Sidebilder
PDF
ePub

At the point A make (E. 1. 23.) the angle BAE equal to the angle c, and at the point B the angle ABK equal to the angle D. Bisect (E. 1. 9.) the angles BAE and A BK, by the lines AF and BF, meeting in F; through F draw (E. 1. 31.) FG parallel to EA, and Fi parallel to KB; then ghF is the triangle required.

For since Fg is parallel to EA, therefore (E. 1. 29.) the angle EAF is equal to the angle Gra; but (by construction) the angle EAF is equal to the angle GAF; therefore the angle GAF is equal to the angle GFA; therefore (E. 1. 6.) GF is equal to GA. In the same manner it may be shown that hy is equal #B; hence the sum of the three sides of the triangle Gaf is equal to the given straight line AB.

Again, since Fg is parallel to E A, and FH to KB, therefore (E. 1. 29.) the angle Fga is equal to the angle E A B, and the angle Fug to the angle KBA ; but (by construction) the angle EAB is equal to the angle c, and the angle KBA is equal to the angle D; therefore also the angle Fgh is equal to the angle c, and the angle FHG to the angle D. Wherefore, &c.

3. To divide a given straight line into any given number of equal parts.

Let ce be a given straight line; it is required to divide it into any given number of equal parts.

From c draw an indefinite line CA making any convenient angle with CB; take any point e in ca, and make er, rs, SA, &c., each equal to ce, so that the number of these equal parts on ca, may be equal to the given number of parts into which CB is to be divided. Join the extremity A of the last part with the extremity B of the given line; and through the points €, r, s, &c. draw (E. 1. 31.) eh, rm, sp, &c. parallel to AB, cutting cb in the points h, m, p, &c. ; then these points will divide cB into the required number of equal parts.

Draw hn, mo, parallel to CA. Because rm is parallel to eh (E. I. 30.) and hn to er, therefore ehnr is a parallelogram, and (E. I. 34.) hn is equal to er or ce; the angle nhm is equal to the ech, and the angle hmn to the angle che (E. 1. 29.); therefore (E. 1. 26.) the triangle nhm is equal to the triangle ech, and hm is equal to ch. In like manner it may be proved that mp is equal hm, and so on.

e

[ocr errors]

m

8

[ocr errors]

A

B

B

E

F

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

Let ABC be the triangle, and p the given point. Bisect bc in E; join AE, PE; from a draw AF parallel to PE; and join PF; then PF will bisect the triangle ABC.

Since AF is parallel to PE, therefore (E. 1. 37.) the triangle PEA is equal to the triangle PEF. To these equals add the triangle BEP, therefore the triangle A BE is equal to the triangle BPF; but since be is equal to EC, therefore (E. 1. 38.) the triangle A BE is equal to half the triangle ABC; therefore the triangle BPF is also equal to half the triangle ABC.

5. Given the base, the sum of the remaining sides, and one of the angles at the base of a triangle, to construct it.

6. Given the base and the vertical angle of an isosceles triangle, to construct it.

7. To construct a triangle from three parts given, of which one, at least, is a side.

8. Upon a given base to describe an isosceles triangle equal to a given rectangle.

9. To divide a triangle into any given number of equal parts by lines drawn from the vertex to the base.

10. Describe a square which shall be equal to the difference between two given squares.

11. Describe a square which shall be equal to the sum of three given squares.

12. Construct a triangle which shall have the magnitude of its angles as the numbers 1, 2, and 3.

13. To find a point within a triangle, so that lines drawn to the angles shall divide the triangle into three equal parts.

14. Draw a line EF parallel to the base Bc of a triangle A BC, so that EF shall be equal to BE.

15. Construct a parallelogram, having given the diagonals and the angle they make with each other.

16. Trisect a given angle, which is the half, or the quarter, or the eighth part, and so on, of a right angle.

17. In a given square to inscribe' an equilateral triangle, having one of its angular points upon one of the angular points of the square, and its two remaining angular points une in each of two adjacent sides of the square.

18. To bisect a parallelogram by a line drawn through a given point in one of its sides.

19. To bisect a trapezium (1st) by a line drawn from one of its angular points, (2d) by a line drawn from a given point in one of its sides.

20. A plane rectilineal figure of any number of sides being given, to find an equal rectilineal figure which shall have the number of its sides less, by one, than that of the given figure.

21. To trisect a parallelogram by lines drawn from a given point in one of its sides.

22. To describe a parallelogram, the surface and perimeter of which shall be respectively equal to the surface and perimeter of a given triangle.

23. To describe a parallelogram which shall be of a given altitude, and equiangular with a given parallelogram, and also equal to it.

