3. When the point D lies beyond B and C.

ANALYSIS.

On BC describe a semicircle, draw the tangent DE, and produce it to meet the perpendicular AG, and join E with the centre F.

Because (III. 36. El.) BD x DC= DE', the ratio of AD to DE' is given, and consequently that of AD to DE. But the angle DEF, being (III. 28. El.) a right angle, is equal to DAG, and the

given, that of AG to EF is also given, and EF, the half of BC, being given, AG and the point G are thence given. Wherefore the tangent GE and its intersection D with AC, are given.

COMPOSITION.

Let M N be the given ratio, and find the mean proportional O; make O: M:: BF: AG, a perpendicular to AC, and draw (III. 30. El.) the tangent GED; then M: N :: AD2: BD x DC.

For join EF. Because the triangles ADG and EDF are similar, AG: AD :: EF: ED, and alternately AG: EF:: AD: ED; but AG: EF:: M: O, and therefore M: 0 :: AD: ED, and M*: O':: AD: ED', that is, MN:: AD': ED', or BDX DC.

PROP. XVII. PROB.

In the same straight line, four points being given, to find a fifth, such that the rectangle under its distances from the first and second points, shall have a given ratio to the rectangle under its distances from the third and fourth.

Let it be required to find a point E, so that AEXEB: DEXEC:: M: N.

1. Let M: N be a ratio of equality.

ANALYSIS.

Because AEEB DE XEC, it is manifest that AE: CE

AC+ BD: BD :: BC: EB; but the ratio of AC + BD to BD is given, whence that of BC to EB, and, therefore, BE and the point E are given.

COMPOSITION.

Make AC+BD: BD:: BC: EB, and E is the point required. For (V. 10. El.) AC : BD :: CE: EB and (V. 19. cor. 1. El.) AE: ED :: CE : EB, and hence (V. 6. El) AEXEB CEXED.

2. Let M:N be a ratio of majority or minority.

ANALYSIS.

Find, by the preceding construction, a point F, such that AFX FB = DFXFC.

EC:: M:M-N; but AEX EB-DEXEC=(AEXEBAFX FB)+(DFX FC-DEX EC), that is, EF (AF+BE) + EF (DF + CE), or = EF (AD + BC.) Wherefore AEX EB: EF (AD+BC):: M:M-N; consequently the point E is assigned by prop. 14 of this Book.

The composition of the problem is thence easily derived, by retracing the steps.

PROP. XVIII. PROB.

In the same straight line, four points being given, to find a fifth, such that the rectangle under its distances from the extreme points shall have a given ratio to the rectangle under its distances from the mean points.

Let it be required to find a point E, so that AE × ED: BEXEC:: M: N.

1. Let ABCD.

ANALYSIS.

Because AE × ED = (AB + BE) (AB + EC), it is evident that AE × ED= AB × AC+ BEXEC, whence

AD, being thus given, the point E is assigned by VI. 20 of the Elements/

COMPOSITION.

Make M-N: M :: AB : P, and (Vİ. 20. El.) cut AD in E or E', such that AE x ED= P x AC; E is the point required. For (V. 7. El.) M: M-N :: P: AB, and hence(V. 24. cor. 2. El.) M: M-N:: PXAC, or AEX ED : AB× AC; consequently M : N :: AE ×ED: AE× ED— BA X AC, or BEXEC.

2. Let AB and CD be unequal.

ANALYSIS.

In AD produced, let the point F be such that AF x FD BF x FC.

=

=

It is evident that BD x DC + FD (CD + BF) BF X FC = AF x FD; whence BD x DC = FD (AF-BF - CD) = FD (AB-CD). But BD × DC is given, and therefore the rectangle FD (AB — CD) is given; and since

ABCD is given, FD and

MH

NH

(BE + AB) (EC + CD) = BE x EC + BE × CD + ABXED=BEXEC+BD×CD-ED × CD + AB × ED;

but BD x CD=FD(AB—CD), and therefore AE × ED = BE x EC + (FD + ED) (ABCD), or AE x ED = BE × EC + EF (AB — CD). Now, since the ratio of AE x ED to BE x EC is given, the ratio of AE x EC to EF (ABCD) is also given; wherefore AB - CD being given, and the points A, C, and F, the point E is given by prop. 14.

Applying that proposition, the construction of the problem is easily obtained.

It yet remains to assign the limitations of this problem.

On AD describe a circle, erect the perpendiculars BI and GCH, join IOH, and, parallel to this, draw KEL through the point of section E, join OG, EG, and IE which produce to the circumference, and join MG and ML.

The point O is evidently given. But the ratio of AE× ED to BE x EC may be considered as compounded of the ratio of AEX ED, or (III. 36 El.) IEXEM, to KE× EL, and of the ratio of KE x EL to BE x EC.

Now, since BK and CL are parallel, KE: EL::BE: EC, or alternately KE: BE:: EL: EC, and therefore (V, 21. El.) KE2: BE':: KE × EL: BE x EC. Again, KE and IO being parallel, KE: 10 :: BE: BO, or alternately KE: BE:: IO: BO, and hence KE: BE:: IO: BO2 Wherefore IO2: BO2 :: KE x EL: BE x EC, and consequently, the ratio of these rectangles is given; let it be that of PQ to ST.

The angle MGL, being equal (III. 20. El.) to MIH in the same segment, is equal (I. 25. El.) to the exterior angle MEL, and consequently (III. 20. cor. El.) the quadrilateral figure MGEL being thus contained in a circle, the angle LME is (III. 20. El.) equal to LGE. Draw LN making the angle MLN equal to EGO, and (I. S4. El.) the exterior angle LNE will be equal to CGO. But the triangles GOC and HOC are obviously equal,