If C and D are equally distant from AB, and on the same perpendicular, then any point in AB may be taken for the point G.

15. Let the angle BAC be bisected by AD, and the angle BAG by AE: the angle DAE shall be a right angle.

Since the angle BAD is half the angle BAC, and the angle BAE is half the angle BAG, the two angles BAD and BAE together are half the two angles BAC and BAG together. But the angles BAC and BAG together are equal to two right angles, by I. 13; therefore the angles BAD and BAE together are equal to a right angle.

16. Let the four straight lines AE, BE, CE, DE meet at the point E, and make the angle AEB equal to the angle CED, and the angle BEC equal to the angle DEA: then shall AE and EC be in one straight line, and also BE and ED in one straight line.

By I. 15, Cor. 2 the four angles AEB, BEC, CED, DEA are together equal to four right angles; but the two angles AEB and BEC are equal to the two angles CED and DEA; therefore the angles AEB and BEC are together equal to two right angles; . therefore AE and EC are in one straight line by I. 14. Similarly it may be shewn that BE and ED are in one straight line.

I. 16 to 26.

17. The angle BDA is greater than the angle CAD, by I. 16; the angle CAD is equal to the angle BAD, by hypothesis: therefore the angle BDA is greater than the angle BAD. Therefore the side BA is greater than the side BD, by I. 19. Similarly it may be shewn that CA is greater than CD.

18. Take any point G in BC, and join AG. The angle AGC is greater than the angle ABC, by I. 16; and the angle AGB is greater than the angle ACB, by I. 16. Therefore the angles ABC and ACB are together less than the angles AGC and AGB together; therefore the angles ABC and ACB are together less than two right angles by I. 13.

19. Join BD. The angle ABD is greater than the angle ADB, and the angle DBC is greater than the angle BDC, by I. 18; therefore the whole angle ABC is greater than the whole angle ADC. Similarly by joining AC we can shew that the angle DCB is greater than the angle DAB.

20. Let ABCD be the square; on BC take any point E; join AE and produce it to meet DC produced at F: then shall AF be greater than AC.

The angle DCA is greater than the angle CFA by I. 16. The angle ACE, which is greater than a right angle, is greater than the angle DCA, which is less than a right angle. Therefore the angle ACF is greater than the angle AFC. Therefore AF is greater than AC, by I. 19.

21. Let AB be the given straight line, O the given point without it. From O draw OC perpendicular to AB; then OC shall be shorter than any other straight line OD drawn from 0 to AB.

For the angle OCD is a right angle; therefore the angle ODC is less than a right angle, by I. 17: therefore OC is less than OD, by I. 19.

Next, let OE be a straight line drawn from 0 to AB, and more remote from OC than OD is: OD shall be less than OE.

For the angle ODC is greater than the angle OEC, by I. 16; and the obtuse angle ODE is greater than the acute angle ODC: therefore the angle ODE is greater than the angle OED. Therefore OD is less than OE, by I. 19.

Lastly, from C on the straight line AB take CF equal to CD, and on the other side of C; join OF. Then OF is equal to OD by I. 4. And no other straight line can be drawn from 0 to AB equal to OD, besides OF.

For if this straight line were nearer to OC than OD or OF is it would be less than OD, and if it were more remote from OC than OD or OF is, it would be greater than OD.

22. Let ABC be a triangle, and O any point. Join OA, OB, OC. Then OA and OB are together greater than AB, OB and OC are together greater than BC, and OC and OA are together greater than CA, by I. 20. Hence twice the sum of OA, OB, and OC is greater than the sum of AB, BC, and CA; therefore the sum of OA, OB, and OC is greater than half the sum of AB, BC, and CA.

23. Let ABCD be a quadrilateral figure. Draw the diagonals AC and BD. Then AB and BC are greater than AC, and AD and DC are greater than AC; therefore the four sides AB, BC, CD, DA are greater than twice AC. Similarly it may be shewn that the four sides are greater than twice BD. Hence twice the sum of the four sides is greater than twice the sum of the diagonals; therefore the sum of the four sides is greater than the sum of the diagonals.

24. Let ABC be a triangle, and D the middle point of the base BC; join AD: then AB and AC together shall be greater than twice AD.

Produce AD to a point E so that DE may be equal to AD; join BE. In the two triangles ADC and EDB the two sides AD, DC are equal to the two sides ED, DB each to each; and the angle ADC is equal to the angle EDB, by I. 15: therefore AC is equal to BE, by I. 4. The two sides AB, BE are greater than AE, by I. 20; therefore the two sides AB, AC are greater than AE, that is greater than twice AD.

