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PROP. XIV. PROB.
TO describe a square that shall be equal to a given rectilineal figure.
a 45. 1.
Let A be the given rectilineal figure; it is required to describe a square that shall be equal to A.
Describe the rectangular parallelogram BCDE, equal to
D Then, because the straight line BF is divided into two equal parts in the point G, and into two unequal in the point E, the rectangle BE.EF together with the square of EG are equal to the square of GF, or GH. But the squares of HE and EG are equal to the square of GH; therefore BE.EF+EGP=HES+EG'. Take away the square of EG, which is common to both, and the remaining rectangle BE.EF is equal to the square of EH. But BD is the rectangle contained by BE and EF, because EF is equal to ED; therefore BD is equal to the square
of EH; and BD is also equal to the rectilineal figure A, therefore the rectilineal figure A is equal to the square of EH. Wherefore a square has been made equal to the given rectilineal figure A, viz. the square described upon EX. Which was to be done.
b 5. 2.
PROP. A, THEOR.
IF one side of a triangle be bisected, the sum of See N. the squares of the other two sides is double of the square of half the side bisected, and of the
of the line drawn from the point of bisection to the opposite angle of the triangle.
Let ABC be a triangle, of which the side BC is bisected
A dicular to BC; then AB2=a
a 47. 1. BE2+AE', and AC2=CE2+ AE2; wherefore AB2+AC= BE2+CE2+2AE. But, because the line BC is cut equally in D, and unequally in E, BE2+CE?=b2BD2 +2DE2;
b 9.2. therefore ABP + AC2=2BD2 B
D E +2DE2+2AE. Now DEP+AEP-a ADand 2DE2+ 2AE2=2AD2; wherefore ABP+AC?=2BD2+2A D. Therefore, &c. Q. E. D.
PROP. B. THEOR.
THE sum of the squares of the diameters of any parallelogram is equal to the sum of the squares of the sides of the parallelogram.
Let ABCD be a parallelogram, of which the diameters are AC and BD ; the sum of the squares of AC and BD is equal to the sum of the squares of AB, BC, CD, DA.
Let AC and BD intersect each other in E. Because the a 15. 1.
vertical angles AED, CEB are equals, and also the alternate b 29. 1.
angles EAD, ECB, the triangles ADE, CEB have two angles in the one equal to two angles in the other, each to each: but the sides AD and BC, which are opposite to equal
angles in these triangles c 34. 1. are also equalc; there А,
fore the other sides which
are opposite to the equal d 26. 1. angles are also equald,
yiz. AE to EC, and ED
B e A. 2. ABP+AD= 2BE2+2AE?; and for the same reason CDP+
BC= 2BE2 + 2EC2 = 2BE2 + 2AE%, because EC=AE.
Therefore ABP + ADP + DC + BC = 4BE? + 4AE. But f2.Cor.8.2. 4BE?=BD’, and 4AE=AC?, because BD and AC are
both bisected in E; therefore ABP + ADP+CD+BC = BD2+AC%. Therefore, the sum of the squares, &c. Q. E. D.
COR. - From this demonstration it is manifest that the diameters of every parallelogram bisect each other.
PROP. C. THEOR.
IF a perpendicular be drawn from any angle of a triangle to the opposite side, the difference of the squares of the sides is equal to the difference of the squares of the segments of the base.
Let ABC be a triangle, having the side AB greater than AC, and AD a perpendicular from the angle A to the opposite side BC; ABAC=BDP_DC.
Because AB=BD2+DA, and AC=CDP+DA,
ABP-AC-=BD-DC?, by taking equals from equals. Therefore, if a perpendicular, &c. Q. E. D.
PROP. D. THEOR.
IF a perpendicular be drawn from any angle of a triangle to the opposite side, the rectangle contained by the sum and difference of the sides is equal to the rectangle contained by the sum and difference of the segments of the base, that is, to the rectangle contained by the base, and the difference or sum of the distances intercepted between the extremities of the base and the perpendicular, according as the perpen. dicular falls within or without the triangle.
a C. 2.
b Cor. 5.2.
Let ABC be a triangle*, having the side AB greater than AC, and AD a perpendicular from the angle A to the opposite side BC; then AB+AC. AB-AC=BD+DC.BD-DC.
* See the last figures.