For. the whole square being divided by the diag. into right angled isosceles triangles, and the parallelograms about the diagonals are equi-angular to those they have one angle in each right, and each of the others half of a right angle, .. the adjacent sides are and also the opposite sides.. the figures are equilat. to each other, and also rectangular.

PROP. 44, THEOR.

To a given right line, to apply a parallelogram equal to a given triangle, and having an angle equal to a given

one.

If the base of the given triangle and the given right line be not in directum; produce the given line until the produced part becomes to the base of the triangle, and on this produced part, construct a triangle to the given one; (cor. prop. 22.) on the half of this base adjacent to the given line, construct a parallelogram to the given triangle and in the given an gle; then complete the parallelogram of which the given line and side of first parallelogram are sides, draw the diagonal of it thro' the angle contained by those sides, and produce it to meet the opposite side of the first formed parm, produced; thro' the point of occourse draw a line par. to the given line, to meet the opposite side of second parm. produced, and produce to meet this par. ́ line, the side of first parm. adjacent to the givèn line, then there is formed on the given line a parm, which with the first formed parm. is complemental of the two about the diagonal of the entire par", and is.. to first parTM. (43, 1,) and it has an angle to the given one, since it is external to an angle, that is internal to one to the given one; the equality of those angles is evident from prop. 29, 1.

PROP. 45, THEOR.

To construct a parallelogram equal to a given rectilinear figure and having an angle equal to a given one.

Resolve the given figure into triangles: construct a to one of them, and having angle to

parallelogram

the given one, on the side of this parm, not about the angle to given one construct a parm to an other of the triangles, and with the given angle; (external to the one to the given one in the first,) and so on until parms, are formed to all the triangles into which the figure is resolved. The figure made up of all these is a parallelogram to the given figure and with the given angle. For the side of the first about the given angle and conterminious with that side of it to which the second is applied is in directum with the conterminious side of the second since they make with the same line at the same point, angles which are equal to two right ones, hence obviously the fig. is a parrallelogram and by construction equal to the given right lined figure and having an angle to the given one.

PROP. 46, PROB.

On a given right line to construct a square.

To either extremity of the given line, draw a perpendicular, and make it to it, (prop. 11 sch. & prop. 3,) through the extremity of this per. (remote from given line) draw a right line parallel to given line, and through the other extremity of given line draw a right line parallel to perpen. to meet the first par. line; those lines form a2.

Because the figure thus formed is a parallelogram and one of its angles being right, the remaining angles must be so, (cor I. prop. 34,) and because the perpendicular and given line (which are adjacent) are their opposite sides, are also to them (prop. 34,).. the figure is equilateral, and it has been proved rectangular, it is .. a 02.

Cor. 1. The squares of equal right lines are equal.

For if a diagonal be drawn in each, a triangle which is half of one'. can be proved to a triangle which is half of the other (prop. 4.).. the squares themselves which are the doubles of those triangles are =

Cor. 2. If two squares be equal, their sides are equal. For if not cut off from conterminous sides of that - which is supposed to have the greater sides, parts = to the

supposed lesser sides, and connect their extremities, the triangle formed by those cut off parts and con. line is = to a triangle which is half the other and .. to a triangle of which itself is a part, which is absurd.

PROP. 47, THEOR.

2

In a right angled triangle, the squares of the side subtending the right angle is equal to the sum of the squares of the sides which contain the right angle.

Describe squares on the sides of the triangle, and from the right angle draw a right line dividing the 2 on the hypothenuse into two parallelograms. It can be proved that each of those parallelograms is respectively to the on that side of the triangle which is adjacent to it. For if the acute angle contained by that side and the hypothenuse be added to the angles of their 'rs which are adjacent to it, two obtuse angles will be formed to one another, the legs about which are also respectively, if then those legs be joined the triangles thus formed shall be,but one of these triangles is half of one of the parallelograms into which the on the hypothenuse is divided, and the other is half of the on the side adjacent to this par. (prop. 41. I.)... this entire parm. is to the entire on that side in like manner the other parm, can be to of side adjacent to it.. the whole

2

2

proved hypothenuse, &c.

