PROP. XLV. If, in the diagram of the Pythagorean Theorem, a line be drawn joining the angles H and E, G and D, the triangles HBE and GAD will be equal. For since HB = CB, and BE AB, and Zs HBE, CBA are together equal to two P C F E R H B PROP. XLVI. If in a right-angled triangle (ABC) a perpendicular (AF) be let fall on the hypotenuse, from the right angle, the squares of the segments into which it divides the hypotenuse, will be equal to the square of the difference of those segments, together with twice the square of the perpendicular. Bisect the hypotenuse BC in D, and join AD. Then, since AD=BD (s. Prop. 16), 4 AD2 = BC2 (S. Prop. 41) = AB2 + AC2 (i. Prop. 47). = 4 AD2 4 AF2 + 4 DF2 (i. Prop. 47); and therefore, AB2 + AC2= 4 AF2 + 4 DF2. Take from both sides 2 AF, and then BF2+FC2 = 2 AF2+4 DF2. But DF is half of the difference of the segments, and therefore 4 DF2 is equal to the square of that difference (s. Prop. 41). Consequently, the squares of the segments BF, FC are together equal to twice the square of the perpendicular AF, and the square of the difference of the segments. PROP. XLVII. In a rhomb the squares of all the sides are together equal to the squares of the diagonals. For in a rhomb the diagonals intersect at right angles (S. Prop. 32), and mutually bisect (s. Prop. 28). Therefore, the squares of the sides of the rhomb are equal to four times the squares of the segments of the diagonals (i. Prop. 47), and are, consequently, equal to the squares of the diagonals (s. Prop. 41). PROP. XLVIII. If on two sides (BA, CA) of a triangle (ABC) any parallelograms be described, and the remote sides (GF, ED) of those parallelograms be produced till they meet, and if a line (HA) be drawn joining that point of meeting and the vertex of the angle on the legs of which were constructed the parallelograms, these parallelograms are together equal to a parallelogram on the third side (BC) of the triangle; the other side (BK) of which parallelogram is equal and parallel to that joining line (HA). H F N Produce HA to L, and KB to N. Then BN = AH (i. Prop. 34), and therefore BN=BK, and the parallelograms BH and BL are equal (i. Prop. 36); but the parallelograms BH and BF are likewise equal (i. Prop. 35); therefore BL and BF are equal. In like manner the parallelograms LC and CD may be shown to be equal, and therefore the G B whole parallelogram KC is equal to the parallelograms BF and CD. SECOND BOOK. PROP. XLIX. Given the hypotenuse and the sum of the sides of a right-angled triangle, to find the sides. square of Take the square of the hypotenuse from the the sum of the sides; and again, take twice the remainder from the square of the sum of the sides: add the square root of the last remainder to the sum of the sides, and half of the result will be the greater side, which, deducted from the sum of the sides, will leave the less side. For the square of the sum of the sides is equal to the squares of the sides together with twice the rectangle contained by the sides (ii. Prop. 4). But the square of the hypotenuse being equal to the squares of the sides (i. Prop. 47), if it be taken from the square of the sum of the sides, the remainder will be equal to twice the rectangle under the sides; and if the double of this remainder, or four times the rectangle under the sides, be taken from the square of the sum of the sides, there will remain the square of the difference of the sides (ii. Prop. 8, and App. C, p. 166). But the sum and the difference of the sides added together, are equal to twice the greater side (App. C, p. 165), which is thus found, and being deducted from the sum leaves the less. PROP. L. Given the area of a right-angled triangle and its altitude (or perpendicular from the vertex of the right angle to the hypotenuse), to find the sides. Divide twice the area by the altitude, and add the square of the quotient to four times the area. The square root of this sum will be equal to the sum of the sides, from which, and the above-mentioned quotient, which is equal to the hypotenuse, the sides of the triangle may be found. For the area of the triangle is equal to half of the rectangle under the hypotenuse and altitude (i. Prop. 41), and therefore twice that area divided by the altitude gives the hypotenuse. But the area is also equal to half of the rectangle under the sides of the triangle (i. Prop. 34); and therefore four times the area, or twice the rectangle under the sides, and the square of the hypotenuse, which is equal to the squares of the sides (i. Prop. 47), are together equal to the square of the sum of the sides (ii. Prop. 4). But the sum of the sides being known, and also the hypotenuse, the sides of the triangle may be found by the preceding Proposition. PROP. LI. Given the segments of the hypotenuse made by the perpendicular from the right angle, to find the sides. Add the rectangle under the segments (or, in numbers, their product) to the square of each segment, and the square roots of the sums so obtained will be equal to the sides. For the rectangle under the segments of the hypotenuse is equal to the square of the altitude (ii. Prop. 14), which |