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the point A to the centre C, and make AC the diam-
fore similar, and give the proportion,
Two chords which intersect, divide each other into parts reciprocally proportional. 126. If these chords are perpendicular to each other, and one of them (as AB fig. 68) is a diameter, it will Fig. 6S. bisect the other in E (109); and therefore in the proportion, instead of ED, we may use its equal EC. This will gi AE : that is, EC i ional will give EC T TE * at 1S, is a mean proportiona between AE and EB. Suppose we draw also the chord BC; since AB is a diameter, the angle ACB inscribed in a semicircle, is a right angle ; and CE is a perpendicular drawn from the vertex of the right-angle of the right-angled triangle ABC, to the hypothen use, dividing the triangle into two smaller triangles. If we examine the triangles into which the triangle ACB is divided, we shall perceive that the triangle ACE has a right angle and the angle A in common with the larger triangle ; the triangle BCE has a right angle, and the angle EBC in common with the larger triangle. Each of these partial triangles is therefore equiangular with the larger ; they are consequently similar to it and similar to each other. If we compare each of them with the l have th ‘oporti AE AC e larger we have the proportions AG = AE # - . that is, AC is a mean proportional between
AB and AE ; and CB is a mean proportional between
If, therefore, we consider these lines AC, AB, CB, and CE, in their connexion with the circle, we should enunciate in this manner the truths just proved ; (1). A perpendicular drawn from any point in the circumference of a circle to the diameter, is a mean proportional between the two segments of the diameter; (2). If, from the vertex of an angle inscribed in a semicircle a perpendicular be drawn to the diameter, each of the chords which form this angle, is a mean proportional between the whole diameter and the segment adjacent to that chord. --
The above result affords us the means of giving the solution which we promised (101) of the problem–To jind a line which shall be a mean proportional between two given lines.
[The learner will perceive that there are here two methods of solving this problem ; he can give examples of each. 127. If, however, we regard the triangle ACB, independently of its connexion with the circle, these truths will be enunciated thus: If from the vertex of the right-angle in a right-angled triangle, a perpendicular be drawn to the hypothenuse,_ (1). This perpendicular will be a mean proportional between the two segments of the hypothen use; and will divide the triangle into two triangles similar to each other and similar to the whole. (2). Each of the sides will be a mean proportional between the entire hypothenuse and the segment adjacent to that side. [These propositions were differently obtained in articles 99 and 100.] 12S. If these lines instead of meeting within the circumference, were to meet beyond the circumference, as in figure 69 ; by drawing the chords CA and BD, we have also, two equiangular triangles AEC and DEB ; the angle E is common to the two triangles, and the two angles at A and D, are measured by half the same arc ; consequently they are equal, and the other two angles are equal, and the triangles similar ; they will give us EA ED * EC T EB meet beyond the circumference, the entire seconts will be in the inverse ratio of the parts without the circle. 129. As the truth of this last proposition does not depend upon any particular position of the secant, we may suppose the secant DE (fig. 70) to revolve about the point E till the interior part CD vanishes and the sescant becomes a tangent ; the proposition will hold true to the vanishing point of the secant ; but at that point the exterior part becomes equal to the whole ; and we
shall have EF T EB '
By drawing the two chords AF and BF, the two similar triangles AEF and BEF, will give the same proportion. So that we are sure of the accuracy of this result.
this proportion ; that is—When two secants
that is—If, from the same point
Why are the triangles AEF and BEF similar !
130. If we make the tangent AB equal to the diameter of the circle (fig. 71), draw the secant through the centre, and take upon the tangent the part AF equal to the exterior part of the secant, the last article will give the proportion
AE AB . (81 AE — AB AB – AD AB = AD 3 (S1) →E→ = −AD−. But AE– AB is AD, and AB — AD is BF ; the protion therefore be AF BF and bw inver portion ineretor comes AB = AF : y sion (7S) we have #. F #. that is, the line AB is di
vided at F, so that the part AF is a mean proportional
Fig. 72. Fig. 73.
Of Polygons inscribed in a Circle, and circumscribed about a Circle.
132. IN figure 68, we have the right-angled triangle ACB, whose vertices are in the circumference of a circle of which the diameter forms the hypothenuse of the triangle ; and as the vertices of any triangle are three points not in the same straight line (110), it is manifest that a circumference may be described passing through these vertices. Under such circumstances the triangle is said to be inscribed in the circle.
133. If we examine the inscribed triangle ABC (fig. 73), we shall perceive, that as each angle is an inscribed angle, it will have for its measure half the opposite arc, of which the side opposite to this angle is the chord ; and as the greater chord subtends the greater arc (107), we have an additional proof that the greater angle in a triangle is opposite to the greater side (47). And as equal chords have equal arcs, and equal arcs subtend equal angles (106), an equilateral triangle must have its angles equal (46); and the equal sides in an isosceles triangle, must be opposite to equal angles (45).
We have also a new proof that the sum of the angles in every triangle, is equal to two right angles (35); for as each of the angles is measured by half of the opposite arc (116), the sum of three angles will have for their measure half the sum of the three arcs, or half the circumference ; which is the measure of two right angles 27).
#1. To inscribe a circle in a triangle ; bisect two of the angles, as A and B (fig. 74); from the point of meeting of the two bisecting lines, draw perpendiculars OD, OE, OF, to the sides of the triangle; the point O will be the centre, and these perpendiculars will be the radii of the circle. The two triangles AOE, AOF, have each a right-angle, and their angles at A equal; their other angles will therefore be equal ; and having the common side AO, they will be equal by the third case of equal triangles; OE and OF are therefore equal. In the same way it may be shown that OE and OD are equal. The circle, therefore, which has O for its centre and OE for its radius, will have each side of the triangle ABC tangent to its circumference. The circle is said