69. Let BAC be a right angle. On AB describe an equilateral triangle ADB; bisect the angle BAD by the straight line AE: then will the angles BAE, EAD, DAC be all equal.

For the angle BAC is a right angle; the angle BAD is one third of two right angles, that is two thirds of one right angle, by I. 32; therefore the angle CAD is one third of a right angle. And as the angle BAD is bisected by AE the angle BAE is equal to the angle EAD, each being one third of a right angle. Hence the angles BAE, EAD, DAC are all equal, each being one third of a right angle.

70. Let AB be the given straight line. On AB describe an equilateral triangle ABC. Bisect the angle CAB by the straight line AD, and bisect the angle CBA by the straight line BD. Through D draw a straight line parallel to CA, meeting AB at E; and through D draw a straight line DF parallel to CB, meeting AB at F. Then will AE, EF, FB all be equal.

Since DE is parallel to CA the angle EDA is equal to the angle DAC, by I. 29; but the angle DAE is equal to the angle DAC, by construction; therefore the angles EDA and DAE are equal: therefore AE is equal to ED, by I. 6. Similarly it may be shewn that BF is equal to FD.

Because DE is parallel to CA the angle DEF is equal to the angle CAB, by I. 29. Similarly the angle DFE is equal to the angle CBA. Therefore the angle EDF is equal to the angle ACB, by I. 32. Thus the triangle EDF is equiangular; and therefore it is equilateral by I. 6.

Now AE was shewn to be equal to ED; therefore AE is equal to EF. Similarly BF is equal to FE. Thus AE, EF, FB are all equal.

71. Let LM and PQ be the parallel straight lines; and A the given point. Suppose A to be between the parallel straight lines. Through A draw a straight line perpendicular to one of the parallel straight lines, and therefore also perpendicular to the other by I. 29. Let this straight line meet LM at B, and PQ at C. From B on LM take BD equal to AC; and from C on PQ take CE equal to AB, and on the same side of BC as BD is. Join AD and AE: these will be the required straight lines.

For the triangles BAD, CEA are equal in all respects by I. 4; so that AD is equal to AE, and the angle CAE is equal to the angle BDA. But the angles BAD and BDA are together equal to a right angle, by I. 32: therefore the angles BAD and CAE are together equal to a right angle. Therefore the angle EAD is a right angle, by I. 13.

If A is not between the parallel straight lines, CE and BD must be taken on opposite sides of BC.

72. Let ABC be the given triangle, and DE the given perimeter. At the point D make the angle LDE equal to half the angle ABC; and at the point E, on the same side of ED, make the angle MED equal to half the angle ACB. Let DL and EM meet at F. From F draw FG, meeting DE at G, making the angle DFG equal to the angle FDG; and from F draw FH, meeting ED at H, making the angle EFH equal to the angle FEH. Then FGH will be the triangle required.

For FG is equal to DG, and FH is equal to HE, by I. 6. Therefore the sum of the sides FG, GH, HF is equal to DE, the given perimeter.

Also the angle FGH is equal to the sum of the angles FDG, DFG, by I. 32; that is to twice the angle FDG, that is to the angle ABC. Similarly, the angle FHG is equal to the angle ACB. Therefore the angle GFH is equal to the angle BAC, by I. 32.

I. 33, 34.

73. Let ABCD be a quadrilateral having AB parallel to DC, and AD equal to BC. Suppose AB less than DC; from A draw a straight line parallel to BC, meeting DC at E. Then ABCE is a parallelogram. Therefore the angle ABC is equal to the angle AEC, and the side AE is equal to the side BC, I. 34. Therefore AD is equal to AE, and the angle ADE is equal to the angle AED. Therefore the angles ABC and ADE are equal to the angles AEC and AED; that is they are together equal to two right angles, by I. 13.

In the same manner it may be shewn that the angles BAD and BCD are together equal to two right angles.

74. Suppose that AB and CD are equal but not parallel, and that the angle ABD is equal to the angle CDB. Produce AB and CD to meet at E. Since the angles ABD and CDB are equal EB is equal to ED. Therefore also EA is equal to EC; and the angle ECA is equal to the angle EAC. The two angles EBD and EDB are together equal to the two angles EAC and ECA, by I. 32. Therefore the angle EBD is equal to the angle EAC. Therefore BD is parallel to AC, by I. 28.

75. Let ABC be a triangle; let E be any point in AC, and D any point in BC: then AD and BE will not bisect each other.

Let AD and BE intersect at F.

If possible suppose that AF is equal to FD, and BF equal to FE. Then the triangles AFE and DFB are equal in all respects, by I. 4; so that the angle EAF is equal to the angle BDF. Therefore AE is parallel to BD, by I. 27. But this is impossible, since BD and AE, when produced, meet at C.

