles reciprocallyproportional.2. Two triangles, which have one angle in one equal to one angle in theother, and the sides about these angles reciprocally proportional, are equal inarea.
Dem.—1. Let the equal angles be so placed as to be vertically opposite, and thatAC,CDmay form one right line; then it may be demonstrated, as in the last Proposition, thatBC,CEform one right line. JoinBD.
Now since the trianglesACB,DCEare equal,
that is,the sides about the equal angles are reciprocally proportional.
2. IfAC:CD::EC:CB, to prove the triangleACBequal toDCE.
Dem.—Let the same construction be made, then we have
Therefore the triangle
ACB : BCD :: DCE : BCD.
Hence the triangleACB=DCE[V.ix.]—that is,the triangles are equal.
This Proposition might have been appended as aCor.to the preceding, since the triangles arethe halves of equiangular parallelograms, or it may be proved by joiningAE, and showing that it isparallel toBD.
PROP.XVI.—Theorem.
PROP.XVI.—Theorem.
1. If four right lines(AB,CD,E,F)be proportional, the rectangle(AB.F)contained by the extremes is equal to the rectangle(CD.E)contained by themeans.2. If the rectangle contained by the extremes of four right lines be equal to therectangle contained by the means, the four lines are proportional.
Dem.—1. ErectAH,CIat right angles toABandCD, and equal toFandErespectively, and complete the rectangles. Then becauseAB:CD::E:F(hyp.), and thatEis equal toCI, andFtoAH(const.), we haveAB:CD::CI:AH. Hence the parallelogramsAG,CKare equiangular, and have the sides about their equal angles reciprocally proportional. Therefore they are [xiv.] equal; but sinceAHis equal toF,AGis equal to the rectangleAB.F. In like manner,CKis equal to the rectangleCD.E. HenceAB.F=CD.E; that is,therectangle contained by the extremes is equal to the rectangle contained by themeans.
2. IfAB.F=CD.E, to proveAB:CD::E:F.
The same construction being made, becauseAB.F=CD.E, and thatFis equal toAH, andEtoCI, we have the parallelogramAG=CK; and since these parallelograms are equiangular, the sides about their equal angles are reciprocally proportional. Therefore
AB : CD :: CI : AH; that is, AB : CD :: E : F.
Or thus:Place the four lines in a concurrent position so that the extremes may form onecontinuous line, and the means another. Let the four lines so placed beAO,BO,OD,OC. JoinAB,CD. Then becauseAO:OB::OD:OC, and the angleAOB=DOC, the trianglesAOB,CODare equiangular. Hence the four pointsA,B,C,Dare concyclic. Therefore [III.xxxv.]AO.OC=BO.OD.
PROP.XVII.—Theorem
PROP.XVII.—Theorem
1. If three right lines(A,B,C)be proportional, the rectangle(A.C)contained bythe extremes is equal to the square(B2)of the mean.2. If the rectangle contained by the extremes of three right lines be equal to thesquare of the mean, the three lines are proportional.
Dem.—1. Assume a lineD=B; then becauseA:B::B:C, we haveA:B::D:C. Therefore [xvi.]AC=BD; butBD=B2. ThereforeAC=B2; that is,the rectangle contained by the extremes is equal to the square of themean.
2. The same construction being made, sinceAC=B2, we haveA.C=B.D; thereforeA:B::D:C; butD=B. HenceA:B::B:C; that is,the three lines areproportionals.
This Proposition may be inferred as aCor.to the last, which is one of the fundamentalPropositions in Mathematics.
Exercises.
Exercises.
1.If a lineCDbisect the vertical angleCof any triangleACB, its square added to therectangleAD.DBcontained by the segments of the base is equal to the rectangle contained by thesides.
Dem.—Describe a circle about the triangle, and produceCDto meet it inE; then it is easy tosee that the trianglesACD,ECBare equiangular. Hence [iv.]AC:CD::CE:CB; thereforeAC.CB=CE.CD=CD2+CD.DE=CD2+AD.DB[III.xxxv.].
2.If the lineCD′bisect the external vertical angle of any triangleACB, its square subtractedfrom the rectangleAD′.D′Bis equal toAC.CB.
3.The rectangle contained by the diameter of the circumscribed circle, and the radius of theinscribed circle of any triangle, is equal to the rectangle contained by the segments of any chord ofthe circumscribed circle passing through the centre of the inscribed.
