gram; the angle A, being a right angle, all the other angles are right angles; and the sides AB, and AD, being equal to the given straight line E; all the sides of the quadrilateral are equal, and the figure is a square described with the straight line E; (definition 24,) as was required to be done.

$36. THEOREM. In a right angled triangle, the square described upon the side opposite to the right angle, is equal to the sum of the squares described upon

the other two sides.

APPLIC. Let ABC, (fig. 41,) be a triangle, right angled in A, the square described upon BC, the line opposite to the angle A, shall be equal to the sum of the squares described upon AB, and AC, the two other sides.

Demonstr. Describe upon the three sides respectively the squares BH, BE, and CF; then the line EAC, will be one continued straight line parallel to DB, because it makes with the AB, at the point A, two angles together, equal to two right angles; and for the same reason, BAF, is one continued straight line, and parallel to CG; the angles at A, being in each case formed by the right angle of the triangle, and the one of the right angles of the respective square. Through the point A, draw AK, parallel to BI, and CH; join CD, and AI, and also BG, and AH. The triangle BDC, and the square BDEA, have the side BD, common, and are between the same parallels BD, and EC; the triangle BDC, is, therefore, half the square BDEA; the triangle BAI, and the parallelogram BIKL, have the same base BI, and are between the same parallels AK and BI; the triangle ABI, is, therefore, half the parallelogram BK. In the two triangles CBD, and ABI, the angle DBC, is equal to the angle ABI, being each of them equal to the sum of one right angle of one of the squares, and the angle ABC, of the triangle; the sides BD, and BC, which include the angle DBC, are equal respectively, to the sides AB, and BI, including the equal angle ABI; in the two triangles DBC, and ABI, we have, there

fore, two sides and the included angle in the one, equal to two sides and the included angle in the other; therefore, these two triangles are equal to one another; therefore, also, their doubles are equal; that is, the square ABDE, is equal to the parallelogram BLKI. Exactly in the same manner it is proved: that the triangle BCG, is equal to the triangle ACH, and the square ACGF, equal to the parallelogram CHKL; the two parallelograms BK, and KC, are, therefore, together, equal to the two squares EB, and CF, taken together; but the two parallelograms BK, and KC, together, form the square BH, described upon the side of the right angled triangle; that is, opposite to the right angle; therefore, the sum of the two squares BE, and CF, described upon the two other sides AB, and AC, including the right angle, are, together, equal to the square upon BC, opposite to the right angle; as was to be demonstrated.

Scholium 1. By the preceding will be found the side of a square, which shall be equal to the sum of two squares, the sides of which are given: by a right angle being formed, upon the two sides, of which the two sides of the given squares will be laid off; the line joining the extremities of these lines, will be equal to the side of the square, that will be equal to the sum of the two other squares.

Scholium 2. The side of the square which shall be equal to the sum of any number of squares, may therefore also be found by continuing the preceding operation, between the line obtained at first, and the other lines successively, until they have all been employed in the operation.

Scholium 3. The side of the square equal to the difference of two squares, will therefore also be obtained by taking on the one side, including the right angle, a line equal to`the side of the smaller square, and with the side of the larger square, describing a circle, the intersection of which with the other line about the right angle, will cut off a line equal to the side of the square, that will be equal to the difference of the two given squares.

$ 37. THEOREM. When the sum of the squares of the two sides of a triangle, is equal to the square of the

third side, the angle of the triangle contained between the two smaller sides, is a right angle.

APPLIC. Let ABC, (fig. 42,) be a triangle, in which the square upon AB, is equal to the sum of the squares upon AC, and BC, the angle at C, will be a right angle.

