of S and S' together are equal to the squares of s, s, and twice the square of P. But, by (IV), the square of H is equal to the squares of s and s' together with twice the rectangle s × 5. Hence, &c.

2. The square of S = H × s. For the square of S = the squares of P and s; but the square of P = 8 × s' •.• by (III), &c. 3. In like manner the square of S′= H × 8.

4. H x P = S x S', for each is twice the area (186). (278) The converses of these properties may be easily established, scil. that a triangle, having any of these properties, must be right angled.

1. If the square of P=s × s', let twice that square be added to the sum of the squares of s and s', and we shall, by (XLVII, Book I.), obtain a magnitude equal to the sum of the squares of S and S'; and, since the square of P is equal to twice the rectangle s × s', we shall also have the same magnitude (IV) equal to the square of H. Hence, by (XLVIII, Book I.), the angle opposite to H is right.

2. If the square of S = H x 8, we have also the square of S = the squares of P and s. Take the square of s from both, and we have the square of P = s × s'; therefore, by the last case, the angle opposite to H is right.

In these cases, the perpendicular P is supposed to fall within the side H; if not, the propositions are not necessarily true.

3. If H × P = S x S. In this case S x S' is twice the area, and also S x the perpendicular on it from the opposite angle is twice the area; ·.· S' is equal to that perpendicular, and therefore must be the perpendicular itself, since no line equal to it could be drawn from the same point.

PROPOSITION XI. PROBLEM.

(279) To divide a given finite right line (A B) so that the rectangle under the whole line and one segment shall be equal to the square of the other segment.

C K D

From the point A erect AC perpendicular and equal to the given line AB, bisect it in E, join E B, produce CA until E F is equal to E B, and in the given line A B take AH equal to A F; the square of A H is equal to the rectangle under the other segment H B and the whole line A B.

E

A

H

B

Complete the square of A B, draw through H the right line G K parallel to A C, and through F the right line FG parallel to A B.

Because CA is bisected in E, and A F is added to it, the rectangle under C F and F A together with the square of E A is equal to the square of E F (VI), or to the square of E B which is equal to E F (const.), and therefore to the squares of E A and AB (XLVII, Book I.); take away the common square of E A, and the

G

rectangle under C F and F A is equal to the square of AB: but because A F and FG are equal, CG is the rectangle under C F and F A, therefore C G and A D are equal, and if the common rectangle C H be taken away A G and H D are equal; but A G is the square of A H, for A H and A F are equal (const.), and the angle A is a right angle, and H D is the rectangle under A B and H B, for B D is equal to A B.

A line divided, as in this proposition, is said (vide Book VI.) to be cut in extreme and mean ratio.'

(280) COR, 1.-By attending to the solution of this problem, it will appear that, in order to cut a line in extreme and mean ratio, it is first necessary to produce it in extreme and mean ratio; that is, to produce it so that the rectangle under the whole produced line and produced part shall be equal to the square of the line itself. In the demonstration of the proposition, it appears that the rectangle CFX FA is equal to the square of C A, and therefore CA has been produced to F in this way, and C A is equal to the given line A B. (281) COR. 2.-Considering C F as a line cut in extreme and mean ratio at A, it will easily appear that the rectangle under the greater segment, and the difference of the segments, is equal to the square of the lesser segment; for A C is the greater segment, and is equal to A B, A F, which is equal to A H, is the less, and therefore H B is the difference of the segments. But by the demonstration of the proposition A B x H B is equal to the square of A H.

Hence it appears, that if a line be cut in extreme and mean ratio, the greater segment will be cut in the same manner, by taking on it a part equal to the less. And the less will be similarly cut, by taking on it a part equal to the difference, and so on.

(282) We have here taken for granted that if the rectangle CF X FA = the square of C A, that CA is greater than A F. This is, in fact, also taken for granted in the demonstration of the proposition itself. It is, however, easily proved. The rectangle CF XFA is equal to the rectangle CA x A F, together with the square of A F, ·.· the square of CA exceeds the square of A F by the rectangle CA × A F, and ... the line C A must be greater than Å F

(283) COR. 3.-Hence it also appears, that when a line is cut in extreme and mean ratio, the rectangle under its segments is equal to the difference between their squares.

Let A be a line cut in extreme and mean ratio, and G its greater segment, L its lesser segment, D the difference of its segments. The student will find no difficulty in establishing the following properties. (284) 1. The sum of the squares of A and L is equal to three times the square of G.

(285) 2. The square of the sum of A and L is equal to five times the square of G.

(286) 3. A x D = GX L.

(287) 4. The square of L·

GX D.

It may also be shown, that a line cut so as to have any of these properties will be cut in extreme and mean ratio.

(288) In any obtuse angled triangle (A B C), the square of the side (A B) subtending the obtuse angle exceeds the sum of the squares of the sides (BC and C A) which contain the obtuse angle by double the rectangle under either of these sides (BC), and the external segment (CD) between the obtuse angle and the perpendicular drawn from the opposite angle.

