« ForrigeFortsett »
in the convex circumference, such that the straight lines CA, CB make equal angles with CE, the tangent at C: CA, CB are together less than the sum of two straight lines DA, DB drawn from any other point D in the circumference to A, B.
Let DA cut the tangent in E, and join EB. Then ED, DB are greater than EB. Add AE; then AD, DB are greater than A E, EB. But (App. I. 12.) AE, EB are greater than AC, CB, because AC, CB make equal angles with CE: therefore AC, CB are less than AD, DB.
Cor. Since the tangent makes equal angles with the radius, the angles FCA, FCB are equal, when the sum of AC, CB is a minimum.
PROP. XIV. THEOR. The squares of the straight lines drawn from any point to two opposite angles of a rectangle, are together equal to the squares of those drawn to the other angles.
Let ABCD be a rectangle, and E a point, either within the figure or without it; join EA, EC ; EB, ED: the squares of EA, EC are together equal to the squares of EB, ED.
Draw the diagonals intersecting in F, and join EF. Then, in the triangles ABC, DCB, AB, DC are equal, BC common, and the right angles at B and C equal; therefore (I. 4.) the diagonals AC, BD are equal : and, since (II. B. cor.) the diagonals bisect each other, the lines AF, BF, CF, DF are all equal. Now (II. A.) the squares of AE, EC are double of the squares of AF, FE. For the same reason, the squares of BE, ED are double of the squares of BF, FE, or of AF, FE. Therefore (I. ax. 1.) the squares of AE, EC are equal to the squares of BÈ, ED.
Schol. This proposition holds, whether the point E be in the plane of the rectangle or not; and the demonstration is the same in both cases.
PROP. XV. THEOR. The perpendiculars drawn from any given point within an equilateral figure to the several sides, or the sides produced, are together equal to the sum of those drawn from any other point within the figure to the sides, or their continuations.
Let A be any point within the equilateral figure BCDE, and let
AF, AG, AH, AK be drawn perpendicular to the sides : the sum of these perpendiculars is equal to the sum of the perpendiculars drawn from any other point within the figure to the sides.
For, if straight lines were drawn from A to the several angles, the rectangle BC.AF would be double of the triangle ABC; the rectangle CD.AG, or BC.AG, double of CAD, &c. Hence (II. 1.) the rectangle under BC and the sum of the perpendiculars AF, AG, AH, &c., is double of the whole figure BCDE. In the same manner, it would be shown, that the rectangle under BC and the sum of the perpendiculars drawn from any other point within the figure to the sides, is double of the figure BCDE. Hence the rectangle under BC and the sum of the one system of perpendiculars, is equal to the rectangle under BC and the sum of the others. The sums therefore are equal; for if they were unequal, the rectangle under BC and one of them could not be equal to the rectangle under BC and the other.*
Cor. 1. Since the rectangle under the side and perpendicular of an equilateral triangle is double of the triangle, it follows that the perpendicular is equal to the sum of the three perpendiculars drawn from any point within the triangle to the sides.
Cor. 2. In an equilateral and equiangalar figure, the sum of the perpendiculars drawn from any point within it to the sides, is equal to the radius of the inscribed circle taken as often as the figure has sides.
Cor. 3. The sum of the two perpendiculars drawn from any point in the base of an isosceles triangle to the other sides, is equal to the sum of those drawn from any other point in the base to those sides, or to the perpendicular drawn from one of the equal angles to the opposite side. The method of proof is the same as that of the proposition, if the vertex and the point in the base be joined.
PROP. XVI. THEOR. The squares of two straight lines drawn from any point in a diameter of a circle, or in its continuation, to the extremities of a chord parallel to it, are together equal to the squares of the segments between the point and the extremities of the diameter.
Let BDC be a circle, A a point in the diameter BC, or its con
* If the given point were without the equilateral figure, it would be shown in a siinilar manner, that if the sum of the perpendiculars which fall on sides lying between the point and the figure, be taken from the sum of the others, the remainder is equal to the sum of the perpendiculars to the sides from a point within the figure. It is easy to see how this will modify the corollaries to this proposition.
tinuation, and DE a chord parallel to BC; join AD, AE: the squares of AD, AE are together equal to the squares of AB, AC.
Throngh the centre F draw EFG, and join FD, AG. Then the angles AFD, AFG are equal, because (I. 29.) they are equal
to the equal angles FDE, FED. Therefore, in the triangles AFD, AFG, since FD is equal to FG, FA common, and the contained angles equal; AD, AG (1. 4.) are equal. Now, (II. A.) because GF, FE are equal, the squares of GA, AE, or of DA, AE, are equal to twice the square of GF or BF, and twice the square of AF.
But (II. 9. or 10.) the squares of BA, AC are equal to twice the squares of BF, AF. Therefore (I. ax, 1.) the squares of DA, AE are equal to the squares of BA, AC.
