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1st,. When the indefinite straight lines both pass through the centre of the circle.

2dly, When they are parallel to one another.

3dly, When they are not parallel, but are equidistant from the centre.

Let AB be a given circle, and CD a given

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straight line; and first let KQ and KR be two given straight lines, of indefinite length, passing through the centre K of the circle: It is required to draw a straight line, touching the circle AB, so that the part of it intercepted between KQ and KR, shall be equal to CD.

Upon CD describe (S. 61. 3.) a ▲ CED, having its vertical CED equal to the given ▲ QKR, and its altitude EH = KB, the semi-diameter of the given circle; from KQ cut off KF = EC; and from F draw (E. 17. 3.) the tangent FBG to the given circle: Then, the tangent FG = CD.

For let FG touch the circle in B, and join K, B: And since (E. 18. 3.) the KBF is a right, as

is also (constr.) the

EC KF, and EH

EHC, and that (constr.)

KB, ... (S. 73. 1.) the ECH = 2 KFB; but (constr.) the CED L FKG; and the side EC of the A ECD, is equal to the side KF, of the A KFG; .. (E. 26. 1.) FGCD.

Secondly, let AB be the given circle, and PQ,

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RS, two indefinite but parallel straight lines: It is required to draw a tangent to the circle AB, such that the part of it intercepted between PQ and RS shall be equal to the given straight line CD...

Take any point E in either of the two parallel straight lines, as PQ, and from the centre E, at a distance equal to CD, describe a circle cutting RS

in F; join E, F ; .. EF = CD; lastly, draw (S. 8. 3.) the straight line GH, touching the circle AB, and parallel to EF; since, ..., EGHF is a □, GH (E. 34. 1.) = EF; and EF was made equal to CD; .. the tangent GH=CD.

Thirdly, let the two indefinite straight lines PQ,

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PR, which meet in P, be equi-distant from the centre K, of the circle AB: It is required to draw a straight line touching the circle AB, so that the part of it intercepted between PQ and PR shall be equal to the given straight line CD.

Join P, K; upon CD describe (E. 33. 3.) a segment of a circle CED,

equal to the given circle CEDG; from D

capable of containing an QPR, and complete the

draw (E. 11. 1.) DHL to

CD, and make DH KB the semi-diameter of the given circle AB; through H draw (E. 31. 1.)

HIF parallel to DC; bisect (E. 30. 3.) FGI in G;

from G draw (S. 75. 3.) GLE, so that LE=KP; join E, C and E, D; from PQ cut off PM = EC; and from M draw (E. 17. 3.) MBN touching the circle in B: The tangent MN CD.

L

For join C, L, and D, L, and K, M, and K,N, and K, B; and draw (E. 12. 1.) LT1 to CD; ... LTDH is a□, and (E. 34. 1.) LT=HD; and HD (constr.) =KB; .'.LT=KB: Again, because (constr. and S. 35. 3.) CG=DG, .:. (E. 27. 3.) the CEG ≤ = DEG;... the CEL is the half of the ▲ CED; and because (hyp.) PQ and PR are equi-distant from the centre K of the circle AB,.. the 2 QPK, or MPK, is the half of the QPR, which (constr.) is equal to the 4 CED; .. the ▲ MPK = ≤ CEL, and the two sides MP, PK, of the ▲ PKM, are equal (constr.) to the two sides CE, EL of the A ELC, each to each; .. (E. 4. 1.) KM=LC, and the ▲ PMK, = ▲ ECL; and because in the two right-angled KBM, LTC, KM=LC, and KB= LT, .. (S. 74. 1.) the 2 KMB LCT; and it has been shewn that the PMK= 2 ECL; .. the whole PMN is equal to the whole ECD; also (constr.) the MPN=CED, and the side PM, of the APMN, is equal to the side EC, of the A ECD; .. (E. 26. 1.) MN=CD.

PROP. LXXXV.

110. THEOREM. If from the intersection of any two tangents to a circle, any straight line be drawn, cutting the chord which joins the two points of contact and again meeting the circumference, it shall be divided by the circumference and the chord into three segments, such, that the rectangle contained by the whole line and the middle part, shall be equal to the rectangle contained by the extreme parts.

From the intersection A of two straight lines

B

R

AB and AC which touch the circle BCR in the points B and C, let there be drawn any straight line APR, cutting the circumference of the circle in P and R, and BC in Q: Then AR × PQ=

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