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4. A circle is described about a rectili
neal figure or circumscribes it, when the cir
cumference passes through all the angular points of the figure.
5. Polygons are equilateral, when their sides, in the same order, are respectively equal: They are equiangular, if an equality obtains between their corresponding angles.
6. Polygons are said to be regular, when all their sides and their angles are equal.
PROP, I. PROB.
Given an isosceles triangle, to construct another on the same base, but with only half the vertical angle.
Let ABC be an isosceles triangle standing on AC; it is required, on the same base, to construct another isosceles triangle, that shall have its vertical angle equal to half of the angle ABC.
Bisect AC in D (I. 7.), join DB, which produce till BE be equal to BA or BC, and join AE, CE: AEC is the isosceles triangle required.
For, the straight line BE being equal to BA and BC, the point B is the centre of a circle which passes through the points A, E, and C ; and consequently the angle ABC is the double of AEC at the circumference (III. 15.), or the vertical angle AEC is half of ABC. But the triangles AED and CED, having the side DA equal to DC, the side DE common to both, and the right angle ADE (III. 4.) equal to CDE are (I. 3.) equal, and consequently AE is equal to CE. Wherefore the triangle AEC is likewise isosceles.
PROP. II. PROB. Given an acute-angled isosceles triangle, to construct another on the same base, which shall have
double the vertical angle.
Let ABC be an acute-angled isos- . P celes triangle; it is required, on the base AC, to construct another isosceles triangle, having its vertical angle /~\ double of the angle ABC. A.
Describe a circle through the three points A, B, and C (III. 9. cor.), and draw AD, CD to the centre D; the triangle ADC is the isosceles triangle
required. For the angle ADC, being at the centre of the
circle, is (III. 15.) double of ABC, the angle at the circumference.
PROP. III. THEOR.
If an isosceles triangle have each angle at the base double of the vertical angle, its base will be equal to the greater segment of one of its sides divided by a medial section.
Let ABC be an isosceles triangle which has each of the angles BAC, BCA double of the vertical angle ABC; the base AC is equal to the greater segment of the side BA formed by a medial section.
For draw CD to bisect the angle BCA (I. 5.), and
about the triangle BDC describe a circle (III. 9. cor.). Because the angle BCA is double of ABC and has been bisected by P CD, the angles ACD, BCD are each of them equal to CBD, and
ID consequently the side BD is equal to CD (I.1.1.). But the triangles BAC and DAC, having the angle * C
ACD equal to ABC, and the angle at A common to both, must have also (I. 30.) the remaining angle CDA equal to BCA or CAD ; whence (I. 11.) the triangle DAC is likewise isosceles, and the side AC equal to CD ; but CD being equal to BD, therefore AC is also equal to it. And since the angle ACD is equal to CBD in the alternate segment of the circle, the straight line AC touches the circumference at C (III. 21. cor.); wherefore the rectangle contained by AB and AD (III. 26. cor. 2.) is equivalent to the square of AC, or the square of BD. Consequently the base AC of this isosceles triangle is equal to the greater segment BD of the side AB cut by a medial section.
Cor. Hence the interior triangle ACD is likewise isosceles and of the same nature with ABC, having the greater segment of AB for its side, and the smaller segment for its base.
PROP. Iv. PROB.
Given either one of the sides, or the base, to construct an isosceles triangle, so that each of the angles at the base may be double of its vertical angle.
First, let one of the sides AB be given, to construct such an isosceles triangle.
Divide AB by a medial section at C (II. 19.), and on CB, as a base with the distance AB for each of the sides, describe an isosceles triangle (I.1.)
Next, let the base AB be given, A. g. B to construct an isosceles triangle of K-B-E, this nature. . Produce AB to C, such that the rectangle AC, CB be equal to the square of AB (II. 19. cor. 2.), and on the base AB, with the distance AC for each of the sides, describe an isosceles triangle. These isosceles triangles will fulfil the conditions required. For it is evident, from the last Proposition, that isosceles triangles constituted on CB or AB, with each of the angles at the base double the vertical angle, would , have AB or AC for their sides, and consequently (I. 2.) must coincide with the triangles now described. Cor. Hence of such an isosceles triangle the vertical angle is equal to the fifth part of two right angles; for each of the angles at the base being double of the vertical angle, they are both equal to four times it, and consequently this vertical angle is the fifth part of all the angles of the triangle, or of two right angles.
PROP. V. PROB.
On a given finite straight line, to describe a regular pentagon.
Let AB be the straight line, on which it is required to describe a regular pentagon.
On AB erect (IV. 4.) the isosceles triangle ACB, having each of the angles at its base double of its vertical angle, from the centre A with the distance AB describe an