Fig. 262.

This problem, which is the 47th of the First Book of Euclid, is said to have been demonstrated first by Pythagoras. It is stated (but the story is of doubtful authority) that as a thank-offering for its discovery he sacrificed a hundred oxen to the gods. From this circumstance it is sometimes called the hecatomb problem. It is of great value in the exact sciences, more especially in Mensuration and Astronomy, in which many otherwise intricate calculations are by it made easy of solution.

## 354. - Proposition

In an equilateral octagon the semi-diagonal of a circumscribed square, having its sides coincident with four of the sides of the octagon, equals the distance along a side of the square from its corner to the more remote angle of the octagon occurring on that side of the square. Let Fig. 263 represent the square referred to; in which O is the centre of each; then A O equals A D. To prove this, it need only be shown that the triangle A O D is an isosceles triangle having its sides A O and A D equal. The octagon being equilateral, it is also equiangular, therefore the angles B CO, ECO, A D O, etc., are all equal. Of the right-angled triangle FEC,FC and FE being equal, the two angles FE C and FCE, are equal (Art. 338), and are therefore (Art. 347) each equal to half a right angle. In like manner it may be shown that FA B and FB A are also each equal to half a right angle. And since FE C and FA B are equal angles, therefore the lines E C and A B are parallel (Art. 331,) and hence the angles ECO and A OD are equal. These being equal, and the angles E C O and A D O being equal by construction, as before shown, therefore the angles A OD and ADO are equal, and consequently the lines A 0 and A D are equal. (Art. 338.)

Fig. 263.

## 355. - Proposition

An angle at the circumference of a circle is measured by half the arc that subtends it; that is, the angle ABC (Fig. 264) is equal to half the angle A D C. Through the centre D draw the diameter BE. The triangle A B D is an isosceles triangle, A Z and B D being radii, and therefore equal; hence, the two angles at F and G are equal (Art. 338), and the sum of these two angles is equal to the angle at H (Art. 345), and therefore one of them, G, is equal to the half of H. The angles at H and at G (or ABE) are both subtended by the arc A E. Now, since the angle at H is measured by the arc A E, which subtends it, therefore the half of the angle at H would be measured by the half of the arc A E; and since G is equal to the half of H, therefore G or A B E is measured by the half of the arc A E. It may be shown in like manner that the angle EBC is measured by half the arc E C, and hence it follows that the angle A B C is measured by half the arc, A C, that subtends it.

Fig. 264

## 356. - Proposition

In a circle all the inscribed angles, A, B, and C (Fig. 265), which stand upon the same side of the chord DE are equal. For each angle is measured by half the arc DFE (Art. 355). Hence the angles are all equal.

## 357. - Corollary

Equal chords, in the same circle, subtend equal angles.

Fig. 265.

## 368. - Proposition

The angle formed by a chord and tangent is equal to any inscribed angle in the opposite segment of the circle; that is, the angle D (Fig. 266) equals the angle A. Let HF be the chord, and E G the tangent; draw the diameter JH; then JHG is a right angle, also JFH is a right angle. (Art. 352.) The angles A and B together equal a right angle (Art. 346); also the angles B and D together equal a right angle (equal to the angle JHG); therefore, the sum of A and B equals the sum of B and D. From each of these two equals, taking the like quantity B, the remainders A and D are equal. Thus, it is proved for the angle at A; it is also true for any other angle; for, since all other inscribed angles on that side of the chord line HF equal the angle A (Art. 356), therefore the angle formed by a chord and tangent equals any angle in the opposite segment of the circle. This being proved for the acute angle D, it is also true for the obtuse angle EHF; for, from any point, K (Fig. 267) in the are H K F, draw lines to J F and H; now, if it can be proved that the angle E HF equals the angle FKH, the entire proposition is proved, for the angle FKH equals any of all the inscribed angles that can be drawn on that side of the chord. (Art. 356.) To prove, then, that EHF equals HKF: the angle EHF equals the sum of the angles A and B; also the angle HKF equals the sum of the angles C and D. The angles B and D, being inscribed angles on the same chord, J F, are equal. The angles C and A, being right angles (Art. 352), are likewise equal. Now, since A equals C and B equals D, therefore the sum of A and B equals the sum of C and D - or the angle E HF equals the angle HKF.

Fig. 266.

Fig. 267.

## 359. - Proposition

The areas of parallelograms of equal altitude are to each other as the bases of the parallelograms. In Fig. 268 the areas of the rectangles A B CD and B E DF are to each other as the bases CD and DF. For, putting the two bases in form of a fraction and reducing this fraction to its lowest terms, then the numerator and denominator of the reduced fraction will be the numbers of equal parts into which the two bases respectively may be divided. For example, let the two given bases be 12 and 9 feet respectively, then 12/9 = 4/3, and this gives four parts for the larger base and three parts for the smaller one. So, in Fig. 268, divide the base CD into four equal parts, and the base DF into three equal parts; then the length of any one of the parts in CD will equal the length of any one of the parts in D F. Now, parallel with A C, draw lines from each point of division to the line A E. These lines will evidently divide the whole figure into seven equal parts, four of them occupying the area A B CD, and three of them occupying the area B E D F. Now it is evident that the areas of the two rectangles are in proportion as the number of parts respectively into which the base-lines are divided, or that - .