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through the points P, F, I, shall also pass through R since the opposite angles of the figure RPFI may be proved together to two right angles: then because REPT is a quadrilateral figure inscribed in a circle, the angles RTI,REP are, but RTI is to RLI ... REF is to RLI.. the angles REF,RLF are together to two right angles, .. the circle passing through E,F,L shall also pass through R, ... &c.

PROP. 98. THEOR.

If from any point [B] without a circle there be drawn two right lines [BI BC] one of which [BC]_cuts the circle and the other [BI] is at right angles to the diameter, the square of the perpendicular is equal to the rectangle under the whole secant and its external segment [BD] together with the rectangle contained by the segments of the diameter [LI IK) made by the perpendicular.

Fig. 41.

T.

For let the secant BFH be drawn through the centre

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Then the of BT is to the 2rs of BI IT, also BT2 is to HB X BF + TF, but TF2 is to the rectangle LI IK + IT2 ··· TB2, is to HB × BF+ LIX IK+IT2, also BT is to BI+IT, if therefore IT be taken away, from both the rectangle HB,BF or CB, BD+ the rectangle LI IK is to IB2.

PROP. 99. PROB.

Through a given point (A) to draw a right line [AC] to subtend a given angle [ABC] so that drawing through C a line parallel to a right line given in position; the difference between AC and the said line CD terminated by AB, in D, may be given.

Fig. 42.

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Through the given angular point B draw BF parallel to the line given in position and to the given difference. Also through F draw BQ parallel to AB meeting CB produced in P, join PA, to which apply BR = BF parallel to which draw AC, it is the line required.

For, draw CDQ parallel to BF, then because BR BE; CA being paralled to BR must be to CQ, which is parallel to BF and DQ — CQ — CD ; .. CA — CD = BF the given difference (by construction).

PROP. 100. PROB.

From two given points [P and E]the one P in the chord AC of a given circle ABCF and the other E in the same chord AC produced until CE equals CP, to draw two right lines to the same point B in the periphery and cutting it in F and D so that the rectangles under BP PF and BD DE may have a given difference.

Fig. 94.

At the point C draw CG at right angles and to CE or CP, draw GE and on it describe a semicircle, in which from G apply GH = the side of a 2 that is half the given difference of the rectangles, then to the circle ABF apply ED EH produce it to B which is the point required.

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For draw BPF and EH then the rectangle AEC = (the rectangle BED = BD X DE + ED2 = BD x DE EH2 add the common square GH and AEX EC + GH2 = BD × DE † EG2 = BD × DE + EC)=BD × DE + PEX EC; take away the common rectangle PEC then AP X PC or BP x PF + FH2 the rectangle BDE.

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From a given point P in one side CD, of a given rectangle ABCD produced; to draw a right line PS cutting the side AC in V, AB in R and DB produced in S so that the internal triangle ARV may be equal to the sum of the two external ones PCV and BRS.

Fig. 96.

From the given point P draw PH parallel to AC and HS to AB making the parallelogram HPDS to twice the parallelogram ABDC,draw PS,and the thing is done.

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For the triangles HSP, DPS being each of them the given rectangle ABDC, therefore by taking away DBRVC common, there will remain the triangle RAV SBR + VCP.

The construction is the same for any parallelogram.

PROP. 102. PROB.

To find a point P in an indefinite right line PQR touching a given circle in Q, at which erecting PD at right angles to PR cutting the circle in E and D, bisecting ED in I, and drawing IF parallel to PE meeting the circle in F, so that drawing DF PF, the triangle DPF so formed may be a maximum. Fig. 45.

Draw a tangent RCA at right angles to RP, then from R through the centre of the given circle draw RD, let fall DP at right angles to RQ then P is the point required.

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For, draw a tangent ADB and complete the ☐2 CP which is the greatest rectangle that can be inscribed in the right angled triangle ARB (prop. on b. 1.) and .. much greater than any other rectangle, having its angular point in the periphery of the given circle, because any rectangle so formed, is less than its similar one inscribed in the triangle, whence it is evident that the triangle PFD being half the square PC is the greatest possible.

PROB. 103. PROB.

To determine the locus of the point P so that if from. a given point E, EP be drawn and EL perpendicular to it to meet the right line MR given in position the angle ELP may be always equal to a given acute angle.

Fig. 98.

Draw EQ at right angles and ET parallel to MR and make QEN to the given angle (N being in MR) through N draw a right line parallel to EQ, and it shall be the required locus.

In NT assume any point P, and draw PL according to the proposition; on PL as diameter, describe a circle which must pass through E and N, because the angles at those points are right, and the angle QENENP; but the angles ENP, ELP are = (prop. 23. b. 3. Elr.) .. ENPELP QEN the given angle, (by construction) and the point P is always in the same right line TN given by position.

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THE

Вентил

ELEMENTS

OF

PLANE GEOMETRY,

COMPRISING

THE DEFINITIONS OF THE FIFTH BOOK, AND THE SIXTH BOOK IN GENERAL TERMS.

WITH NOTES AND OBSERVATIONS.

ALSO,

A COLLECTION OF

THEOREMS, LOCI, PORISMS, AND PROBLEMS.

PART III.

BY JAMES LUBY.

DUBLIN:

PRINTED BY R. GRAISBERRY,

FOR HODGES AND M'ARTHUR, 21, COLLEGE-GREEN.

1824,

1831. e. 18 (2)

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