24. From the circumference of a given circle to draw to a straight line given in position, a line which shall be equal and parallel to a given straight line.

25. To describe a triangle which shall be equal to a given equilateral pentagon, and of the same altitude.

26. To inscribe a square in a given right-angled isosceles triangle.

27. To inscribe a square in a given equilateral four-sided figure. 28. To inscribe a square in a given quadrant of a circle.

29. Given one angle, a side opposite to it, and the difference of the other two sides, to construct the triangle.

30. Given the perimeter and the vertical angle of an isosceles triangle, to construct it.

31. From one of the angles of a parallelogram to draw a line to the opposite side, which shall be equal to that side together with the segment of it which is intercepted between the line and the opposite angle.

32. Two straight lines are given in position, without being produced to meet; it is required to draw a line which would (if produced) bisect the angle between the two straight lines.

33. Given the vertical angle, the difference of the angles at the base, and the perpendicular height of a triangle, to construct it.

34. In a given triangle to construct an equilateral parallelogram, which shall have one of its angles coinciding with one of the angles of the triangle, and the opposite angular point situated in one of the sides of the triangle.

A

E

F

P

B

D

THEOREMS. No. II. 1. The line joining the vertex and the middle of the base of a triangle, bisects every line that is drawn parallel to the base, and is terminated by the two remaining sides of the triangle.

Let PQ be any straight line, drawn parallel to the base bc of the triangle ABC; and let Ad, joining the vertex and the middle D of BC, cut PQ in G; then PQ is bisected in the point G.

If PG be not equal to GQ, one of them is the greater: let pg be the greater; and join DQ, and DP.

Since (hyp.) Bd is equal to Dc, and PQ is parallel to BC, therefore (E. 1. 38.) the triangle BDP is equal to the triangle DCQ; also because BD is equal to DC, therefore the . triangle BDA is equal to the triangle CDA; therefore, taking the former equal triangles from the latter, there remains the triangle DPA equal to the triangle DQA. But since Pg is greater than GQ, the triangle APG is greater than AGQ, and the triangle God is greater than GQP; therefore, the whole triangle DPA is greater than the whole triangle DQA; but it has been shown that the triangle DPA is equal to the triangle DQA; and it is also greater, which is absurd: hence neither of the two lines PG, GQ, can be greater than the other ; therefore PG must be equal to GQ.

2. The two sides of a triangle are together greater than the double of the line joining the vertex and the middle of the base.

Let ABC be a triangle, and AD the straight line joining the vertex A, and the middle D, of the base BC; then AB and AC are together greater than twice AD. Produce AD to making DQ equal to AD; and join BQ.

Since (hyp.) bd is equal to Dc, and (constr.) DQ is equal to AD, and (E. 1. 15.)

B

D

the angle BDQ is equal to ADC; therefore (E. 1. 4.) BQ is equal to Ac. But (E. 1. 20.) AB, BQ are together greater than AQ; but AC has been proved to be equal to BQ, and AQ (by const.) is the double of AD; therefore AB, AC are together greater than the double of AD.

3. If either diagonal of a parallelogram be equal to one of the sides, the other diagonal will be greater than any side of the parallelogram.

4. A trapezium, having two of its sides parallel, is equal to half of a rectangle between the same parallels, and having its base equal to the sum of the two parallel sides of the trapezium.

5. If the side of a square be equal to the diagonal of another square, the former square is double of the latter.

6. The two sides of a triangle are together greater than the double of the line drawn from the vertex to the base, bisecting the vertical angle.

7. If in the sides of a square, at equal distances from the four angles, four other points be taken, one in each side ; the figure contained by the straight lines which join them shall also be a square.

8. If four straight lines cut each other, without including a space, but so as to make three internal angles, towards the same parts, together less than four right angles, the two lines which are not joined shall meet, if produced.

9. If two opposite sides of a parallelogram be bisected, and lines be drawn from these two points of bisection to the opposite angles, the lines will be parallel, and will trisect the diagonal.

10. In Prob. 4. No. 1. the sum of the lines CE and ED is less than the sum of any other two lines which can be drawn from the given points c and d to meet the line A H.

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

12. If from the angular points of the squares described upon the sides of a right-angled triangle perpendiculars be let fall upon the hypotenuse produced, they will cut off equal segments, and the perpendiculars will together be equal to the hypotenuse.

13. If a line PQ be drawn parallel to the base BC of a triangle ABC (see fig. Theo. 1. No. 2.) through the point g where the lines cg and BG bisecting the angles of the base

« ForrigeFortsett »