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circle. It is to the quotient obtained by dividing the of the side by the altitude.

Note 3. In Cor. 3, if the given A be obtuse, and the perpendicular be let fall from one of the acutes, it will meet the opposite side produced. The extremity of the diameter being connected with the extremity of the subtending side adjacent to the perpendicular, the proof of the similarity of the right angled As will be the same as when the perpen. falls within the given A. But if it be connected with that extremity of the subtending side, remote from the perpendicular, then the external of the L given ▲ is to the contained by the diam. and connecting line (Prop. 22. 3.)

Note 4. If any angle of a ▲ be bisected by a right line produced to the periphery of the circumscribing circle, and a perpendicular be let fall from the vertex of this on the opposite side; then the rectangle under the whole bisecting line and the internal part of it will be to the' rectangle under the diameter and perpendicular. For, by Cor. 2. The rectangle under the whole produced bisecting line, and the internal part of it is to the rect. under the sides; and by Cor. 3, the rectangle under the diameter and perpendicular is to the same rect. under the sides; .. &c.

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Note 5. The right line bisecting an of a ▲, can never pass through the centre of the circumscribing circle, except when the triangle is isosceles. For when the bisecting line passes through the centre, the rectangle under the diameter and perpendicular is to the rect. under the diam. and internal part of it, .. the internal part is at right angles to the side opposite the bisected and the triangle is isosceles, and consequently the base is bisected; but if the line bisecting the base is at rights to it, it must pass through the centre; for the right line bisecting any chord of a circle at right angles passes through the centre.

Note 6. If from the point where the line bisecting angle of a triangle meets the circumference of the circumscribing circle, a diameter be drawn, it will be at right angles to the side subtending that angle. For the right line bisecting any of a triangle, inscribed in a circle, also bisects the opposite part of the circumference; and if a diameter be drawn from the point of bi

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section of any segment of a circle, it will be at rights to the chord of that segment, .. &c.

Note 7. The 2d and 3d Cors. are cases of the same theorem, scil. If from an angle of any triangle inscribed in a circle two right lines be drawn, one to the side subtending that angle, and the other cutting from the circle a segment which shall contain an angle equal to the angle contained by the first drawn line, and the side which it meets; the rectangle under the sides of the triangle is equal to the rectangle under these lines. In Cor. 2. these lines coincide.

Note 8. If in the A ABC (Fig. 21 of Elrington) ABX AC=BD× DC+AD2 the line AD will divide the angle BAC into two equal parts. If not draw a line AKF bisecting the angle, then ABX BC=BK × KC+ AK... AD×ÄE=AK+AF .. the quadrilateral EDKF would be inscribable in a circle, .. the sum of the angles FKE, FED, would be equal to two right angles; .. the angle AKC FED FEA.. the arc AC arc AB which only takes place when the ▲ is isosceles; ... unless in this case, the angle is bisected.

PROP. XVII. THEOR.

If three right lines be proportional, the rectangle under the extremes is equal to the square of the mean. And if the rectangle under the extremes be equal to the square of the mean, the three right lines are proportional.

Part 1. Assume a right line to the mean; then there are four right lines proportional, and therefore the rectangle under the extremes is to the rectangle under the means (16. 6), but the means are,.. the rectangle under them is to the square of the given mean, ... &c.

Part 2. Assume as before; then the rectangle under the assumed and given mean is = to a 2 of the given mean, and since the rectangle under the extremes is = to the rectangle under the means, the four lines are proportional (16. 6.); .. the three given lines are proportional, since the assumed mean is to the given mean.

Cor. 1. If three right lines be proportional, any pa

rallelogram under the extremes is equal to an equiangular parallelogram under the mean.

Cor. 2. Any given right line may be divided, so that the rectangle under the segments shall be equal to a given

square.

By dividing it, so that the side of the square shall be a mean proportional between the segments (Cor. 2. 13. 6). However it is evident that this cannot be done if that side is greater than half the given line.

Cor. 3. Hence it is evident that the rectangle, under any two lines, is a mean proportional between their

squares.

