Exercises.
Exercises.
1.The extremities of the base of an isosceles triangle are equally distant from any point in theperpendicular from the vertical angle on the base.
2.If the line which bisects the vertical angle of a triangle also bisects the base, the triangle isisosceles.
3.The locus of a point which is equally distant from two fixed lines is the pair of lines whichbisect the angles made by the fixed lines.
4.In a given right line find a point such that the perpendiculars from it on two given lines maybe equal. State also the number of solutions.
5.If two right-angled triangles have equal hypotenuses, and an acute angle of one equal to anacute angle of the other, they are congruent.
6.If two right-angled triangles have equal hypotenuses, and a side of one equal to a side of theother, they are congruent.
7.The bisectors of the three internal angles of a triangle are concurrent.
8.The bisectors of two external angles and the bisector of the third internal angle areconcurrent.
9.Through a given point draw a right line, such that perpendiculars on it from two given pointson opposite sides may be equal to each other.
10.Through a given point draw a right line intersecting two given lines, and forming anisosceles triangle with them.
ParallelLines.
ParallelLines.
Def.i.—If two right lines in the same plane be such that, when producedindefinitely, they do not meet at any finite distance, they are said to beparallel.
Def.ii.—Aparallelogramis a quadrilateral, both pairs of whose opposite sides are parallel.
Def.iii.—The right line joining either pair of opposite angles of a quadrilateral is called adiagonal.
Def.iv.—If both pairs of opposite sides of a quadrilateral be produced to meet, the right line joining their points of intersection is called itsthirddiagonal.
Def.v.—A quadrilateral which has one pair of opposite sides parallel is called atrapezium.
Def.vi.—If from the extremities of one right line perpendiculars be drawn to another, the intercept between their feet is called theprojectionof the first line on the second.
Def.vii.—When a right line intersects two other right lines in two distinct points it makes with them eight angles, which have received special names in relation to one another. Thus, in the figure—1, 2; 7, 8 are calledexteriorangles; 3, 4; 5, 6,interiorangles. Again, 4; 6; 3, 5 are calledalternateangles; lastly, 1, 5; 2, 6; 3, 8; 4, 7 are calledcorrespondingangles.
PROP.XXVII.—Theorem.If a right line(EF)intersecting two right lines(AB,CD)makes thealternate angles(AEF,EFD)equal to each other, these lines are parallel.
PROP.XXVII.—Theorem.If a right line(EF)intersecting two right lines(AB,CD)makes thealternate angles(AEF,EFD)equal to each other, these lines are parallel.
Dem.—IfABandCDare not parallel they must meet, if produced, at some finite distance: if possible let them meet inG; then the figureEGFis a triangle, and the angleAEFis an exterior angle, andEFDa non-adjacent interior angle. Hence [xvi.]AEFis greater thanEFD; but it is also equal to it (hyp.), that is, both equal and greater, which is absurd.HenceABandCDareparallel.
Or thus:BisectEFinO; turn the whole figure roundOas a centre, so thatEFshall fall on itself; then becauseOE=OF, the pointEshall fall onF; and because the angleAEFis equal to the angleEFD, the lineEAwill occupy the place ofFD, and the lineFDthe place ofEA; therefore the linesAB,CDinterchange places, and the figure is symmetrical with respect to the pointO. Hence, ifAB,CDmeet on one side ofO, they must also meet on the other side; but two right lines cannot enclose a space (Axiomx.); therefore they do not meet at either side.Hence they areparallel.
PROP.XXVIII.—Theorem.
PROP.XXVIII.—Theorem.
If a right line(EF)intersecting two right lines(AB,CD)makes the exterior angle(EGB)equal to its corresponding interior angle(GHD), or makes two interiorangles(BGH,GHD)on the same side equal to two right angles, the two right linesare parallel.
Dem.—1. Since the linesAB,EFintersect, the angleAGHis equal toEGB[xv.]; butEGBis equal toGHD(hyp.); thereforeAGHis equal toGHD, and they are alternate angles.Hence[xxvii.]ABis parallel toCD.
