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straight lines which, with another straight line, make the interior angles upon the same side less than two right angles, do meet (12.Ax.) if produced far enough; therefore HB, FE shall meet if produced;

Let them meet in K, and through K draw KL parallel to EA or FH (31. 1.) and produce HA, GB to the points L, M: then HLKF is a parallelogram, of which the diameter is HK, and AG, ME are the parallelograms about HK; and LB, BF are the complements; therefore LB is equal (43. 1.) to BF:

But BF is equal to the triangle C; wherefore LB is equal (1 Ax.) to the triangle C; and because the angle GBE is equal (15. 1.) to the angle ABM, and likewise (Constr.) to the angle D; the angle ABM is equal (1 Ax.) to the angle D. Therefore the parallelogram LB is applied to the straight line AB, is equal to the triangle C, and has the angle ABM equal to the angle D. Which was to be done.

PROPOSITION XLV.

PROB. To describe a parallelogram equal to a given rectilineal figure, and having an angle equal to a given rectilineal angle.

Let ABCD be the given rectilineal figure, and E the given rectilineal angle. It is required to describe a parallelogram equal to ABCD, and having an angle equal to E.

B

F C L

ск H M

Join DB, and describe (42. 1.) the parallelogram FH equal to the triangle ADB, and having the angle A HKF equal to the angle E; and to the straight line GH apply (44. 1.) the parallelogram GM equal to the triangle DBC, having the angle GHM equal to the angle E; and because the angle E is equal (Constr.) to each of the angles FKH, GHM, the angle FKH is equal (1 Ax.) to GHM: add to each of these the angle KHG; therefore the angles FKH, KHG are equal (2 Ax.) to the angles KHG, GHM; but FKH, KHG are equal (29. 1.) to two right angles; therefore also KHG, GHM are equal (1 Ax.) to two right angles and because at the point H in the straight line GH, the two straight lines KH, HM upon the opposite sides of it make the adjacent angles equal to two right angles, KH is in the same straight line (14. 1.) with HM:

And because the straight line HG meets the parallels KM, FG, the alternate angles MHG, HGF (29. 1.) are equal: add to each of these the angle HGL; therefore the angles MHG, HGL are equal (2 Ax.) to the angles HGF, HGL: but the angles MHG, HGL are equal (29. 1.) to two right angles; wherefore also the angles HGF, HGL are equal (1 Ax.) to two right angles, and FG is therefore in the same straight line (14. 1.) with GL;

And because KF is parallel to HG (Constr.), and HG to ML; KF is parallel (30. 1.) to ML; and KM, FL are parallels; wherefore KFLM is (Def. 34. 1.) a parallelogram; and because the triangle ABD is equal to the parallelogram HF (Constr.), and the triangle DBC to the parallelogram GM; the whole rectilineal figure ABCD is equal to the whole parallelogram KFLM (2 Ax.); therefore the parallelogram KFLM has been described equal to the given rectilineal

figure ABCD, having the angle FKM equal to the given angle E. Which was to be done.

COR. From this it is manifest how to a given straight line to apply a parallelogram, which shall have an angle equal to a given rectilineal angle, and shall be equal to a given rectilineal figure, viz., by applying (44. 1.) to the given straight line a parallelogram equal to the first triangle ABD, and having an angle equal to the given angle.

PROPOSITION XLVI.

PROB. To describe a square upon a given straight line.

Let AB be the given straight line; it is required to describe a square upon AB.

From the point A draw (11. 1.) AC at right angles to AB; and make (3. 1.) AD equal to AB, and through the point D draw DE parallel (31. 1.) to AB, and through B draw BE parallel to AD; therefore ADEB is a (Def. 34. 1.) parallelogram: whence AB is equal (34. 1.) to DE, and AD to BE; but BA is equal to AD; therefore the four straight lines BA, AD, DE, EB are equal (1 Ax.) to one another, and the parallelogram ADEB is equilateral:

E

Likewise all its angles are right angles; because the straight line AD meeting the parallels AB, DE, the angles BAD, ADE are equal (29. 1.) to two right angles: but BAD D is a (Constr.) right angle; therefore also ADE is a right angle; but the opposite angles of parallelograms are equal (34. 1.); therefore each of the opposite angles ABE, BED is a (1 Ax.) right angle; wherefore A

B

the figure ADEB is rectangular; and it has been demonstrated that it is equilateral; it is therefore a (30 Def.) square, and it is described upon the given straight line AB. Which was to be done.