76. Let ABCD be a quadrilateral, having AB equal to DC, and AD equal to BC: the figure shall be a parallelogram.

Join AC; then in the two triangles ABC and CDA, the sides BA, AC are equal to the sides DC, CA each to each; and the base BC is equal to the base DA: therefore the angle BAC is equal to the angle DCA. Therefore AB is parallel to DC, by I. 27.

Similarly it may be shewn that AD is parallel to BC.

77. Let ABCD be a quadrilateral such that the angle A is equal to the angle C, and the angle B equal to the angle D: then the figure shall be a parallelogram.

The angle A is equal to the angle C, and the angle B is equal to the angle D; therefore the two angles A and B together are equal to the two angles C and D together. But the four angles A, B, C, D together are equal to four right angles, by I. 32; since the quadrilateral may be divided into two triangles by drawing AC or BD. Thus the angles A and B are together equal to two right angles. Therefore AD is parallel to BC, by I. 28. Similarly AB is parallel to DC.

78. Let ABCD be a parallelogram. Draw the diagonals AC and BD intersecting at E: then AC and BD shall be bisected at E.

In the triangles AED and CEB the sides AD and BC are equal, by I. 34; the angles ADE and CBE are equal, by I. 29; and the angles DAE and BCE are equal, by I. 29. Therefore the triangles are equal in all respects by I. 26. Thus AE is equal to CE, and DE is equal to BE.

79. Let ABCD be a quadrilateral figure, and let AC and BD intersect at E; suppose that AC and BD are bisected at E: then ABCD will be a parallelogram.

In the triangles AED and CEB the sides AE, ED are equal to the sides CE, EB each to each, by supposition; the angle AED is equal to the angle CEB, by I. 15; therefore the triangles are equal in all respects, by I. 4. Thus the angle ADE is equal to the angle CBE; therefore AD is parallel to CB, by I. 27.

Similarly AB is parallel to DC.

80.

Let ABCD be a parallelogram, and suppose that the straight line AC bisects the angles at A and C: then the four sides of the parallelogram will be equal.

For in the triangles ACB and ACD the side AC is common, the angle ACB is equal to the angle ACD, and the angle BAC is equal to the angle DAC; therefore the side BC is equal to the side DC, and the side BA to the side DA, by I. 26. But AB is equal to DC, and AD is equal to BC, by I. 34. Therefore the four sides AB, BC, CD, DA are all equal.

81. Let AB and CD be the parallel straight lines, and O the given point. In AB take any point E, and from E as centre, with radius equal to the given length, describe a circle meeting CD at F; join EF. Through O draw a straight line parallel to EF, meeting AB at G, and CD at H. Then EGHF is a parallelogram, by construction; therefore GH is equal to EF, by I. 34: thus GH is equal to the given length.

82. Let ABCD be a parallelogram; let straight lines bisecting the angles A and B meet at E: then AEB will be a right angle.

The angles EAB and EBA are together half of the angles DAB and ABC together. But the angles DAB and ABC are together equal to two right angles, by I. 29. Therefore the angles EAB and EBA are together equal to a right angle. Therefore the angle AEB is a right angle, by I. 32.

83. Let ABCD be a parallelogram. Suppose that the straight lines which bisect the angles A and C are not coincident: then they shall be parallel.

Let the straight line which bisects the angle A meet BC at E; and let the straight line which bisects the angle C meet DA at F. Then, by I. 29, the angle BEA is equal to the angle DAE, that is to the half of the angle DAB, that is to the half of the angle DCB, by I. 34. Thus the angle BEA is equal to the angle BCF. Therefore EA is parallel to CF, by I. 28.

84. Let ABCD be a parallelogram, and suppose that the diagonals AC and BD are equal: then all the angles of the parallelogram will be equal.

In the two triangles ABC and BAD the side AB is common; AD is equal to BC, by I. 34; and AC is equal to BD by supposition: therefore the angle ABC is equal to the angle BAD. Similarly it may be shewn that any other two adjacent angles are equal; so that all the four angles are equal.

85. Let AB and CD be the given straight lines; suppose that the required point is to be at a distance equal to E from AB, and at a distance equal to F from CD.

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Draw a straight line parallel to AB, and at a distance E from it; also draw a straight line parallel to CD, and at a distance F from it; let the two straight lines thus drawn meet at O: then O will be the required point.

For the distance of O from AB will be equal to E, and the distance of O from CD will be equal to F, by I. 34.

Two straight lines can be drawn parallel to AB, and at the required distance from it, namely, one on each side of it; and in like manner two straight lines can be drawn parallel to CD, and at the required distance from it: hence four points can be found which will satisfy the conditions of the problem, assuming that AB and CD are not parallel.

86. Let AB and CD be the two given straight lines in which the required straight line is to be terminated. Let E be a straight line to which the required straight line is to be equal, and F that to which the required straight line is to be parallel.

From A draw a straight line parallel to F, by I. 31; and cut off from it AG equal to E, by I. 3. Through G draw a straight line parallel to AB, Through H draw a straight line parallel to AG, Then HK is the required straight line.

and let it meet CD at H. and let it meet AB at K.

