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The parallels FG and AI are cut proportionally by the diverging lines CA, CH, and CI (VI. 1. El.); but FO is equal to OG, and consequently AH is equal to HI. Wherefore (II. 4. El.) HK is parallel to IB, and the angle AKH is equal to ABI (I. 25. El.); and since the angle ABI or ABC is equal to ALC (III. 20. El.), the angle AKH is equal to ALC or ALH, and hence (III. 20. cor. El.) the quadrilateral figure AHKL is contained in a circle. Consequently (III. 20. El.) the angle HAK is equal to HLK; but HAK is e

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circle. Wherefore (III. 20. El.) the angle MLO is equal to MKO; but MKO is a right angle, and consequently MLO is likewise a right angle, and thence (III. 28. El.) ML is a tangent. But the point M, being the concourse of ED and BA, is given, and, therefore, the tangent ML to the given circle is given (III. 30. El.); whence the diameter LC, and the point C, are given.


Produce ED and BA to meet in M, draw the tangent ML and the diameter LC; the straight lines AC and BC will cut off from the centre equal portions, OF and OG, of the given diameter ED.

For draw AI parallel to DE, and OK perpendicular to AB, and join LK and KH.

Because ML is a tangent, MLO is a right angle, and, therefore, equal to MKO; consequently (III. 20. El.) MKL

is equal to MOL, that is, (I. 25. El.) to AHL. Wherefore the quadrilateral figure AHKL is contained in a circle, and hence (III. 20. El.) the angle ALH is equal to AKH; but, for the same reason, ALH or ALC is equal to ABC or ABI, and consequently AKH is equal to ABI, and (I. 25. El.) KH parallel to BI. Now since AK is equal to KB, it follows that AH is equal to HI, and hence that FO is equal to OG.


Through a given point to draw a straight line, so that the rectangle under its segments, intercepted by two straight lines given in position, shall be equal to a given space.

Let AB, AC be two straight lines, and D a point, through which it is required to draw EF, such that the rectangle under its segments ED, DF shall be equal to a given space.

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and consequently (V. 6. El.) AD.DG ED.DF. But the rectangle ED, DF is given, and therefore also the rectangle AD, DG; and since AD is given in position and magnitude, DG and the point G are given. Again, the angle DFG, being equal to DAC, is given, and thence (III. 31. El.) the segment of the circle which contains it; wherefore the contact or intersection of that arc with the straight line AB is given, and consequently the position of EF or E'F' is likewise given.





Join AD, make the rectangle AD, DG equal to the given space, and on DG describe (III. 31. El.) an arc containing an angle equal to DAC, and meeting AB in F or F; EDF or EDF' is the straight line required.

For the triangles ADE and FDG are similar, and conse quently (VI. 13. El.) AD: ED :: DF: DG; whence (V. 6. El.) ED.DF ≈ AD.DG; but the rectangle AD.DG is equal to the given space, and therefore the rectangle ED.DF is also equal to that space.


Two straight lines being given, to draw, through a given point, another straight line, cutting off segments which are together equal to a given straight line.

Let AB, AC be two straight lines, and D a given point, through which it is required to draw a straight line EF, so as to cut off the segments AE and AF, that are together equal to ON.

The point D may lie either within or without the angle formed by the straight lines AB and AC.

1. Let D have an internal position.


Draw DG and DH (I. 26. El.) parallel to AB and AC. Because the point D is given, and AB,AC are given in po

sition, the parallelo

gram AGDH is giyen. And since the triangles EDG and DFH are evidently similar, EG: GD :: DH: HF, and therefore EG.HF= GD.DH. But AG and AH, or DH and GD, being given, the rectangle GD, DH is given, and, therefore, EG.HF is given. Make FK

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EG, and the rectangle HF, FK is hence given; but HK, being the excess of AF and AE above GD and DH, is given, and consequently its point of section F or F', and the straight line EDF or E'DF, are given.


Draw the parallels DG and DH. From ON, the sum of the two segments AE and AF, cut off OPAG + AH, and make HK PN. On HK describe a semicircle,


from the extremities of the diameter erect the perpendiculars HI and KL equal to AH and AG, join IL, and at right angles to this, and from the point or points where it meets the circumference, draw MF or MF; EDF or E'DF is the straight line required.


For (VI. 20. El.) HI.KL = HF.FK, and consequently AH.AG HF.FK. But, from the similar triangles EGD and DHF, EG: GD, or AH:: DH, or AG: HF, and therefore (V. 6. El.) AH.AG = HF.EG; whence HF.FK = HF.EG, and FK EG. And since AG + AH = OP, and HF+EG HK PN, it follows that AG + EG +AH + HF, or AE + AF


2. Let the point D have an external position with respect to the straight lines AB and AC.


Draw DG parallel to AB, and DH parallel to AC and meeting AB produced. The triangles EDG and DHF being similar, EG: DG:: DH: HF, and (V. 6. El.) EG.HF =DG.DH; but DG and DH are both given, and hence the rectangle under EG and HF is given. Make FK

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