Dem.—LetObe the centre of the inscribed circle. JoinOB(seeforegoing fig.); let fall theperpendicularOG, draw the diameterEFof the circumscribed circle. Now the angleABE=ECB[III.xxvii.], andABO=OBC; thereforeEBO= sum ofOCB,OBC=EOB.HenceEB=EO. Again, the trianglesEBF,OGCare equiangular, becauseEFB,ECBare equal, andEBF,OGCare each right. Therefore,EF:EB::OC:OG; thereforeEF.OG=EB.OC=EO.OC.
4.Ex.3 may be extended to each of the escribed circles of the triangleACB.
5.The rectangle contained by two sides of a triangle is equal to the rectangle contained by theperpendicular and the diameter of the circumscribed circle. For, letCEbe the diameter. JoinAE.Then the trianglesACE,DCBare equiangular; henceAC:CE::CD:CB; thereforeAC.CB=CD.CE.
6.If a circle passing through one of the anglesAof a parallelogramABCDintersectthe two sidesAB,ADagain in the pointsE,Gand the diagonalACagain inF; thenAB.AE+AD.AG=AC.AF.
Dem.—JoinEF,FG, and make the angleABH=AFE. Then the trianglesABH,AFEareequiangular. ThereforeAB:AH::AF:AE. HenceAB.AE=AF.AH. Again, it is easy to seethat the trianglesBCH,FAGare equiangular; thereforeBC:CH::AF:AG; henceBC.AG=AF.CH, orAD.AG=AF.CH; but we have provedAB.AE=AF.AH. HenceAD.AG+AB.AE=AF.AC.
7.IfDE,DFbe parallels to the sides of a triangleABCfrom any pointDin the base, thenAB.AE+AC.AF=AD2+BD.DC. This is an easy deduction from 6.
8.If through a pointOwithin a triangleABCparallelsEF,GH,IKto the sides be drawn, thesum of the rectangles of their segments is equal to the rectangle contained by the segments of anychord of the circumscribing circle passing throughO.
9.The rectangle contained by the side of an inscribed square standing on the baseof a triangle, and the sum of the base and altitude, is equal to twice the area of thetriangle.
10.The rectangle contained by the side of an escribed square standing on the base of atriangle, and the difference between the base and altitude, is equal to twice the area of thetriangle.
11.If from any pointPin the circumference of a circle a perpendicular be drawn to any chord,its square is equal to the rectangle contained by the perpendiculars from the extremities of the chordon the tangent atP.
12.IfObe the point of intersection of the diagonals of a cyclic quadrilateralABCD, the fourrectanglesAB.BC,BD.CD,CD.DA,DA.AB, are proportional to the four linesBO,CO,DO,AO.
13.The sum of the rectangles of the opposite sides of a cyclic quadrilateralABCDis equal tothe rectangle contained by its diagonals.
Dem.—Make the angleDAO=CAB; then the trianglesDAO,CABare equiangular;thereforeAD:DO::AC:CB; thereforeAD.BC=AC.DO. Again, the trianglesDAC,OABare equiangular, andCD:AC::BO:AB; thereforeAC.CD=AC.BO. HenceAD.BC+AB.CD=AC.BD.33This Proposition is known as Ptolemy’s theorem.
14.If the quadrilateralABCDis not cyclic, prove that the three rectanglesAB.CD,BC.AD,AC.BDare proportional to the three sides of a triangle which has an angle equal to the sum of apair of opposite angles of the quadrilateral.
15.Prove by using Theorem 11 that if perpendiculars be let fall on the sides and diagonals of acyclic quadrilateral, from any point in the circumference of the circumscribed circle, the rectanglecontained by the perpendiculars on the diagonals is equal to the rectangle contained by theperpendiculars on either pair of opposite sides.
16.IfABbe the diameter of a semicircle, andPA,PBchords from any pointPin thecircumference, and if a perpendicular to AB from any pointCmeetPA,PBinDandE, and thesemicircle inF,CFis a mean proportional betweenCDandCE.
PROP.XVIII.—Problem.
PROP.XVIII.—Problem.
On a given right line(AB)to construct a rectilineal figure similar to agiven one(CDEFG), and similarly placed as regards any side(CD)of thelatter.
Def.—Similar figures are said to besimilarly described upon given right lines, when these lines are homologous sides of the figures.