AC. Therefore, the difference of the squares of AB and AC is equal to the rectangle under their sum and difference.

(49.) This proposition corresponds with the following algebraic formula: (a+b)× (a−b)=a2—b2.

Thus, let AB=a; AC=b.

If a=9, and b=3, we shall have

(9+3)x(9-3)=12×6=92-32-81-9=72.

PROPOSITION VIII.

THEOREM. In any right-angled triangle, the square of the hypothenuse is equal to the sum of the squares of the other two sides.

Let ABC be a right-angled triangle, having the rightangle C; then will the square of the hypothenuse AB be equal to the sum of the squares of the other two sides AC, CB. Or,

AB AC+BC2.

D

For, on AB describe the square AF, and on AC, CB, the squares AH, BK; then draw CM parallel to AD or BF, and join AL, BC, CD, CF. AC meets the two CH, CB, so

Now, because the line as to make two right

angles, these two form one straight line HB, (B. I, Prop. 1, Cor. 1;) and because the angle GAC is equal to the angle DAB, being each a right-angle, or the angle of a

square; if to each of these angles we add the common angle BAC, then will the whole angle or sum GAB be equal to the whole angle or sum CAD. But the line GA is equal to the line AC, and the line AB to the line AD, being sides of the same square so that the two sides GA, AB, and their included angle GAB, are equal to the two sides CA, AD, and the contained angle CAD, each to each. Therefore, the whole triangle AGB is equal to the whole triangle ACD, (B. I, Prop. III.) But the square AH is double the triangle AGB, on the same base GA, and between the same parallels GA, HB, (B. II, Prop. III, Cor. :) in like manner the parallelogram AM is double the triangle ACD, on the same base AD, and between the same parallels AD, CM; and since the doubles of equal things are equal, (Ax. VI,) it follows that the square AH is equal to the parallelogram AM.

In like manner, the other square BK may be shown to be equal to the parallelogram BM. Consequently the two squares AH and BK together are equal to the two parallelograms AM and BM together, or to the whole square AF. That is, the sum of the squares on the two lesser sides is equal to the square on the hypothenuse.

Cor. 1. Hence, the square of either of the two lesser sides is equal to the difference of the squares of the hypothenuse and the other side, (Ax. III;) or equal to the rectangle contained by the sum and difference of the hypothenuse and other side, (B. II, Prop. VII.)

Cor. 2. Hence, also, if two right-angled triangles have two sides of the one equal to two corresponding sides of the other, their third sides will be equal, and the triangles themselves equal.

Cor. 3. The square on the diagonal of a square is double the square itself.

Cor. 4. The sum of the squares of the sides of a rectangle is equal to the sum of the squares of the diagonals. (50.) This theorem may be demonstrated in the following man

ner:

Let ABC be a right-angled triangle, rightangled at C; then will the square upon the hypothenuse AB be equal to the sum of the squares upon the sides AC and BC.

B

Upon AB construct the square ABDF, so that the triangle ABC may be included within it. Produce the shorter side BC, and from the angle D draw to it the perpendicular DG; which, being produced, draw from the angle F the line FH perpendicular to DH, and produce FH until it meet the longer side of the triangle at the point K. We have thus divided the square ABDF into five portions, namely, four triangles each equal to the original triangle ABC, and the square KCGH, the side of which is equal to KC the difference between AC and BC; consequently the area of this square is

(AC-BC)2=AC2-2AC.BC+BC2.

[B. II, Prop. vi.]

The area of the triangle ABC is equal to AC. BC; therefore the area of the four triangles is equal to 2 AC.BC, which, added to the area of the internal square, gives AC2+BC2 for the area of the entire square ABDF; but the area of this square is denoted by AB2, and therefore AB2=AC2+BC2.

(51.) The following is another simple demonstration:

B

The angles LAB and DAC are each right: taking from each the angle LAC, we have the remaining angles CAB and DAL equal; the angles ACB and ADL are also equal, each being a right-angle. We also have DA equal to AC, being sides of the same square; therefore, the two triangles ACB and ADL are equal, (B. I, Prop. Iv.) In a similar manner, it may be shown that the triangle BGM is equal to the triangle ACB: hence, AL, AB, BM are all equal; and since the angle BAL is right, the figure ALMB is a square. Again, since the sides KL, LM are respectively parallel to CA, AB, and lie in the same direction, the angle KLM is equal to CAB, (B. 1, Prop. xxII :) for a similar reason, the angle KML is equal to CBA, and the side LM is equal to the side AB; hence, the triangle LKM is equal to the triangle ACB, (B. I, Prop. iv.) Now, since the three triangles ADL, LKM, MGB, are each equal to the triangle ACB, they are equal to each other; consequently the square on AB is equal to the whole space ADKGB diminished by three times the triangle ACB. Again, the rectangle CFKH, contained by CF and CH, which are respectively equal to AC and CB, is double the triangle ACB, (B. II, Prop. 11 :) hence, the squares AF and BH are together also equal to the whole space ADKGB diminished by three times the triangle ACB, and consequently these two squares are equivalent to the square upon AB.

(52.) The algebraic condition (p2+q2)2=( p2 —q2)2+(2 pq)2, enables us to find numerical values for the sides of triangles which shall be right-angled.

Thus, p2+q=hypothenuse; p2-q2 and 2 pq=the sides.
If p=2, and q=1, we shall find

(53.) From this we know, that if a triangle is constructed, having 6, 8, and 10 for its sides, it must be right-angled. This relation was formerly very extensively employed by carpenters, in framing by the old scribe rule. They referred to it under the name of "six, eight, and ten." When they wished to ascertain whether the angle formed by the meeting of two sticks of timber was a right-angle, they measured from the angular point on the one stick 6 feet, and on the other 8 feet; then if the ten-foot pole would reach across from the one point to the other, the angle was right.

(54.) There is a rule used among carpenters for cutting braces, so as to give 17 inches in length for every 12 inches of run and rise; that is, they proceed on the supposition that the diagonal of a square 12 inches on a side, is 17 inches. Now this is a little too great; since 122+122=288, while 17-289. Notwithstanding this small error, it is found, in the case of short braces, and when the timber is not very firm, to answer very well in practice.

(55.) We may remark, in this place, that our algebraic formula ( p3+q2)2=( p2 — q2)2+(2 pq)2, readily indicates that it is impossible to find the diagonal of a square in a rational expression; for, in the case of a square, we must have p2-q2=2 pq, or, which is the same thing, p2-2 pq=q2. Adding q2 to each side of this equation, we have p2-2 pq+q=2 q'. Extracting the square root of each member, we find p-q=q√2, where the right-hand member is irrational so long as q is rational.

(56.) PROBLEM. There are two columns, in the ruins of Persepolis, left standing upright: one is 64 feet above the plane, the other 50 feet. In a direct line between these, stands an ancient statue, the head of which is 97 feet from the top of the higher column, and 86 feet from the top of the lower column. The distance from the base of the statue to the base of the lower column is 76 feet. Required the distance between the tops of the columns?

Solution. Let AB, CD (next page,) represent the columns standing on the horizontal plane, denoted by AC; let G denote the base, and F the head of the statue; then we shall have

AB=64 feet, CD=50, BF=97, DF=86, and CG=76.