AB2 = BC × BD; AC2 = BC × DC; from which, by adding member to member, that is, AB2 + AC2 = BC × (BD + DC) = BC × BC; BC2. Cor. When AB = AC, we have BC2=2AB2; that is, in any right-angled isosceles triangle, the square of the hypotenuse is double the square of one of the sides. Consequently, the second power of the numerical value of the diagonal of a square is double that of the side. And the diagonal and side must be to each other in the ratio of ✓2 to 1; that is, they are incommensurable, as has already been shown (B. II., P. XIV.). Scholium I. From the two equations, AB2 = BC × BD, AC2= BC × DC, we deduce the proportion, AB2: AC2:: BC × BD : BC × DC, or, :: BD: DC. In a similar manner, by combining the identical equation BC2 = BC2, with the same two equations above, we have and BC2: AB2:: BC × BC: BC × BD, or, :: BC: BD; BC2: AC2 :: BC × BC : BC × DC, or, : : BC: DC. From this we see that, in any right-angled triangle, the squares or second powers of the sides containing the right angle and of the hypotenuse, are proportional to the segments of this hypotenuse, and the hypotenuse itself. That is to say, we have AB2: AC2: BC2:: BD:DC: BC. Scholium II. The segments BD, DC, formed by the perpendicular AD on the hypotenuse BC, are called the projections of the two sides on the hypotenuse. In general, the projection of any definite straight line MN, on another indefinite straight line AB, is the distance PQ of the intersection of the perpendiculars drawn through the extremities of the first line to the second. M A P N R B If through M we draw MR parallel to PQ, we shall have MR = PQ, and the right-angled triangle MNR will give MN2 = MR2 + NR2 = PQ2 + (NQ — MP)2. That is-The square of a straight line of determinate length is equal to the square of its projection on another line, plus the square of the difference of the perpendiculars which determine this projection. THEOREM XV. In any obtuse-angled triangle, the square of the side opposite the obtuse angle is equal to the sum of the squares of the other two sides, increased by twice the product of either of the sides containing the obtuse angle into the projection of the other side on the prolongation of the first. If the triangle ABC is obtuse-angled at A, we shall have BC2 AB2 + AC2+2AB × AD. = For, since BD = AB+ AD, if we square both members of this equation, using the algebraic formula, we shall have 0 BD2 = AB2 + AD2 + 2AB × AD. Adding now, to each member of this, CD2, and observing that BD2+CD2 = BC2, and also that AD2 + CD2 = AC2, we shall obtain BC2= AB2+AC2+2AB × AD. THEOREM XVI. In any triangle, the square of the side opposite an acute angle is equal to the sum of the squares of the other two sides, diminished by twice the product of one of these sides, by the projection of the other on the preceding one, produced if necessary. agram we use; but in both cases BD is the difference of AB and AD; consequently, if we take the squares of both members, by the well-known algebraic formula, we shall have BD2 = AB2 + AD2 — 2AB × AD. Adding to each member of this DC, and observing that BD2+CD2 = BC, and also that AD2+ DC2= AC2, we shall obtain BC2 = AB2 + AC2 — 2AB × AD. This result will hold good even when the perpendicular CD coincides with the side CB, in which case AD=AB, and the above will become BC-AC-AB2, or, AC2= AB2+ BC2, AB2+BC2, which we know to be true, since the triangle ABC is, in this case, right-angled at B (T. XIV.). Scholium. A triangle is right-angled, acute-angled, or obtuseangled, according as the square of the longest side is equal, less, or greater than the sum of the squares of the other two sides. If a triangle has 3, 4, and 5 for the numerical values of its sides, it will be right-angled, since 52 = 32 +42. One having 4, 5, and 6 for its sides, is acute-angled, since 62 <42 + 52. One having 2, 3, and 5 for its sides, is obtuse-angled, since 52 > 22+ 32. Thus we are furnished with a second very simple method for determining the kind of triangle, when its sides are given (B. I., T. XXXV., S.). THEOREM XVII. In any triangle, the sum of the squares of any two sides is equal to twice the square of half the third side, increased by twice the square of the line drawn from the middle of this third side to the opposite angle. If CD is drawn bisecting AB, we shall have AC2+BC2 = 2AD2 + 2CD2. For, the two triangles ADC, CDB, the C one obtuse-angled and the other acute-angled at D, give, by the two preceding Theorems, AC2= AD2+ CD2+2AD × DE, BC2 BD2+CD2 - 2DB +DE; adding these, and observing that AD = DB, we have AC2+BC2 = 2AD2 + 2CD2. Cor. The sum of the squares of the four sides of any parallelogram is equal to the sum of the squares of its diagonals. This proposition is easily deduced from the above. DETERMINATION OF AREAS. THEOREM XVIII Two parallelograms having equal bases and the same altitude, are equivalent. We will suppose the one LF DI E CF figure placed upon the other, so that their inferior bases, which are equal, may coincide. Let ABCD be the A L B E L' first parallelogram, and ABEF or ABE'F' the second. Since these parallelograms have the same altitude, their superior bases will be situated in the same indefinite line LL', parallel to the lower base. If now, we confine our attention to the two parallelograms ABCD, ABEF, we see that the triangles BCE, ADF are equal, having equal angles CBE and DAF (B. I., T. III.) included between equal sides, namely, BE = AF, BC=AD (B. I., T. XXXI.). If, now, from the quadrilateral ABED we subtract the triangle BEC, we shall have left the parallelogram ABCD; but if from the same quadrilateral we subtract the equal triangle AFD, we shall have left the parallelogram ABEF. Hence the two parallelograms are equivalent. We may in the same way prove that the parallelograms ABCD, ABE'F' are equivalent. Cor. A parallelogram is equivalent to a rectangle having the same base and the same altitude. Two rectangles of the same base and the same altitude are equal, and consequently equivalent. THEOREM XIX. A triangle is equivalent to half of a parallelogram having the same base and the same altitude. For, through B and C, draw BD parallel to AC, and CD parallel to AB; we thus form a parallelogram having the same base and the same altitude as the triangle. Α C B D The triangle is one half this parallelogram (B. I., T. XXXI., Cor I.). Cor. Two triangles of the same base, or of equal bases, and the same altitude, are equivalent. THEOREM XX. Two rectangles of the same altitude are to each other as their bases. Let ABCD, AFGD be the two rectangles having the common altitude AD; then will they be to each other as their bases AB, AF. D G C A F B We will first suppose the bases AB, AF to be commensurable; as, for example, suppose they are to each other as 7 to 4. If we divide AB into seven equal parts, AF will contain four of these parts; at each point of division draw lines perpendicular to the base, thus forming seven partial rectangles, which will be equal, since they have equal bases and the same altitude (T. XVIII.). Now, as the rectangle ABCD contains seven of these equal rectangles, while AFGD contains only four, it follows, that but therefore, ABCD: AFGD::7:4; AB: AF::7:4; ABCD: AFGD::AB: AF. |