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2. If in the legs AC,AD of an acute angle, any two points D, C be assumed, and perpendiculars be let fall from them on the opposite sides, the rectangles under the assumed parts and their respective intercepts between the acute angle and perpendiculars are equal.

Fig. 4.

Join CD, then the of CD is less than the sum of the 'rs of CA,AD by two rectangles under DA and AB and also by two rectangles CA and AE, ... those rectangles are =.

3. If an obtuse (DAC) and an acute angle BAC be contained by lines DA,AC and BA,AC respectively equal, and the excess of the one above a right angle be equal to the excess of a right angle above the other, the square of the side DC subtending the obtuse angle exceeds the square of BC subtending the acute by four rectangles under either of them A C and the intercept between the perpendicular and angle.

Fig. 5.

Let fall the perpendiculars DE,BF, then EA is evidently to AF, and the of DC is to the sum of the 'rs of DA and AC with two rectangles under CA and AE, and the of BC is less than the rs of DA,AC by two rectangles CAE,.. BC is less than DC2 by four rectangles under CA,AE or CA,AF.

Cor. 2. Hence it is evident the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of the sides.

PROP. 14. PROB.

To describe a square equal to a given rectilineal figure.

Make a rectangle to it, if its adjacent sides be the problem is done.

If not produce either side and make the produced part to the adjacent side, bisect the whole produced side and describe a semicircle on it, and produce the other side to meet the circumference, this produced part is the side of the square sought.

=

. For connect the point of bisection with its extremity at the periphery; the square of the intercept is to the difference between the of the connecting line and intermediate part of the diameter, (prop. 47, 1,) but the

of this part is to the rectangle under the unequal parts of the whole produced side, (prop. 5, 2,) i. e. to the rectangle made to the given figure.

THE

THIRD BOOK,

IN

GENERAL TERMS.

PROP. 1. PROB.

To find the centre of a given circle.

Draw any right line within the circle, let it be terminated by the circumference, bisect it and from the point of bisection draw a line at right angles to it and produce this line to meet the circumference, bisect it; its point of bisection shall be the centre of the circle.

For if it be possible, let any other point be the centre of the given circle, the right lines drawn from it to the extremity of the assumed line are to one another, also the segments of this line are and the line drawn to its point of bisection common, .. the triangles formed by those are in every way,.. the angles made by the line. drawn from the assumed point to the point of bisection are right; then one of them is to an angle made by the perpendicular on this line of which it is a part, which is absurd,.. this assumed point is not the centre in like manner it can be proved that no other but the point of bisection of the line drawn at right angles from the point of bisection of the first drawn line can be the centre of the circle.

Schol. Hence it is evident that if any line terminated in a circle be bisected by a perpendicular, that perpendicular if produced shall pass through the centre.

PROP. 2, THEOR.

If any two points be taken in the circumference of a circle, the right line which joins them falls within the

circle.

For if possible let the right line joining them be such that it has a point outside the circle; connect the centre of the circle with its extremities and with this point.

Then in the triangle of which the line connecting the assumed points is the base, the angles at the base are =, but one of the angles made by the line drawn from the centre to the point outside the circle, being external to one of those, is greater than it, and .. than the other, .. the side opposite this angle, viz. the line drawn to the extremity from the centre, is greater than this drawn to the point outside the circle, but this drawn to the extremity is a radius of the circle, and this drawn to the point outside it greater than a radius,.. it is not less than the line to the extremity, .. this line connecting the assumed points is not a right line, in like manner it can be demonstrated that if any other point but the extremities be in the circumference it cannot be a right line; .. every point of the right line falls within the circle.

PROP. 3, THEOR.

If a right line drawn through the centre of a circle, bisect a right line which does not pass through the centre, it is perpendicular to it. And if it cuts it at right angles it bisests it.

PART 1. Connect the centre of the circle with the extremities of the bisected line; then there are two triangles formed, having the half line and connecting line in one respectively to the other half line and connecting line in the other, and the intercept between the centre and point of bisection common, .. (prop. 8, 1,) the angles at the point of bisection are =, and are .. right angles, (prop. 13, 1.)

PART 2. Because the triangle formed by the line not passing through the centre, and the lines from the centre to its extremity is isosceles, the angles at this line are

and also those at the point of bisection (by constr.) and the lines from the centre (viz. those opposite those

angles as being radii of the same circle) are, .. the two triangles are, and their other sides, viz. the parts of the line not passing through the centre, .. this line is bisected.

PROP. 4. THEOR.
lines

If in a circle two right es cut one another, which do not both pass through the centre, they do not bi

sect one another.

If one of them passes through the centre it is evident that it cannot be bisected by the other which does not pass through the centre.

But if neither of the lines passes through the centre, draw a right line from the centre to their intersection, if one of them is bisected by this, it is at right angles to it, and if the other is bisected by it, it is also at right angles to it,.. an angle made by this line from the centre with one of the given lines is to an angle made by it with the other which is a part of it, which is absurd,.. the lines do not bisect one another.

PROP. 5. THEOR.

If two circles cut one another they have not the same

centre.

For if possible let them have a common centre and from it draw two right lines, one cutting both circunferences, and the other to a point of bisection.

Then since those lines are radii of one circle they are , and one of them drawn to the intersection is to part of the other, (viz. of the one cutting the circles,) as being radii of the same circle, (def. 15, b. 1,) ... this part is to the whole line cutting the circles, which is absurd; ... this point is not the centre of both circles, and in the same manner it can be proved that no other point is the centre of both.

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