25. Let ABC be a triangle in which the angle C is equal to the sum of the angles A and B. At the point C in the straight line AC make the angle ACG equal to the angle A, and let CG meet AB at D. Then the triangle ACD is isosceles, by I. 6. Also as the angle ACB is equal to the sum of the angles A and B, and the angle ACG is equal to the angle A, the angle BCG is equal to the angle B: therefore the triangle BCD is isosceles, by I. 6.

//tri

26. It is shewn in the preceding Exercise that ACD is an isosceles triangle, having AD equal CD; also that BCD is an isosceles triangle having BD equal to CD. Hence, as AD and BD are each equal to CD, the point D is the middle point of AB, and AB is equal to twice CD.

27. Let AB be the given base; at the point A make the angle DAB equal to the given angle. From AD cut off AE equal to the given sum of the sides; join EB. At the point B make the angle EBF equal to the angle AEB, and on the same side of EB; let BF meet AE at C: then ACB will be the triangle required.

For, since the angle EBC is equal to the angle BEC, the sides EC and BC are equal, by I. 6. Therefore the sum of AC and CB is equal to the sum

of AC and CE, that is to the given sum of the sides. Also the base AB and the angle BAC have the required values.

28. From any point D in the straight line bisecting the angle A of a triangle draw DE perpendicular to the side AB, and DF perpendicular to the side AC. Then in the triangles DAE and DAF, the side DA is common; the angle DAE is equal to the angle DAF, by hypothesis; and the right angles DEA and DFA are equal: therefore DE is equal to DF by I. 26.

29. Let AB be the given straight line in which the point is to be found. Let CD and EF be the other two straight lines, and let them meet, produced if necessary, at 0. Through O draw a straight line bisecting the angle between CD and EF; and let AB, produced if necessary, meet this straight line at K; then K will be such a point as is required: for the perpendiculars from K on the straight lines CD and EF may be shewn to be equal in the manner of the preceding Exercise.

Two straight lines can be drawn through O bisecting angles formed by the given straight lines, so that in general two solutions of the problem can be obtained, but there will be only one solution if AB is parallel to either of the bisecting straight lines.

If CD and EF are parallel the construction fails. We must then draw a straight line KL parallel to CD and EF, and midway between them: the intersection of this straight line with AB, produced if necessary, will be the required point. But if KL is parallel to AB there will be no solution.

30. Suppose A the given point through which the straight line is to be drawn, and B and C the other given points from which perpendiculars are to be drawn. Join BC, and bisect it at D; join AD: this shall be the required straight line.

For draw BE and CF perpendicular to AD, produced if necessary. Then in the triangles BDE, CDF the sides BD and CD are equal; the angles BDE and CDF are equal, by I. 15; and the right angles BED and CFD are equal therefore BE is equal to CF, by I. 26.

31. In the triangles ADB, ADE the side AD is common; the angles BAD and EAD are equal by hypothesis; and the right angles ADB and ADE are equal: therefore BD is equal to ED, by I. 26.

32. Bisect the angle BAC by the straight line AD; from P draw PG perpendicular to AD, and produce PG both ways to meet AB at E, and AC at F: then AE will be equal to AF.

For in the two triangles AGE, AGF the side AG is common; the angles EAG and FAG are equal, by construction; and the right angles EGA and FGA are equal: therefore AE is equal to AF, by I. 26.

33. Let ABC be a triangle having the angle B a right angle, and let DEF be a triangle having the angle E a right angle; also let AC be equal to DF, and AB equal to DE: then shall the triangles ABC and DEF be equal in all respects.

Produce CB to G, so that BG may be equal to EF; and join AG. Then the angle ABG is a right angle, by I. 13, and is therefore equal to the angle DEF; also the sides AB, BG are equal to the sides DE, EF each to each: therefore the triangles ABG, DEF are equal in all respects, so that AG is

equal to DF. But AC is equal to DF by hypothesis; therefore AC is equal to AG, and the angle ACG is equal to the angle AGC. Therefore the two triangles ABC, ABG are equal in all respects, by I. 26. But the triangles ABG, DEF were shewn to be equal in all respects; therefore the triangles ABC, DEF are equal in all respects.

I. 27 to 31.

34. Let ABC be a triangle having the sides AB and AC equal. Draw any straight line parallel to BC, meeting AB at D, and AC at E; then the angle ADE shall be equal to the angle AED.

For the angle ADE is equal to the angle ABC, and the angle AED is equal to the angle ACB, by I. 29. But the angle ABC is equal to the angle ACB, by I. 5. Therefore the angle ADE is equal to the angle AED.

35. Let the straight lines A and B meet at K, let the straight lines C and D meet at L, and let the straight lines A and D meet at M. The acute angle at K is equal to the acute angle at M, by I. 29; and the acute angle at M is equal to the acute angle at L also, by I. 29. Therefore the acute angle at K is equal to the acute angle at L.