B

The square of BA is equal to the sum of the squares of A D and DB (XLVII, Book I); but the square of DB is equal to the squares of DC and C B together with double the rectangle under DC and CB (IV); therefore the square of AB is equal to the squares of A D, D C, and C B together with double the rectangle under DC and CB; but the square of AC is equal to the squares of AD and DC (XLVII, Book I.;; and therefore the square of AB is equal to the squares of AC and CB together with double the rectangle under BC and CD, therefore the square of A B exceeds the sum of the squares of A C and C B by double the rectangle under DC and C B.

It is evident that if the perpendicular were drawn from B to AC produced, it would in like manner be proved that double the rectangle under A C and its production would be equal to the excess of the square of A B above the squares of A B and B C. And hence it follows, that the rectangle B C × C D is equal to the rectangle under AC and its produced part.

(289) In any triangle (A B C) the square of the side (A B) subtending an acute angle (C) is less than the sum of the squares of the sides (A C and C B) containing that angle, by twice the rectangle under either of them (A C) and the segment between the acute angle and the perpendicular (B F), let fall from the opposite angle.

B

T A C

B

A F

C

The squares of AC and C F are equal to twice the rectangle under AC and CF together with the square of A F (VII), and if the square of the perpendicular BF be added to both, the squares of AC, CF, and BF are equal to twice the rectangle under A C and C F together with the squares of BF and AF, or with the square of AB, which is equal to them; but the squares of BF and CF are equal to the square of BC, and therefore the squares of BC and AC are equal to twice the rectangle under AC and C F together with the square of AB; therefore the square of A B is less than the sum of the squares of A C and CB by twice the rectangle under AC and CF.

If the angle A happen to be a right angle, the perpendicular BF will coincide with B A, and the points F and A will be the same, but the demonstration remains unchanged.

(290) If the angle A be right the double rectangle A CXC F becomes equal to twice the square of A C, and the proposition becomes equivalent to the forty-seventh of the first book.

(291) This proposition and the twelfth may be reduced to one, thus: The difference between the square of one side of a triangle, and the sum of the squares of the other two sides, is equal to twice the rectangle under either of these two sides and the intercept between the perpendicular on it and the angle included by the sides."

(292) COR. 1.-If a perpendicular to B C be drawn from the angle A, the rectangle under the side B C and the part intercepted between this perpendicular and C is equal to the rectangle ACX CF. For each of these rectangles is half the difference between the square of A B and the squares of B C and A C.

(293) COR. 2.-If the three sides of a triangle be given in numbers, its area may be found by these principles. Find half the difference between the square of any side and the sum of the squares of the other two sides. This is the rectangle under either of those other two sides and the intercept between the perpendicular and the included angle Let this then be divided by either of the other sides, and the quote will be that intercept. Take its square from the square of the other side, and the remainder is the square of the perpendicular, the square root of which is the perpendicular itself. This multiplied by half the divisor gives the area.

If it happen that the square of one side be equal to the sum of the squares of the other two, the angle included by those two must be right, and in that case the area may at once be found by taking half their product.

(294) To construct a square equal to a given rectilinear figure (Z).

Construct a rectangle CI equal to the given rectilinear figure (XLV, Book I.); if the adjacent sides be equal, the problem is solved.

Ꮓ

B

If not, produce either side I A, and make the produced part A L equal to the adjacent side A C; bisect IL in O, and from the centre O with the radius OL describe a semicircle LB I, and produce CA till it meet the periphery in B; the square of A B is equal to the given rectilinear figure.

L

For draw O B, and because I L is bisected in O and cut unequally in A, the rectangle under I A and A L together with the square of OA is equal to the square of O L (V), or of OB, which is equal to O L, and therefore to the squares of OA and AB (XLVII, Book I.); take away from both the square of O A, and the rectangle under I A and AL is equal to the square of A B; but the rectangle under I A and AL is equal to IC, for AL and A Care equal (const.); therefore the square of A B is equal to the rectangle IC, and therefore to the given rectilinear figure Z.

(295) Schol. From this proposition it appears, that if a perpendicular BA be drawn from any point in a semicircle to the diameter, the square of the perpendicular is equal to the rectangle under the segments into which it divides the diameter.

(296) The following is a selection from some of the most useful and remarkable theorems and problems which may be inferred from the second book.

(297) To divide a line internally so that the rectangle under its segments shall have a given magnitude.

Let the given magnitude be equal to the square of the line A, and let B C be the given line.

On B C describe a semicircle, and through B draw BD A perpendicular to B C and equal to A.

Draw D E parallel

Then B C is cut

to BD and E F perpendicular to B C.
as required at F. This appears from (295).

It is evident that if A were greater than half of B C, the parallel D E would not meet the semicircle, and the problem would be impossible; and since, in general, the pa

B

D

rallel meets the circle at two points, there are two points at which B C may be cut as required, and these points are at equal distances from its middle point.