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For it is to the tween them, but the

of the mean proportional be's. of proportional right lines are themselves proportional, as will be shewn in Prop. 22.

PROP. XVIII. PROB.

On a given right line to construct a rectilineal figure similar to a given one, and similarly posited.

Resolve the given figure into As, by drawing right lines from any of it to all others except those two adjacent to the one from which the lines are drawn. On the given right line construct a A equiangular with that which stands on the base of the given figure; and construct a ▲ equiangular with the second ▲ of the given figure (standing upon another side of the first A) on an homologous side of the first constructed A, and so on till you have as many As constructed as the given figure had been resolved into.

Then the As of which the given fig. and constructed one consist, being respectively similar, it is evident that the figures themselves are equiangular, and that the sides about the - s of the figures, which are also s of the triangles that compose them, are proportional; the proportionality of the other sides about the s follows ex aquo, from the proportionality of the sides about the s composing them.

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PROP. XIX. THEOR.

Similar triangles are to one another in the duplicate ratio of their homologous sides.

Assume on the greater base from either extremity a third proportional to that base and the homologous side of the other A, and join the extremity of the assumed part with the vertex of the opposite

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Then the A standing on the assumed part is to the other given A, for they have an in each =, and the sides about the = s reciprocally proportional, (the side conterminous with the greater base being to its homologous side of the other given triangle as the greater base is to the lesser base), but the greater base the lesser base the lesser base the assumed part, .. the side conterminous with the greater base: its homologous side of the other given :: Tesser base: the assumed part. But the whole the ▲ on the assumed part, or to the other given triangle the greater base: the assumed part (1. 6.) and the greater base the assumed part in the duplicate ratio of the greater to lesser base (Def. 10. 6.) .. the given As are to one another in the duplicate ratio of those bases, which bases are homologous sides.

Note. We can then cut from a given ▲ a A similar to it, and that shall be any submultiple of it. By assuming from the vertex on one of the sides, a part that shall be the same submultiple of that side that the required A is of the given one, and by drawing through the extremity of this part a line parallel to the base.

PROP. XX. THEOR.

Similar polygons may be divided into similar triangles; equal in number and proportional to the polygons; and the polygons are to each other in the duplicate ratio of their homologons sides.

PART 1. Divide the polygons into As by lines drawn from the vertices of correspondings. The As are evidently in number; and those As in each polygon to which an entire of its respective polygon belongs, are

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similar, since those Ls are and the sides about them proportional (6. 6); the As adjacent to each of those will also be similar, having one in each = (viz. the excesses of 4s of the foregoing As above corresponding 4s of the polygons) and the sides about those s proportional, which is inferred ex æquali; therefore those As are similar; and so on, the As into which one polygon is divided are respectively similar to those of the other.

PART 2. One pair of those similar As are to one another in the duplicate ratio of their homologous sides (19. 6); also the pair of similar As adjacent to those are to one another in the duplicate ratio of the same sides, (they being common to those two pair of As), ... one pair of As are to one another as the adjacent pair, and so on; ... as one of the antecedents: one of the consequents :: the sum of the antecedents: the sum of the consequents, i. e. so is one polygon to the other.

PART 3. Since one of the polygons: the other :: one of the similar As: its corresponding one; and since those As are to one another in the duplicate ratio of their homologous sides, .. the polygons are to one another in the duplicate ratio of those sides.

Cor. 1. Hence if three right lines be proportional, any rectilineal figure on the first, is to a similar rectilineal figure on the second, as the first to the third.

Cor. 2. Hence a rectilineal figure may be described which will be to a given rectilineal figure in any given ratio.

Find a mean proportional between a side of the given figure and a right line which is to that side in the given ratio; and on this mean proportional let a figure be described similar to the given one and similarly posited.

Note. Since the two polygons are in the duplicate ratio of the first given right line to the second, it is necessary that these right lines should lie between = <s, and ..that the polygons should be similarly described on them. For on the given right line as many polygons of different magnitudes may be constructed, similar to the given polygon as there are sides of the given polygon of different lengths; for let a quadrilateral figure be given, whose sides are proportional to the numbers 1, 2, 4 and

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