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also the right angle AFG is equal to the right angle EFG, and the side FG is common to the two triangles AFG, EFG,

therefore AG is equal to EG, and AF to FE. Hence the point E being known, the point G is determined by the intersection of CD and .

Synthesis. From A draw AF perpendicular to CD, and produce it to E, making FE equal to AF, and join BE cutting CD in G.

Join also ÀG. Then AG and BG make equal angles with CD. For since AF is equal to FE, and FG is common to the two triangles AGF, EGF, and the included angles AFG, EFG are equal;

therefore the base AG is equal to the base EG,

and the angle AGF to the angle EGF;
but the angle EGF' is equal to the vertical angle BGD,

therefore the angle AGF is equal to the angle BGD; that is, the straight lines AG and BG make equal angles with the straight line CD.

Also the sum of the lines AG, GB is a minimum. For take any other point H in CD, and join EH, HB, AH. Then since any two sides of a triangle are greater than the third side, therefore EH, HB are greater than EB in the triangle EHB.

But EG is equal to AG, and EH to AH;

therefore AH, HB are greater than AG, GB. That is, AG, GB are less than any other two lines which can be drawn from A, B, to any other point H in the line CD.

By means of this Proposition may be found the shortest path from one given point to another, subject to the condition, that it shall meet two given lines.

PROPOSITION V. PROBLEM. Given one angle, a side opposite to it, and the sum of the other two sides, construct the triangle.

Analysis. Suppose BAC the triangle required, having BC equal to the given side, BAC equal to the given angle opposite to BC, also BD equal to the sum of the other two sides.

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Join DC. Then since the two sides BA, AC are equal to BD, by taking BA from these equals, the remainder AC is equal to the remainder AD.

Hence the triangle ACD is isosceles, and therefore the angle ADC is equal to the angle ACD.

But the exterior angle BAC of the triangle ADC is equal to the two interior and opposite angles ACD and ADC:

Wherefore the angle BAC is double the angle BDC, and BDC is the half of the angle BAC.

Hence the synthesis.

At the point D. in BD, make the angle BDC equal to half the given angle,

and from B the other extremity of BD, draw BC equal to the

given side, and meeting DC in C,
at C in CD make the angle DCA equal to the angle CDA, so

that CA may meet BD in the point A.
Then the triangle ABC shall have the required conditions.

PROPOSITION VI. PROBLEM.
To bisect a triangle by a line drawn from a given point in one of the sides.

Analysis. Let ABC be the given triangle, and D the given point in the side AB.

E F с
Suppose DF the line drawn from D which bisects the triangle;
therefore the triangle DBF is half of the triangle ABC.

Bisect BC'in E, and join AE, DE, AF,
then the triangle ABE is half of the triangle ABC:
hence the triangle ABE is equal to the triangle DBF;

take away from these equals the triangle DBE, therefore the remainder ADE is equal to the remainder DEF.

But ADE, DEF are equal triangles upon the same base DE, and on the same side of it, they are therefore between the same parallels, (1. 39.)

that is, AF is parallel to DE,

therefore the point Fis determined. Synthesis. Bisect the base B C in E, join DE, from A, draw AF parallel to DE, and join DF.

Then because DE is parallel to AF, therefore the triangle ADE is equal to the triangle DEF;(I. 37.)

to each of these equals, add the triangle BDE, therefore the whole triangle ABE is equal to the whole DBF,

but ABE is half of the whole triangle ABC'; therefore DBF is also half of the triangle ABC.

PROPOSITION VII. THEOREM. If from a point without a parallelogram lines be drawn to the extremities of two adjacent sides, and of the diagonal which they include; of the triangles thus formed, that, whose base is the diagonal, is equal to the sum of the other two.

Let ABCD be a parallelogram of which AC is one of the diagonals, and let P be any point without it: and let AP, PC, BP, PD be joined.

Then the triangles APD, APB are together equivalent to the triangle APC.

EC Draw PGE parallel to AD or BC, and meeting AB in G, and DC in E; and join DG, GC.

Then the triangles CBP, CBG are equal: (1. 37.)
and taking the common part CBH from each,

the remainders PHB, CHG are equal. Again, the triangles DAP, DAG are equal; (1. 37.) also the triangles DAG, AGC are equal, being on the same base AG, and between the same parallels , DC:

therefore the triangle DAP is equal to the triangle AGC:

but the triangle PHB is equal to the triangle CHG, wherefore the triangles PHB, DAP are equal to AGC, CHG, or

ACH, add to these equals the triangle APH, therefore the triangles APH, PHB, DAP are equal to APH, ACH, that is, the triangles APB, DAP are together equal to the triangle

РАС. If the point P be within the parallelogram, then the difference of the triangles APB, DAP may be proved to be equal to the triangle PAC.

8. Describe an isosceles triangle upon a given base and having each of the sides double of the base, without using any proposition of the Elements subsequent to the first three. If the base and sides be given, what condition must be fulfilled with regard to the magnitude of each of the equal sides in order that an isosceles triangle may be constructed ?

9. In the fig. Euc. I. 5. If FC and BG meet in H, then prove that AH bisects the angle BAC.

10. In the fig. Euc. I. 5. If the angle FBG be equal to the angle ABC, and BG, CF, intersect in 0; the angle BOF is equal to twice the angle BAC.

