whence the ratio of AD to DE is given. Join EF, and produce ED to meet AG a perpendicular to AC. The triangles ADG and EDF are evidently similar, and therefore AD:AG::DE: EF, or alternately AD: DE ::

DB

M

N

ᅡ

AG: EF; and since the ratio AD to DE is given, the ratio of AG to EF is also given, and the radius EF 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 to these find (VI. 18. El.) a mean proportional O, on BC describe a semicircle, make O : M :: BF: AG, a perpendicular erected from A, and (III. 30. El.) draw the tangent GDE; the intersection D is the point required.

For, the triangles DAG, and DEF being similar, AD:AG::DE:EF, and alternately AD: DE::AG: EF, or M: 0; wherefore (V. 21. cor. 1. El.) AD2 : DE' :: M2 : 0', that is, (V. 23. El.) M: N ; but (III. 36. El.) DE2=CD× DB, and consequently AD': CD× DB:: M: N.

2. When D lies between the points B and C.

ANALYSIS.

On BC describe a semicircle, draw DF perpendicular to the diameter, and meeting the circumference in F, and join AF.

Because (III, 36.) BD × DC=DF, the ratio of AD' to DF* is given, and consequently that of AD to DF; but

fore the intersection For F', the perpendicular FD, or F'D', and the point D, or D', are all given.

COMPOSITION.

Let M:N express the given ratio, and to these find (V. 18. El.) a mean proportional O, make (VI. 3. El.) M to O as AC to the perpendicular CE, join AE meeting the circumference of a semicircle described on BC in the point F or F, and let fall the perpendicular FD or F'D'; then M: NAD: BDx DC, or AD: BD x D'C.

For the triangle ACE is evidently similar to ADF or ADF', and therefore AC: CE :: AD : DF, and AC2 : CE:: AD*: DF; but (V. 23. El.) M: N :: M: O3, or as AC: CE, and consequently AD: DF, that is, BD× DC :: M: N.

This problem evidently requires limitation; for, if AE should diverge too much from AC, it will not meet the circumference at all. Hence an extreme case will occur, when AE touches the circle. But the ratio of AC to CE, or of AD to DF, will then be the same as that of a tangent from A is to the radius HB; and consequently the limiting ratio is the duplicate of this,-or the ratio of M to N can. never approach nearer to the ratio of equality than that of AB × AC, or AH' — HB2, to HB'.

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

angle at Dis common to the triangles DGA and DFE, which are therefore similar, and hence AD: AG :: DE : EF, or alternately AD: DE

::AG: EF. And

since the ratio of

AD to DE is

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 MN 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 M2 : Oa:: AD2 : ED', that is, M:N :: AD: ED', or BD x 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.

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.

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=CEX ED.

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

ANALYSIS.

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

Because AEXEB

: DE x EC ::

M: N, it follows

that AE × EB: A

AEX EB-DEX

M

N

EC:: M:M-N; but AEX EB-DEXEC=(AEX EBAF× FB)+(DF × FC-DE× 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 AB-CD.

ANALYSIS.

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