36. Let the straight line AB be terminated by two parallel straight lines. Let C be the middle point of AB; through C draw any straight line DCE, terminated by the same parallel straight lines as AB, so that AD is parallel to EB. Then will ED be bisected at C.

Fór in the two triangles ACD, BCE the two sides AC, BC are equal by hypothesis; the angles ACD, BCE are equal by I. 15; and the angles CAD, CBE are equal by I. 29: therefore the triangles are equal in all respects by I. 26, so that CD is equal to CE.

37. Let O be a point equidistant from two parallel straight lines: ; through O draw one straight line AOB terminated by the parallels, and also another straight line COD terminated by the parallels, so that A and C are on one of the parallels, and B and D on the other: then will AC be equal to BD.

Since the given straight lines are parallel, a straight line can be drawn through O to meet the parallels at right angles, and this straight line will be bisected at O because O is equidistant from the parallels, by hypothesis. Therefore by Exercise 36 the straight lines AB and DC are bisected at 0. Thus in the two triangles AOC, BOD the two sides AO, OC are equal to the two sides BO, OD each to each; and the angle AOC is equal to the angle BOD, by I. 15: therefore AC is equal to BD, by I. 4.

38. Let ABC be a triangle: produce BA to D; suppose that AE bisects the angle DAC, and that it is parallel to BC: then will ABC be an isosceles triangle.

For since AE is parallel to BC the angle DAE is equal to the angle ABC, by I. 29, and also the angle CAE is equal to the angle ACB, by I. 29. But the angle DAE is equal to the angle EAC, by hypothesis; therefore the angle ABC is equal to the angle ACB: therefore the side AB is equal to the side AC, by I. 6.

39. Take any point E in DC, and at the point E make the angle CEF equal to the given angle. Through A draw a straight line parallel to FE, and meeting CD at B: then B is the required point.

For the angle ABC is equal to the angle FEC by I. 29; and therefore the angle ABC is equal to the given angle.

40. Let ABC be a triangle; let a straight line be drawn bisecting the angle A, and meeting BC at D. From D draw a straight line parallel to AB, meeting AC at F, and also a straight line parallel to AC, meeting AB at E: then DE shall be equal to DF.

For in the triangles AED, AFD the side AD is common; the angle EAD is equal to the angle FAD, by construction; the angle EDA is equal to the angle DAF, and the angle FDA to the angle DAE, by I. 29; so that the angle EDA is equal to the angle FDA; hence the two triangles are equal in all respects by I. 26. Thus DE is equal to DF.

41. The angle FEC is equal to the angle ECB, by I. 29; the angle ECB is equal to the angle ECF, by hypothesis; therefore the angle FEC is equal to the angle ECF. Therefore EF is equal to FC, by I. 6. Again, the angle FGC is equal to the angle GCD, by I. 29; the angle GCD is equal to the angle FCG, by hypothesis; therefore the angle FCG is equal to the angle FGC. Therefore FG is equal FC, by I. 6. And it has been shewn that FE is equal to FC; therefore EF is equal to FG.

42. Bisect the angle ABC by a straight line meeting AC at E; through E draw a straight line parallel to CB meeting AB at D: then D shall be the point required.

For the angle DEB is equal to the angle EBC, by I. 29; the angle DBE is equal to the angle EBC, by construction; therefore the angle DEB is equal to the angle DBE; therefore DB is equal to DE, by I. 6. And the angle DEA is equal to the angle BCA by I. 29, and is therefore a right angle; so that DE is perpendicular to AC.

43. Bisect the angle ABC by a straight line, meeting AC at E; through E draw a straight line parallel to BC, meeting AB at D. Then will BD, DE, EC be all equal.

For the angle DEB is equal to the angle EBC, by I. 29; the angle EBC is equal to the angle DBE by construction; therefore the angle DEB is equal to the angle DBE. Therefore DB is equal to DE, by I. 6. Again, the angle ADE is equal to the angle ABC, and the angle AED is equal to the angle ACB, by I. 29; also the angle ABC is equal to the angle ACB: therefore the angle ADE is equal to the angle AED. Therefore AD is equal to AE, by I. 6. But the whole AB is equal to the whole AC; therefore DB is equal to EC. Thus BD, DE, EC are all equal.

44. From A draw a straight line bisecting the angle BAC, and meeting BC at F. Then the triangles BAF and CAF are equal in all respects, by I. 4; so that the angles AFB, AFC are equal, and therefore each of them is a right angle. Therefore AF is parallel to ED, by I. 28. The angle AED is equal to the angle CAF, and the angle EDA is equal to the angle BAF, by I. 29. But the angle BAF is equal to the angle CAF, by construction; therefore the angle AED is equal to the angle ADE: therefore AE is equal to AD, by I. 6.