COR. Hence every parallelogram that has one right angle has all its angles right angles.

PROPOSITION XLVII.

THEOR. In any right-angled triangle, the square which is described upon the side subtending the right angle is equal to the squares described upon the sides which contain the right angle.

Let ABC be a right-angled triangle having the right angle BAC; the square described upon the side BC is equal to the squares described

upon BA, AC.

A

H

K

On BC describe (46. 1.) the square BDEC, and on BA, AC the squares GB, HC; and through A draw (31. 1.) AL parallel to BD or CE, and join AD, FC. Then, because each of the angles BAC (Hyp.), BAG is a right angle (30 Def.) the two straight lines AC, AG, upon the opposite sides of AB, make with it at the point A the adjacent angles equal to two right angles; therefore CA is in the same straight line (14. 1.) with AG;

For the same reason, AB and AH are in the same straight line;

E

L

And because the angle DBC is equal (11 Ax.) to the angle FBA, each of them being a right angle, add to each the angle ABC, and the whole angle DBA is equal (2 Ax.) to the whole FBC; and because the two sides AB, BD are equal (30 Def.) to the two FB, BC, each to each, and the angle DBA equal to the angle FBC; therefore the base AD is equal (4. 1.) to the base FC, and the triangle ABD to the triangle FBC:

Now the parallelogram BL is double (41. 1.) of the triangle ABD, because they are upon the same base BD, and between the same parallels BD, AL; and the square GB is double of the triangle FBC, because these also are upon the same base FB, and between the same parallels FB, GC. But the doubles of equals are equal (6 Ax.) to one another; therefore the parallelogram BL is equal to the square GB:

And, in the same manner, by joining AE, BK, it is demonstrated, that the parallelogram CL is equal to the square HC;

Therefore the whole square BDEC is equal (2 Ax.) to the two squares GB, HC; and the square BDEC is described upon the straight line BC, and the squares GB, HC upon BA, AC; wherefore the square upon the side BC is equal to the squares upon the sides BA, AC. Therefore, in any right-angled triangle, etc.

PROPOSITION XLVIII.

Q. E. D.

THEOR. If the square described upon one of the sides of a triangle be equal to the squares described upon the other two sides of it, the angle contained by these two sides is a right angle.

If the square described upon BC, one of the sides of the triangle ABC, be equal to the squares upon the other sides BA, AC, the angle BAC is a right angle.

From the point A draw (11. 1.) AD at right angles to AC, and make (3. 1.) AD equal to BA, and join DC: then, because DA is equal to AB, the square of DA is equal to the square of AB; to each of these add the square of AC; therefore the squares of DA, AC are equal (2 Ax.) to the squares of BA, AC:

But the square of DC is equal (47. 1.) to the squares of DA, AC, because DAC is a (Constr.) right angle;

and the square of BC, by hypothesis, is equal to the squares of BA AC; therefore the square of DC is equal (1 Ax.) to the square of and therefore also the side DC is equal to the side BC.

BC;

And because the side DA is equal (Constr.) to AB, and AC common to the two triangles DAC, BAC, the two DA, AC are equal to the two BA, AC; and the base DC is equal to the base BC; therefore the angle DAC is equal (8. 1.) to the angle BAC: but DAC is a right angle; therefore also BAC is a right angle. Therefore, if the square, etc. Q. E. D.

EXERCISES ON BOOK I.

1. If a straight line meet another so as to make two adjacent angles, and each of these angles be bisected, the bisecting lines will be at right angles to one another.

2. If through the vertex of an isosceles triangle a line be drawn parallel to the base, it will bisect the angles at the vertex made by producing the equal sides of the triangle.

3. The difference between any two sides of a triangle is less than the third side.

4. If two straight lines which meet one another be parallel to two other lines which meet one another, the angle contained by the first two will be equal to the angle contained by the other two.