Demonstr. Upon CB, and at the point C, make BCD, equal to a right angle, and CD, equal to CA, join DB; the triangle BCD, being right angled at C, the sum of the squares upon CB, and CD, is equal to the square upon DB; the CD, being equal to CA, the sum of the squares upon CB, and CD, is equal to the sum of the squares upon CB, and CA; but the sum of the squares upon CB, and CD, is equal to the square upon BD, (by § 36,) and the sum of the squares upon CB, and CA, is equal to the square upon AB, by the supposition; therefore, the square upon DB, is equal to the square upon AB; therefore, the DB, is equal to AB, and the two triangles ACB, and DCB, have their three sides equal each to each, and the angles ACB, and DCB, are equal, being opposite to the equal side AB, and DB; the angle DCB, is a right angle by construction; therefore, also, ACB, is a right angle; as was to be demonstrated.

Scholium. A right angle may be constructed according to this proposition, by forming a triangle such: that the sum of the squares of two of its sides be equal to the square of the third side. Thus will be for instance a triangle, the sides of which would be 3, 4, and 5, for 9+16=25, gives to the square of these numbers the property required; and so would their equimultiples do the same.

CHAPTER IV.

Of the Circle, and the straight lines referring to the Circle.

§ 38. THEOREM. A straight line cannot cut the circumference of a circle in more than two points.

APPLIC. Let the circle ABDE, (fig. 43,) the centre of which is F, be cut by the straight line AC, the circumference of the circle shall be cut by it only in the two points A, and B.

Demonstr. If possible, suppose the straight line AB, cut the circumference of the circle also in another point, as D; join BD, then ABD, is a straight line; from the centre of the circle F, draw FG, perpendicular upon AB, and FH, perpendicular upon DB; join also FB, which meeting the supposed straight line ABD, in the point B, must make with it angles on one side that are together equal to two right angles, that is the angles ABF, and FBD, are together equal to two right angles. In the quadrilateral figure FGBH, the four angles together are equal to four right angles; the angles FGB, and FHB, are right angles by construction, therefore together equal to two right angles; and the angles GFH, and ABD, together, must be equal to two right angles; therefore, the GFH, must be less than two right angles, but the GBH, which is the sum of GBF, and FBH, is equal to two right angles, by the supposition; therefore GBH, would be at the same time equal and less than two right angles, which is impossible. Therefore, the straight line AB, does not cut the circumference of the circle ABDE, in the point D, or any other point except A, and B; as was to be demonstrated.

Corol. The two straight lines joining three points in the circumference of a circle make together an angle less than two right angles.

§ 39. PROBLEM. To find the centre of a circle.

APPLIC. Let ABDI, (fig. 44,) be the circumference of a circle, it is required to find the centre of it.

Solution. Draw any line AB, cutting the circumference of the circle in A, and B; bisect AB, in E, and from E, draw the EG, perpendicular to AB; from B, draw any other line in the circle as BD, bisect it in F, and from it draw FH, perpendicular to BD; these two perpendiculars will intersect one another, because the angle ABD, is less than two right angles; their point of intersection C, is the centre of the circle, which is required to be found.

Proof. Join CA, CB, and CD; in the two triangles AEC, and BEC, the sides AE, and EB, are equal by the bisection, the CE, is common to these two triangles, and the angles which these equal sides include at E, are right angles, therefore equal; therefore, the two triangles are equal, and the sides AC, and BC, opposite to the right angles at E, are equal. Exactly in the same manner is proved, that in the two triangles CBF, and CDF, the sides CB, and CD, are equal, therefore the three sides CA, CB, and CD, are all equal to one another; and any number of points may be taken in the circumference of the circle, between which straight lines being drawn and bisected, the perpendiculars drawn at their point of bisection, will meet in the same point C, as they would always form equal triangles with the two parts of the bisected line and the perpendicular, and all their sides opposite to the right angles become equal. This point C, of their common intersection will therefore be equidistant from any point of the circumference of the circle; that is, it will be the centre of it, and the CA, CB, CD, and so forth, will be the equal radii of the circle.

Corol. 1. When more than two equal straight lines can be drawn from a point within the circle to the circumference, that point is the centre of the circle.

Corol. 2. Three points determine the circumference of a circle, for the continuance of the same circle has all the radii equal. Corol. 3. The line drawn perpendicular to the middle of a chord, passes through the centre of the circle.