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.

PROP. XVI. PROB.

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.

D

GE

and consequently (V. 6. El,) AD.DG ED.DF. But the rectangle ED, DF is given, and therefore also the rect, angle 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.

COMPOSITION.

B

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 consequently (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.

PROP. XVII. PROB.

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.

ANALYSIS.

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 parallelogram AGDH is given. 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

= 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. '

COMPOSITION.

Draw the parallels DG and DH. From ON, the sum of the two segments AE and AF, cut off OP AG + 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 M'F'; EDF or E'DF is the straight line required.

For (VI. 20. El.) HI.KL = HF.FK, and consequently AH.AGHF.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 And since AG + AH = OP, and HF EG HK PN, it follows that AG + EG +AH+HF, or AE + AF = ON.

EG.

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

ANALYSIS.

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

EG, and therefore HK = HF EG DG + AF — (DH — AE) = AF + AE (DH-DG); whence HK and the rectangle HF,FK are given, and consequently (VI. 20. El.) the point F is given.

If DFE' intersect the straight lines AB and AC on the other side of their vertex A, the triangles E'DG and DF'H are still similar, and E'G: DG :: DH: HF'; wherefore EG.HF', being equal to DG.DH, is given. Make F'K' EG, and thence HK' E'G HF AE' + DH AF + AE+ (DH- DG); consequently HK' and the rectangle HF.F'K' are given, and therefore (VI. 20. El.) the point F is given.

(DGAF)

COMPOSITION.

Make OP or OP' equal to the difference of the paral lels DH and DG, from H place likewise towards opposite parts HKPN and HK' = P'N, on HK and HK' describe semicircles, from H erect the perpendicular HI equal to DG, and, from K and K', the perpendiculars KL and K'L', each equal to DH, join 1L and IL', and, at right angles to these, from the points of section M and M', draw MF and M'F'; the straight lines DEF and DFE' will cut off segments from AB and AC, which are together equal to ON.

For (VI. 20. E!.) HF.FK = HI.KM = DG.DH; but DG.DH = HF.EG, and consequently HF.EG = HF.FK, or EG FK. Wherefore HK HF EG = AF + AE (DH DG); and since HK = PN = ON (DHDG), it follows that AF + AE =

In like manner, it is shown that E'G

ON.

F'K', and hence

HK' E'G HF = AF + AE+ (DH — DG); but = — HKPN ON + (DH-DG), and consequently AF + AE = ON.