For HK is equal to AG, by I. 34; so that HK is of the required length : and it is parallel to AG, and therefore to F, by I. 30.

87. Let AEB, BFC, CGD be the three equilateral triangles: then will EF be equal to AC, and GF be equal to BD.

In the two triangles ABC and EBF the two sides AB, BC are equal to the two sides EB, BF each to each. The angle FBC is equal to the angle ABE, each being the angle of an equilateral triangle; to each of them add the angle ABF; therefore the angle ABC is equal to the angle EBF. Hence the triangles ABC and EBF are equal in all respects, by I. 4; so that AC is equal to EF.

Similarly it may be shewn that BD is equal to GF.

89. Let ABCD be a parallelogram, and let ABEF be another parallelogram having BE equal to BC, but the angle ABE greater than the angle ABC: then will the diagonal BF be less than the diagonal BD.

The two angles ABC, BCD are together equal to two right angles, and so also are the two angles ABE, BEF, by I. 29; therefore the two angles ABC, BCD are together equal to the two ABE, BEF: but the angle ABE is greater than the angle ABC, by hypothesis; therefore the angle BEF is less than the angle BCD.

In the two triangles BCD, BEF the two sides BC, CD are equal to the two sides BE, EF each to each; but the angle BCD is greater than the angle BEF: therefore the base BD is greater than the base BF, by I. 24.

89. Let AD, BE, CF be the perpendiculars from A, B, C respectively on the straight line: the sum of AD and CF shall be equal to twice BE.

The straight line DF produced does not pass between A and C; suppose that it cuts AC produced through C. Through E draw a straight line parallel to AC, and let it meet AD at G, and CF produced, through F', at H. Thus GE is equal to AB, and EH is equal to BC, by I. 34; therefore GE is equal to EH.

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In the two triangles GED, HEF the angle GED is equal to the angle HEF, by I. 15; the angle DGE is equal to the angle FHE, by I. 29; and the side GE was shewn equal to the side HE: hence these triangles are equal in all respects, by I. 26; so that GD is equal to HF. Therefore AD and CF together are equal to AG and CH together, that is equal to twice BE, by I. 34.

90. Let ABCD be the parallelogram; let O be the point of intersection of the diagonals AC and BD. Then by Exercise 78 the diagonals AC and BD are bisected at O. By Exercise 89 the sum of the perpendiculars from A and C on any straight line outside the parallelogram is twice the perpendicular from 0; and also the sum of the perpendiculars from B and D is twice the perpendicular from 0. Hence the sum of the perpendiculars from A and C is equal to the sum of the perpendiculars from B and D.

91. Let ABCDEF be the six-sided figure. Then AB is by supposition equal and parallel to ED; therefore ABDE is a parallelogram, by I. 33. Therefore AD passes through the middle point of BE, by Exercise 78. Similarly it can be shewn that CF passes through the middle point of BE. Thus AD, BE, and CF meet at a point.

92. Through E draw a straight line parallel to AB, and let it meet AC at F. On FC take FG equal to AF. Join GE and produce it to meet AB at H: then GEH shall be the straight line required.

Through F draw a straight line parallel to GH, and let it meet AB at K. In the triangles AFK and FGE the side AF is equal to the side FG by construction; the angle AFK is equal to the angle FGE, and the angle FAK is equal to the angle GFE, by I. 29: therefore FK is equal to GE, by I. 26. But FK is equal to EH, by I. 34: therefore GE is equal to EH, so that GH is bisected at E.

93. Let ABCD be the given parallelogram, and P the given point on the side AB. On CD take CQ equal to AP; join AC and PQ intersecting at R. Through R draw a straight line at right angles to PQ, meeting AD at S, and CB at T. Then PSQT will be the required rhombus.

In the triangles APR and CQR the sides AP and CQ are equal, by construction; the angle ARP is equal to the angle CRQ, by I. 15; and the angle RAP is equal to the angle RCQ, by I. 29: therefore PR is equal to QR, and AR is equal to CR, by I. 26.

In the triangles PRS and QRS the sides PR and QR are equal; RS is common; and the angles PRS and QRS are equal being right angles: therefore PS is equal to QS, by I. 4.

In the triangles CRT and ARS the sides AR and CR are equal; the angle ARS is equal to the angle CRT, by I. 15; and the angle ASR is equal to the angle CTR, by I. 29: therefore RS is equal to RT, by I. 26.

In the triangles SRP and TRP the sides SR and TR are equal; the side RP is common; and the angles SRP and TRP are equal being right angles : therefore SP is equal to TP, by I. 4.

In the same manner it may be shewn that TQ is equal to SQ, and also equal to TP. Hence PSQT is a rhombus.

The construction fails if the straight line through R at right angles to PQ, instead of meeting AD and CB, meets DC and BA.

T. EX.

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