« ZurückWeiter »
arc of a circle, and from the centre B with the same distance describe another arc, and from C inflect the straight lines CE, CD P E equal to AB: The points C, D, E mark out the pentagon. C3 For it is evident from this construction that BF and AG bisect A. • * B the angles at the base of the triangle ACB, and consequently (IV. 3.) AB is equal to BF and FC, or AG and GC. Again, the triangles BAD and BFC, having the sides AB, BD equal to BF, BC, and the contained angles equal, are themselves equal (I. 3.), and consequently AB is equal to AD, and the angle BAD equal to BFC, or three times ACB. In the same way it is shewn that AB is equal to BE, and that the angles round the figure are
each equal to thrice the vertical angle of the original isosceles triangle. -
PROP. VI. PROB.
On a given finite straight line, to describe a regular hexagon.
Let AB be the given straight line, on which it is required to describe a regular hexagon.
On AB construct (I.1.) the equilateral triangle AOB, and repeat equal triangles about the vertex O'; these triangles will together compose the hexagon required.
Because AOB is an equilateral triangle, each of its ano gles is equal to the third part of two right angles (I. 30. cor. 1.); wherefore the vertical angle AOB is the sixth part of four right angles, or six of such angles may be placed about the a point O. But the bases of the triangles AOB, AOC, COD, A- * DOE, EOF, and BOF are all equal; and so are the angles at the bases, and which, taken by pairs, form the intermal angles of the figure BACDEF. This figure is, therefore, a regular hexagon.
- PROP. VII. PROB.
On a given finite straight line, to describe a regular octagon.
Let AB be the given straight line, on which it is required to describe a regular octagon. /
Bisect AB (I. 7.) by the perpendicular CD, which make equal to CA or CB, join DA and DB, produce CD until DO be equal to DA or DB, draw AO and BO, thus forming C: H (IV. 1.) an angle equal to the half of ADB, and, about the vertex O, repeat the equal triangles AOB, AOE, EOF, FOG, GOH, HOI, IOK, and KOB to compose the octagon.
For the distances AD, BD are evidently equal ; and because CA, CD, and CB are all equal, the angle ADB is contained in a semicircle, and is therefore a right angle (III. 19.). Consequently AOB is equal to the half of a right angle, and eight such angles will adapt themselves about the point O. Whence the figure BAEFGHIK, having eight equal sides and equal angles, is a regular octagon.
PROP. VIII. PROB.
On a given finite straight line, to describe a regular decagon.
Let AB be the straight line, on which it is required to describe a regular decagon.
On AB construct (IV. 4.) an isosceles triangle having each of the angles at its base double of the vertical angle, and, about the point O, place a series of triangles all equal to AOB: A regular decagon will result from this composition.
For the vertical angle AOB of the isosceles triangle is equal to the fifth part of two right angles (IV. 4. cor.), or to the tenth part of four right angles; whence ten such angles may be formed about the point O. The figure BACDEFGHIK, having therefore ten equal sides and equal angles, is a regular decagon.
PROP. IX. PROB.
On a given finite straight line, to describe a regular dodecagon.
Let AB be the straight line, on which it is required to describe a regular twelve-sided figure. On AB construct (I. 1.) the equilateral triangle ACB, and again (IV. 1.) the isosceles triangle AOB, having its vertical angle equal to the half of ACB, and repeat this triangle AOB about the point O; a regular dodecagon will be thus formed. For ACB being an equilateral triangle, each of its angles is the third part of two right angles (I. 30. cor. 1.); consequently the angle AOB is the sixth part of two right angles or the twelfth part of four right angles, and twelve such angles can, therefore, be placed about the vertex O. Scholium. Hence a regular twenty-sided figure may be described on a given straight line, by first constructing on it an isosceles triangle having each of the angles at the base double of the vertical angle, and then erecting another isosceles triangle with its vertical angle equal to the half of this. And, by thus changing the elementary triangle, a regular polygon may be always described, with twice the number of sides. 2.
PROP. X. PROB.
Let ABC be a triangle, it which it is required to inscribe a circle.
Draw AD and CD (I. 5.) to bisect the angles CAB and ACB, and from their point of concourse D, with its distance DE from the base, describe the circle EFG : This circle will touch the triangle internally.
For let fall the perpendiculars DG and DFupon the sides AB and BC (I. 6.). The triangles ADE, ADG, having the an- IB, gle DAE equal to DAG, the right angle DEA equal to DGA, and the interjacent side AD common, are equal (I. 20.), and therefore the side DE is equal to DG. In the same manner, it is proved, from the equality of the triangles CDE, CDF, that DE is equal to DF; consequently DG is equal to DF, and the circle passes through the three points E, G, and F. But it also touches (III. 20.) the sides of the triangle in those points, for the angles DEA, DGA, and DFC are all of them right angles.
PROP. XI. PROB.
In a given circle, to inscribe a triangle equiangular to a given triangle.
Let GDH be a circle, in which it is required to inscribe