5. Vary the second part of the demonstration of i. 32 so as to be effected without producing either side of the triangle.

6. Prove the second corollary of i. 32 independently of the first, and from it deduce the first. Also,

(a) Show that each angle of an equilateral triangle is two-thirds of a right angle;

(b) That the two acute angles of a right-angled triangle are equal to one right angle;

(c) If a triangle have its exterior angle and likewise one of its interior angles double of those in another triangle, its remaining opposite interior angle is double of the corresponding angle in the other.*

7. The diagonals of a parallelogram bisect one another at their point of intersection, and any line drawn through their intersection bisects the parallelogram.

8. If through the middle of a line intercepted between two parallel lines, any other line be drawn terminating in those parallels, then this line will be bisected at that point.

9. (a) If one of the sides of a triangle be divided into any number of equal parts, and lines be drawn through the several points of division parallel to the base, these parallels will divide the other side into the same number of equal parts. Also (b), the second, third, fourth, etc. of the parallels intercepted between the sides, will be twice, three times, four times, etc. the first intercepted parallel. (c) And if the sides of the triangle be produced either below the base or above the vertex, and parts be taken in these prolongations equal to the former parts, the same relations as in (a) and (b) will hold good.

10. Divide a given line into any given number of equal parts. 11. (a) If two sides of a triangle be bisected, the line which joins the points of bisection will be parallel to the base, and equal to half the base. (b) If lines be drawn from the points of bisection to the opposite angles, they will all meet in one point; and each will be divided at that point into segments such that those drawn to the angles are respectively the doubles of those drawn to the bisecting points.

12. If the three sides of a triangle be bisected and the points of

This is the foundation of the construction of Hadley's sextant.
The construction of the diagonal scale depends upon this property.

bisection two and two be joined, four triangles will be formed equiangular with the first, and each equal to one fourth of that triangle.

13. If one angle of a parallelogram be a right angle, all the angles are right angles: and if likewise any two adjacent sides be equal, it is a square. Prove also that if all the sides of a quadrilateral figure be bisected, and lines be drawn to join the adjacent points of bisection, they will form a parallelogram equal to half the given quadrilateral.

14. If the two diagonals of a parallelogram be equal, all the angles of the parallelogram are right angles.

15. If the four sides of a quadrilateral figure be equal, the figure is a parallelogram, and the diagonals are perpendicular to one another. 16. If the right angle of a right-angled triangle be bisected by a line which also cuts the hypothenuse, and from the point of intersection lines be drawn parallel to the sides, a square will be formed of which the diagonal is the bisecting line.

17. Equal squares have equal sides, and the squares described upon equal lines are equal.

18. In the figure to Prop. 47, if BC be produced both ways, and perpendiculars be drawn to it from F and K, then

(a) These perpendiculars will be respectively equal to the segments of BC made by AL;

(b) They will meet BC at distances from B and C respectively, each equal to the perpendicular from A upon BC;

(c) The lines BK, CF are respectively perpendicular to AE, AD;

(d) The squares described on the sides BA, AC about the right angle are equal to the rectangles under the hypothenuse and the segments of it made by the perpendicular from A upon BC respectively adjacent to BA, AC.

19. Describe a square equal to any number of given squares, and likewise a square equal to the difference between two given squares.

20. In four several triangles there are alike given the base and one of the angles at the base; and the remaining data in the several cases

are:

(a) The perpendicular;

(b) The sum of the sides;

(c) The difference of the sides;

(d) The line drawn from the middle of the base to the vertex. It is required to construct these triangles.

21. (a) Given two straight lines, to exhibit half their sum and half their difference by a construction.

(B) Given two rectilinear angles to construct angles equal to half their sum and half their difference.

22. If an angle of a triangle be two-thirds of a right angle, and the sides containing it equal, the triangle is equilateral.

23. If one acute angle of a right-angled triangle be equal to one of another, the triangles will be equiangular. Will they be equal?

24. If two angles of a triangle be together equal to one right angle, the triangle will be right-angled; if less, obtuse-angled ; and if greater, acute-angled.

25. Let two right-angled triangles have any one of the three sides of the first equal to a corresponding one of the second; and any one

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