11. From the extremities of the base of an isosceles triangle straight lines are drawn perpendicular to the sides, the angles made by them with the base are each equal to half the vertical angle.

12. A line drawn bisecting the angle contained by the two equal sides of an isosceles triangle, bisects the third side at right angles.

13. If a straight line drawn bisecting the vertical angle of a triangle also bisect the base, the triangle is isosceles.

14. Given two points one on each side of a given straight line ; find a point in the line such that the angle contained by two lines drawn to the given points may be bisected by the given line.

15. In the fig. Éuc. 1. 5, let F and G be the points in the sides AB and AC produced, and let lines FII and GK be drawn perpendicular and equal to FC and GB respectively: also if BII, CK, or these lines produced meet in 0; prove that BH is equal to CK, and BO to CO.

16. From every point of a given straight line, the straight lines drawn to each of two given points on opposite sides of the line are equal: prove that the line joining the given points will cut the given line at right angles.

17. If A be the vertex of an isosceles triangle ABC, and B A be produced so that AD is equal to BA, and DC be drawn; shew that BCD is a right angle.

18. The straight line EDF, drawn at right angles to BC the base of an isosceles triangle ABC, cuts the side AB in D, and CA produced in E; shew that AED is an isosceles triangle.

19. In the fig. Euc. I. 1, if AB be produced both ways to meet the circles in D and E, and from C, CD and CE be drawn; the figure CDE is an isosceles triangle having each of the angles at the base, equal to one fourth of the angle at the vertex of the triangle.

20. From a given point, draw two straight lines making equal angles with two given straight lines intersecting one another.

21. From a given point to draw a straight line to a given straight line, that shall be bisected by another given straight line.

22. Place a straight line of given length between two given straight lines which meet, so that it shall be equally inclined to each of them.

23. To determine that point in a straight line from which the straight lines drawn to two other given points shall be equal, provided the line joining the two given points is not perpendicular to the given line.

24. In a given straight line to find a point equally distant from two given straight lines. In what case is this impossible?

25. If a line intercepted between the extremity of the base of an isosceles triangle, and the opposite side (produced if necessary) be equal to a side of the triangle, the angle formed by this line and the base produced, is equal to three times either of the equal angles of the triangle.

26. In the base B C of an isosceles triangle ABC, take a point D, and in CA take CE equal to CD, let ED produced meet AB produced in F; then 3. AEF=2 right angles + AFÉ, or = 4 right angles - AFE.

27. If from the base to the opposite sides of an isosceles triangle, three straight lines be drawn, making equal angles with the base, viz. one from its extremity, the other two from any other point in it, these two shall be together equal to the first.

28. A straight line is drawn, terminated by one of the sides of an isosceles triangle, and by the other side produced, and bisected by the base; prove that the straight lines, thus intercepted between the

vertex of the isosceles triangle, and this straight line, are together equal to the two equal sides of the triangle.

29. In a triangle, if the lines bisecting the angles at the base be equal, the triangle is isosceles, and the angle contained by the bisecting lines is equal to an exterior angle at the base of the triangle.

30. In a triangle, if lines be equal when drawn from the extremities of the base, (1) perpendicular to the sides, (2) bisecting the sides, (3) making equal angles with the sides; the triangle is isosceles : and then these lines which respectively join the intersections of the sides, are parallel to the base.

II.

31. ABC is a triangle right-angled at B, and having the angle A double the angle C; shew that the side BC is less than double the side AB.

32. If one angle of a triangle be equal to the sum of the other two, the greatest side is double of the distance of its middle point from the opposite angle.

33. If from the right angle of a right-angled triangle, two straight lines be drawn, one perpendicular to the base, and the other bisecting it, they will contain an angle equal to the difference of the two acute angles of the triangle.

34. If the vertical angle CAB of a triangle ABC be bisected by AD, to which the perpendiculars CE, BF are drawn from the remaining angles : bisect the base BC in G, join GE, GF, and prove these lines equal to each other.

35. The difference of the angles at the base of any triangle, is double the angle contained by a line drawn from the vertex perpendicular to the base, and another bisecting the angle at the vertex.

36. If one angle at the base of a triangle be double of the other, the less side is equal to the sum or difference of the segments of the base made by the perpendicular from the vertex, according as the angle is greater or less than a right angle.

37. If two exterior angles of a triangle be bisected, and from the point of intersection of the bisecting lines, a line be drawn to the opposite angle of the triangle, it will bisect that angle.

38. From the vertex of a scalene triangle draw a right line to the base, which shall exceed the less side as much as it is exceeded by the greater.

39. Divide a right angle into three equal angles.

40. One of the acute angles of a right-angled triangle is three times as great as the other; trisect the smaller of these.

41. Prove that the sum of the distances of any point within a triangle from the three angles is greater than half the perimeter of the triangle.

42. The perimeter of an isosceles triangle is less than that of any other equal triangle upon the same base.

43. If from the angles of a triangle ABC, straight lines ADE, BDF, CDG be drawn through a point D to the opposite sides, prove that the sides of the triangle are together greater than the three

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