2. SinceAGHandBGHare adjacent angles, their sum is equal to two right angles [xiii.]; but the sum ofBGHandGHDis two right angles (hyp.); therefore rejecting the angleBGHwe haveAGHequalGHD, and they are alternate angles;thereforeABis parallel toCD[xxvii.].
PROP.XXIX.—Theorem.
PROP.XXIX.—Theorem.
If a right line(EF)intersect two parallel right lines(AB,CD), it makes—1.thealternate angles(AGH,GHD)equal to one another;2.the exterior angle(EGB)equal to the corresponding interior angle(GHD);3.the two interior angles(BGH,GHD)on the same side equal to two right angles.
Dem.—If the angleAGHbe not equal toGHD, one must be greater than the other. LetAGHbe the greater; to each addBGH, and we have the sum of the anglesAGH,BGHgreater than the sum of the anglesBGH,GHD; but the sum ofAGH,BGHis two right angles; therefore the sum ofBGH,GHDis less than two right angles, and therefore (Axiomxii.) the linesAB,CD, if produced, will meet at some finite distance: but since they are parallel (hyp.) they cannot meet at any finite distance.Hence the angleAGHis not unequal toGHD—that is, it is equal toit.
2. Since the angleEGBis equal toAGH[xv.], andGHDis equal toAGH(1),EGBis equal toGHD(Axiomi.).
3. SinceAGHis equal toGHD(1), addHGBto each, and we have the sum of the anglesAGH,HGBequal to the sum of the anglesGHD,HGB; but the sum of the anglesAGH,HGB[xiii.] is two right angles;therefore the sum of the anglesBGH,GHDis two right angles.
Exercises.
Exercises.
1.Demonstrate both parts of Prop.xxviii.without using Prop.xxvii.
2.The parts of all perpendiculars to two parallel lines intercepted between them areequal.
3.IfACD,BCDbe adjacent angles, any parallel toABwill meet the bisectors of these anglesin points equally distant from where it meetsCD.
4.If through the middle pointOof any right line terminated by two parallel right lines anyother secant be drawn, the intercept on this line made by the parallels is bisected inO.
5.Two right lines passing through a point equidistant from two parallels intercept equalportions on the parallels.
6.The perimeter of the parallelogram, formed by drawing parallels to two sides of an equilateraltriangle from any point in the third side, is equal to twice the side.
7.If the opposite sides of a hexagon be equal and parallel, its diagonals are concurrent.
8.If two intersecting right lines be respectively parallel to two others, the anglebetween the former is equal to the angle between the latter. For ifAB,ACbe respectivelyparallel toDE,DF, and ifAC,DEmeet inG, the anglesA,Dare each equal toG[xxix.].
PROP.XXX.—Theorem.If two right lines(AB,CD)be parallel to the same right line(EF), they areparallel to one another.
PROP.XXX.—Theorem.If two right lines(AB,CD)be parallel to the same right line(EF), they areparallel to one another.
Dem.—Draw any secantGHK. Then sinceABandEFare parallel, the angleAGHis equal toGHF[xxix.]. In like manner the angleGHFis equal toHKD[xxix.]. Therefore the angleAGKis equal to the angleGKD(Axiomi.).Hence[xxvii.]ABis parallel toCD.
PROP.XXXI.—Problem.Through a given point(C)to draw a right line parallel to a given right line.
PROP.XXXI.—Problem.Through a given point(C)to draw a right line parallel to a given right line.
Sol.—Take any pointDinAB. JoinCD(Post.i.), and make the angleDCFequal to the angleADC[xxiii.].The lineCEis parallel toAB[xxvii.].
Exercises.
Exercises.
1.Given the altitude of a triangle and the base angles, construct it.
2.From a given point draw to a given line a line making with it an angle equal to a given angle.Show that there will be two solutions.
3.Prove the following construction for trisecting a given lineAB:—OnABdescribe anequilateral△ABC. Bisect the anglesA,Bby the linesAD,BD, meeting inD; throughDdraw parallels toAC,BC, meetingABinE,F:E,Fare the points of trisection ofAB.
