to the distance of the given point from the vertex, and BC is equal to the given straight line. Wherefore all the points and lines retain, by this construction, their relative position.

PROP. XXVI. PROB.

Between the side of a given rhombus, and its adjacent side produced, to insert a straight line of a given length, and directed to the opposite cor

ner.

Let ABCD be a rhombus, of which the side BC is produced; it is required, from the opposite corner A, to draw AEF, such that the exterior portion EF shall be equal to a given straight line.

ANALYSIS.

Join AC, and, meeting this produced, draw EG, making

the angle AEG equal to ACF. The triangles CAF and EAG are evidently similar, and AC:CF::AE:EG; but CE being parallel to AB, BC: CF:: AE:EF(VI. 1. El.) ; whence (V. 17. El.) AC: BC:: EF: EG. But AC, BC, and EF being given, EG is (VI. 3. El.) also given. Again, the angle ACD is (1.29 cor. El.) equal to ACB and therefore to FCG; conse

B

KE
L

G

quently adding ECF to each, the whole angle ACF, or AEG, is equal to ECG. Hence the triangles AGE and EGC are similar, and AG: EG:: EG: GC, or AG. GC=EG'. Wherefore the rectangle AG, GC is given, and consequently (VI. 20. El.) the point G, and thence the point E and the straight line AF.

COMPOSITION.

Let the intercepted segment be equal to K, join AC, make AC: BC:: K: L, divide AC in G (VI. 20. El.) so that AG. GC=L', and from G, with the radius L, describe a circle cutting CD in E; AEF is the straight line required.

For since AG. GC = L' = EG, AG: EG :: EG: GC, and therefore the triangles AGE and EGC are similar, and the angle AEG is equal to ECG, or ACF; whence the triangles AFC and AGE are likewise similar, and AC:CF:: AE: EG; but (VI. 1. El.) BC: CF:: AE: EF, and consequently (V. 17. El.) AC: BC:: EF: EG. Now AC: BC:: K: L, or EG; wherefore EF = K.

Otherwise thus.

ANALYSIS.

Draw FG making the angle AFG equal to ADC, cut CE, let fall the perpendicular CM, make MN

off CH

= MA, and join CN, CG, and AH.

The triangles CMA and CMN are evidently equal, and

and AFG are similar, AD: AE:: AF: AG and AD. AG= AE. AF. But the angle ACD, being equal to CAD, is equal to CNA, and consequently the triangles ADC and ACN are similar; whence AN: AC:: AC: AD, and therefore AN. AD AC. Again, because AC bisects the vertical angle HAF (VI. 23. El.) FA.AH=AC' + FC.CH, that is, FA.AE= AC2 + FC.CE.; wherefore FC.CE=FA.AEAC, that is, AG. AD-AN.AD, or NG.AD. But BA and CE being parallel, FC: EF:: AD: AE:: AF: AG, and CE: EF:: AB or AD: AF; consequently (V.21. El.) FC.CE: EF :: AD:AG :: (V.13. El ) NG× AD: NG× AG; since, therefore, FC.CE±NG. and AG, it follows (V.8. and 4. El.) that EFNG. AG. Now NG.AG=(II. 28. El.) MG1 — MA' (II. 29. cor. El.) CG-CA'; wherefore EF =CG1 — CA*, or CG2 = CA' + EF. Hence CG and the point G are given, and the angle AFG, being equal to ADC, is (III. 31. El.) contained in a given segment of a circle; wherefore the intersection F and the inflected line AF, are given,

COMPOSITION.

Let K be equal to the intercepted portion of the straight line which is to be inflected from A, and find (II. 16. El.) L the side of a square equivalent to the squares of K and of the diagonal AC, produce AD, and from C place CG

equal to L, upon AG describe (III. 31. El.) a segment of a circle containing an angle equal to ADC, and join A with the point of intersection F; AF is the straight line required.

=

For let fall the perpendicular CM, make MN MA, and join GF,CN, and AH.

=

The triangles CMA and CMN are evidently equal. But the triangles AHC and AEC are likewise equal; For the angle AFG, being equal to ADC, is equal to the angle adjacent to DAB, and consequently (III. 29. cor. El.) AB touches the circle at A; whence the angle BAH = HFA= DAE,and taking these from the equal angles BAC and DAC, there remains CAH CAE, but the angles ACH and ACE are also equal, and the side AC is common to the two triangles; wherefore AH AE, and CHCE. And because the triangles ADE and AFG are similar, AD: AE:: AF: AG, and AD. AG = AE. AF. Again, the triangles ANC and ACD being similar, AN: AC:: AC: AD, and AN. AD AC. But FC: EF:: AD: AE:: AF: AG, and CE: EF :: AB, or AD; AF; consequently FC×CE: EF2 ::AD: AG:: NG x AD: NG X AG; and since AC bisects the angle FAH, FC.CE + AC'=FA.AH FA.AE =AG.AD AN.AD+NG.AD, it follows that FC.CH, or FC. CE NG. AD, and hence EF NG. AG. Now K =CG-AC NG. AG; wherefore EF K, and EF =K.

=

=

=

=

PROP. XXVII. PROB.

Through two given points, to describe a circle touching a straight line given in position.

Let it be required to describe a circle through the points A, B, and touching the straight line CD.

It is evident that CD must either be parallel or inclined to the straight line which joins the points A and B.

1. Let CD be parallel to AB.

ANALYSIS.

E

D

From the point of contact E, draw (I. 6. El.) EG perpendicular to CD. Hence (III. 28. cor. El.) EG passes through the centre of the circle, and since it is also perpendicular to AB (I. 25. El.) it bisects that chord at right angles (III. 5. El.) the point G is, therefore, given, and the perpendicular GE; consequently the three points A, E, and B being thus given, the circle AEB is given.

COMPOSITION.

B

G A

Draw (I. 7. El.) GE bisecting AB at right angles, and (III. 11. cor. El.) through the points, A, E and B describe a circle; this will touch the straight line CD.

For (III. 6. El.) GE must pass through the centre of the circle, and (I. 25. El ) it meets the parallels CD and AB at right angles; whence (III. 28. El.) CD is a tangent to the circle.

2. Let CD be inclined to AB.

Y