2

2

2 of

Cor. 1. Given in numbers any two sides of a right angled triangle, the third side can be found, for it is the square root of the sum or difference of the squares of the given lines, according as the given sides contain the right angle or not.

2

Cor. 2. To find a square equal to the sum of two or more squares. Assume lines respectively to the sides of any two'rs, with these form a right angle and connect their extremities, the of the joining line is to the sum of the 'rs of the assumed lines. Then with this joining line and a side of another form a right angle, the of the line joining their extremities is = to the sum of the threers, and so on.

Cor. 3. To find a line whose square is equal to the difference between two squares.

Produce the greater of the lines 'till the produced part becomes to the lesser, from their common extremity as centre and the greater as radius, describe a circle and from the other extremity of produced part erect a perpendicular to meet its circumference, the of this perpendicular is to the required difference, because the of it with the 2 of the lesser is to the of the greater, ... &c.

2

2

Cor. 4. If from any angle of a triangle a perpendicular be let fall on the opposite side, the difference between the squares of the sides which contain that angle, is equal to the difference of the squares of the segments of the sides on which the perpendicular falls.

2

For the of one side is to of adjacent seg. and of per,, and the of the other side is

of adjacent seg. and

2

to

of per. If the ' of perpendicular be taken away, it is evident that the dif. between the 'rs of the sides must be to the dif. between the O'rs of the segments.

Cor. 5. The excess of the square of either side, above the square of the conterminous seg. of third side is the same. For it is the of the perpendicular.

PROP. 48. THEOR.

If the square of one side of a triangle be equal to the sum of the squares of the other two sides, the angle opposite to that side is a right angle.

From the vertex of this angle raise a perpendicular to either of the sides containing it and to the other side, connect the extremity of this perpendicular with the extremity of the side to which it is a perpendi cular, the square of this connecting line is to the square of the third side of the given triangle. Therefore these two triangles are equilateral to one another and therefore equi-angular. Therefore the angle required to be proved so is a right angle.

=

OBSERVATIONS.

1st. In the fifth, the equality of the angles below the base seems to be introduced merely to prove that of the angles at the base; but the equality of the angles at the base can be proved without producing the sides. FIG. 1. PLATE 1.

=

On the sides AB, AC assume portions AD, AE, join DC, EB, and DE. Then in the triangles BAE, CAD, the sides BA and AE are respectively to CA and AD and the angle A common.. DC and BE are = and the angle ABE is to ACD. Then in the triangles DBE, ECD the sides DB, BE are respectively to EC and CD and the angle DBE to ECD... the angles CED, BDE are and also the angles BED and CDE... the angles BDC and CEB are—. Also in the triangles BDC, CEB the sides BD and DC are respectively = to CE and EB and the contained angles the angles

DBC and ECB are =.

2nd. In the sixth, (which is the converse of the fifth,) we must cut off the part towards the base, for if we cut off towards the vertex we would not have the necessary sides or angles,

This

3rd. In the seventh, we mean that if one pair of conterminous sides be the other pair cannot be so. proposition is made use of only in proving the eight. It does not follow from this prop. that there cannot be on the same base two triangles whose legs are respectively =; but the ⇒ legs must not be conterminous,

4th. The eighth is the second case of triangles; and is the converse of the fourth.

5th. In the ninth, the equilat. tri. is constructed on the side of the connecting line remote from the given angle, least its vertex should coincide with the given angle.

6th. In the eleventh. If the point be given in the extremity of the line you may first produce the line on that side. Vide Deducibles, Prop. 12.

7th. In the twelfth, we say an indefinite line, for otherwise the point might so lie that we could have no perpendicular on it from the given point.