4.Inscribe a square in a given equilateral triangle, having its base on a given side of thetriangle.
5.Draw a line parallel to the base of a triangle so that it may be—1.equal to the intercept itmakes on one of the sides from the extremity of the base; 2.equal to the sum of the two interceptson the sides from the extremities of the base; 3.equal to their difference. Show that there are twosolutions in each case.
6.Through two given points in two parallel lines draw two lines forming a lozenge with thegiven parallels.
7.Between two lines given in position place a line of given length which shall be parallel to agiven line. Show that there are two solutions.
PROP.XXXII.—Theorem.
PROP.XXXII.—Theorem.
If any side(AB)of a triangle(ABC)be produced(toD), the external angle(CBD)is equal to the sum of the two internalnon-adjacent angles(A,C), and thesum of the three internal angles is equal to two right angles.
Dem.—DrawBEparallel toAC[xxxi.]. Now sinceBCintersects the parallelsBE,AC, the alternate anglesEBC,ACBare equal [xxix.]. Again, sinceABintersects the parallelsBE,AC, the angleEBDis equal toBAC[xxix.]; hence the whole angleCBDis equal to the sum of the two anglesACB,BAC: to each of these add the angleABCand we have the sum ofCBD,ABCequal to the sum of the three anglesACB,BAC,ABC: but the sum ofCBD,ABCis two right angles [xiii.];hence the sum of the three anglesACB,BAC,ABCis two rightangles.
Cor.1.—If a right-angled triangle be isosceles, each base angle is half a right angle.
Cor.2.—If two triangles have two angles in one respectively equal to two angles in the other, their remaining angles are equal.
Cor.3.—Since a quadrilateral can be divided into two triangles, the sum of its angles is equal to four right angles.
Cor.4.—If a figure ofnsides be divided into triangles by drawing diagonals from any one of its angles there will be (n−2) triangles; hence the sum of its angles is equal 2(n−2) right angles.
Cor.5.—If all the sides of any convex polygon be produced, the sum of the external angles is equal to four right angles.
Cor.6.—Each angle of an equilateral triangle is two-thirds of a right angle.
Cor.7.—If one angle of a triangle be equal to the sum of the other two, it is a right angle.
Cor.8.—Every right-angled triangle can be divided into two isosceles triangles by a line drawn from the right angle to the hypotenuse.
Exercises.
Exercises.
1.Trisect a right angle.
2.Any angle of a triangle is obtuse, right, or acute, according as the opposite side is greaterthan, equal to, or less than, twice themediandrawn from that angle.
3.If the sides of a polygon ofnsides be produced, the sum of the angles between each alternatepair is equal to 2(n−4) right angles.
4.If the line which bisects the external vertical angle be parallel to the base, the triangle isisosceles.
5.If two right-angled△sABC,ABDbe on the same hypotenuseAB, and the verticesCandDbe joined, the pair of angles subtended by any side of the quadrilateral thus formed areequal.
6.The three perpendiculars of a triangle are concurrent.
7.The bisectors of two adjacent angles of a parallelogram are at right angles.
8.The bisectors of the external angles of a quadrilateral form a circumscribed quadrilateral, thesum of whose opposite angles is equal to two right angles.
9.If the three sides of one triangle be respectively perpendicular to those of another triangle,the triangles are equiangular.
10.Construct a right-angled triangle, being given the hypotenuse and the sum or difference ofthe sides.
11.The angles made with the base of an isosceles triangle by perpendiculars from its extremitieson the equal sides are each equal to half the vertical angle.
12.The angle included between the internal bisector of one base angle of a triangle and theexternal bisector of the other base angle is equal to half the vertical angle.
13.In the construction of Prop.xviii.prove that the angleDBCis equal to half the differenceof the base angles.
14.IfA,B,Cdenote the angles of a△, prove that12(A+B),12(B+C),12(C+A) will be theangles of a△formed by any side and the bisectors of the external angles between that side and theother sides produced.