Cor. 1. Hence (I. ax. 1.) if there were another chord parallel to DE, the squares of the straight lines drawn from its extremities to A are equal to the squares of AD, AE; and the same holds regarding all other chords parallel to the diameter through A.
Cor. 2. The squares of BA, AC are equal to twice the square of the line drawn from A to the point of bisection of the arc DE. This may be demonstrated in the same manner as the proposition itself: and the line so drawn may be regarded as what AD, AE become, when DE recedes from the diameter so as to be reduced to a point.
PROP. XVII. THEOR.
If the circumference of a circle be cut by two straight lines which are perpendicular to one another, the squares of the four segments between the point of intersection of the two lines and the points in which they meet the circumference, are together equal to the square of the diameter.
Let the straight lines AB, CD cut one another perpendicularly in the point E, and the circumference of the circle ABC in the points A, B, C, D; and draw the diameter AF: the sum of the squares of AE, BE, CE, DE is equal to the square of AF.
Join AC, CF, FB, BD. Then (111. 31.) ACF, ABF are right angles ; and therefore, AEC being a right angle, FB (I. 28.)
is parallel to CD; and (III. 26. cor.) the arcs CF, DB, and (III. 29.) their chords are equal. Now (I. 47.) the square of
BD or CF is equal to the squares of BE, ED; and the square of AC to the squares of AE, EC. For the same reason, the square of AF is equal to the squares of FC, CA, that is, to the squares of the segments EA, EB, EC, ED. Cor. If AD, CB were joined, the sum of their
would be equal to the sum of the squares of AC, DB; the sum of the squares of each pair being equal to the sum of the squares of the four segments EA, EC, EB, ED.
PROP. XVIII. PROB.
From a given equilateral and equiangular figure, to cut off as many triangles as it has sides, so that the remaining figure may be a regular polygon having twice a s many sides as the given figure.
Let ABC be a given equilateral and equiangular figure: it is required to cut off from it as many triangles, as it has sides, so that the remaining figure may be a regular polygon, having twice as many sides as the given one.
In the given figure inscribe (IV. A. cor.) the circle FDE, touching the sides in D, E, F: bisect the arcs DE, EF, FD in the points G, H, K; and through these draw tangents to the circle, cutting the sides in L, M, N, O, P, Q: QLMNOP is the polygon required.
For if straight lines be drawn from the centre R to D, E, F, the quadrilaterals ADRF, BDRE, CERF have each two right angles at the points of contact D, E, F: therefore (1. 32. cor. 1.) the remaining angles A, DRF are equal to two right angles ; and so likewise are the two B, DŘE, and the two C, ERF. Take away the equal angles A, B, C, and the remaining angles DRF, DRE, ERF are equal : and therefore (III. 26.) the arcs FD, DE, EF are equal. Hence also their halves,
KD, DG, &c., are all equal, and (III. 29. and IV. 11. cor. 1.) D, G, E, H, F, K are the angular points of a regular polygon inscribed in the circle, which has evidently twice as many sides as the given figure: and therefore (IV. B.) LMNOPQ is the polygon required, since its sides touch the circle in the points G, E, H, &c.
Schol. In case of an equilateral triangle ABC, it is plain that the construction will be effected simply by trisecting the sides, and joining the points of trisection LM, NO, PQ.
PROP. XIX. THEOR.
The diagonals of a regular pentagon which meet within the figure, divide one another in extreme and mean ratio : and the greater segment of each is equal to a side of the pentagon.
Let ABCDE be a regular pentagon, and AC, BE two of its diagonals which cut one another in F: AC, BE are each divided in extreme and mean ratio in F.
About the pentagon describe (IV. A. cor.) a circle. Then (III. 28.) the arcs AB, BC, &c., are all equal, and AC is parallel to DE, and BE to CD: therefore CDEF is a parallelogram, and FC, FE are each equal to a side of the figure. Now, the triangles ABE, ABF have an angle common, and (III. 27.) AEB is equal to BAF, because they stand on equal arcs: therefore (VI. 4. cor.) BE:BA :: BA: BF, or EB : ET :: EF: FB; whence (VI. def. 3.) BE is divided in extreme and mean ratio in F: and it would be shown in the same manner that AC is also divided similarly in the same point.
Cor. The segment BG is divided similarly in F. For, in the similar triangles BFC, GFA, CF: FB : : AF: FG; or BG: BF :: BF: FG, because BG is equal to FC, and BF to AF.
PROP. XX. PROB. On a given straight line to describe a regular pentagon, hexagon, or quindecagon.
In any circle (IV. 11. 15. or 16.) inscribe a figure of the kind required : and (VI. 18.) on the given line describe one similar to it.
Schol. It is plain, that in the same manner, on a given straight line, a regular polygon may be described similar to any of those, the method of inscribing which in a circle was pointed out in the scholium to the 16th proposition of the fourth book; or, univer