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the two straight lines, so drawn, will be the point which was to be found.

105. COR. If from the point, thus found, any number of straight lines be drawn cutting the three given circles, the rectangles contained by the whole lines, so drawn, and the parts of them without the circles, shall (E. 36. 3. and S. 80. 3.) be equal to one another.


106. PROBLEM. To divide a given straight line into two parts, so that the square of the one shall be equal to the rectangle contained by the other and a given straight line.

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lines: It is required to divide AB into two parts, so that the square of the one shall be equal to the


rectangle contained by the other and by the given

line L.

From B draw (E. 11. 1.) BC i to AB; make BC=L, and (E. 31. 1.) complete the ABCD; produce (S. 73. 3. cor.) CB to E, so that CE X EB may be equal to ABXL; lastly upon BE describe (E. 46. 1.) the square EFGB: Then, AB is divided in G, so BG' = AG X L.

For produce FG to H; then (constr.) the rectangle CE × EB, = AB× L; but CF is the rectangle CE × EB, because EF = EB; and CA is the rectangle AB XL, because CB was made equal to L; .. the rectangle CF CA; take away the common part CG, and there remains BF=HA; and BF is the square of BG, and HA =AG X L, because (constr. and E. 24. 1.) AD= BC, which was made equal to L.


107. THEOREM. If a given circle be cut by any number of circles, which all pass through the same two given points without the given circle, the straight lines, joining the points of each of these intersections, are either all parallel, or all meet when produced in the same point.

Let CDF be a given circle; and, first, let the

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circle ACDB, which passes through the two given points A and B, cut the circle CDF in C and D; let the straight line joining C, D, be parallel to AB; then shall the straight line joining the points, in which any other circle that passes through A, B, cuts the circle CDF, be parallel to AB and CD.

For, find (E. 1. s.) the centre K of the circle CDF, and from K draw (E. 12. 1.) KEX 1 to CD; .. (E. 3. 3.) KX bisects CD at right ; .. (E. 1. 3. cor.) the centre of the circle ACDB is in KX, which (hyp. and E. 29. 1.) cuts AB at right, and .. bisects it; the centres, ..., of all the circles that pass through A and B are (S. 3. 1. cor. 3.) in KX; .. (S. 1. 3.) KX cuts all the straight lines, which join the intersections of these circles, with the given circle CDF, at right ; .. (E.28. 1.) the

straight lines joining the several pairs of intersections are parallel to one another and to AB.

But, secondly, let the circle GLMH, which passes through the two given points G, H, cut the given circle CDF in L and M; and let the straight line joining L and M be not parallel to AB; produce,..., LM to meet GH produced in N; and let any other circle GIFH, passing through G and H, cut the circle CDF in I and F; then are the points I, F and N in the same straight line.

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For join N, F, and if NF, produced, do not pass through I, let it, if it be possible, pass otherwise, as NFPQ: Then (E. 36. 3.

cor.) PN X NF = GN × NH;

LNX NM, and LN x NM

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also QN × NF=


× NF GN X NH; also PN X NF=GN X NH; ... QN × NF = PN X NF; .. QN = PN; that is the less is equal to the greater, which is impossible; .. NF, when produced, cannot pass otherwise than through the point I, so that the three points I, F and N are in the same straight line.


108. THEOREM. If a perpendicular be let fall from the right angle, of a right-angled triangle, on the hypotenuse, the rectangle contained by the. hypotenuse and either of the segments, into which

it is divided by the perpendicular, is equal to the square of the side adjacent to that segment.

Let the BAC, of the AABC, be a right angle,

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and from A let AD be drawn to the hypotenuse BC: Then CBX BD=AB, and BC X CD - AC.

For if upon AC, as a diameter, a circle be described, it will pass (S. 29. 1. cor. 2.) through the point D, because (hyp.) the ADC is a right ≤ ; and (E. 16. 3. cor.) it will touch AB in A, because the CAB is a right '; .. (E. 36. 3.) CBX BD =AB2.

And, in the same manner, it may be shewn that BC XCD = AC.


109. THEOREM. To draw a tangent to a circle, such, that the part of it intercepted between two straight lines, given in position, but of indefinite length, shall be equal to a given finite straight line:

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