PROP.XXXIII.—Theorem.The right lines(AC,BD)which join the adjacent extremities of two equaland parallel right lines(AB,CD)are equal and parallel.
PROP.XXXIII.—Theorem.The right lines(AC,BD)which join the adjacent extremities of two equaland parallel right lines(AB,CD)are equal and parallel.
Dem.—JoinBC. Now sinceABis parallel toCD, andBCintersects them, the angleABCis equal to the alternate angleDCB[xxix.]. Again, sinceABis equal toCD, andBCcommon, the trianglesABC,DCBhave the sidesAB,BCin one respectively equal to the sidesDC,CBin the other, and the anglesABC,DCBcontained by those sides equal; therefore [iv.] the baseACis equal to the baseBD, and the angleACBis equal to the angleCBD; but these are alternate angles; hence [xxvii.]ACis parallel toBD, and it has been proved equal to it.ThereforeACis both equal and parallel toBD.
Exercises.
Exercises.
1.If two right linesAB,BCbe respectively equal and parallel to two other right linesDE,EF,the right lineACjoining the extremities of the former pair is equal to the right lineDFjoining theextremities of the latter.
2.Right lines that are equal and parallel have equal projections on any other right line;and conversely, parallel right lines that have equal projections on another right line areequal.
3.Equal right lines that have equal projections on another right line are parallel.
4.The right lines which join transversely the extremities of two equal and parallel right linesbisect each other.
PROP.XXXIV.—Theorem.
PROP.XXXIV.—Theorem.
The opposite sides(AB,CD;AC,BD)and the opposite angles(A,D;B,C)of a parallelogram are equal to one another, and either diagonal bisects theparallelogram.
Dem.—JoinBC. SinceABis parallel toCD, andBCintersects them, the angleABCis equal to the angleBCD[xxix.]. Again, sinceBCintersects the parallelsAC,BD, the angleACBis equal to the angleCBD; hence the trianglesABC,DCBhave the two anglesABC,ACBin one respectively equal to the two anglesBCD,CBDin the other, and the sideBCcommon.Therefore[xxvi.]ABis equal toCD,andACtoBD; the angleBACto the angleBDC, and the triangleABCto thetriangleBDC.
Again, because the angleACBis equal toCBD, andDCBequal toABC,thewhole angleACDis equal to the whole angleABD.
Cor.1.—If one angle of a parallelogram be a right angle, all its angles are right angles.
Cor.2.—If two adjacent sides of a parallelogram be equal, it is a lozenge.
Cor.3.—If both pairs of opposite sides of a quadrilateral be equal, it is a parallelogram.
Cor.4.—If both pairs of opposite angles of a quadrilateral be equal, it is a parallelogram.
Cor.5.—If the diagonals of a quadrilateral bisect each other, it is a parallelogram.
Cor.6.—If both diagonals of a quadrilateral bisect the quadrilateral, it is a parallelogram.
Cor.7.—If the adjacent sides of a parallelogram be equal, its diagonals bisect its angles.
Cor.8.—If the adjacent sides of a parallelogram be equal, its diagonals intersect at right angles.
Cor.9.—In a right-angled parallelogram the diagonals are equal.
Cor.10.—If the diagonals of a parallelogram be perpendicular to each other, it is a lozenge.
Cor.11.—If a diagonal of a parallelogram bisect the angles whose vertices it joins, the parallelogram is a lozenge.
Exercises.
Exercises.
1.The diagonals of a parallelogram bisect each other.
2.If the diagonals of a parallelogram be equal, all its angles are right angles.
3.Divide a right line into any number of equal parts.
4.The right lines joining the adjacent extremities of two unequal parallel right lines will meet, ifproduced, on the side of the shorter parallel.
5.If two opposite sides of a quadrilateral be parallel but not equal, and the other pair equal butnot parallel, its opposite angles are supplemental.
6.Construct a triangle, being given the middle points of its three sides.
7.The area of a quadrilateral is equal to the area of a triangle, having two sides equal to itsdiagonals, and the contained angle equal to that between the diagonals.