Euclid. # A text-book of Euclid's Elements : for the use of schools : Books I-VI and XI online

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.*. the pentagon GHKLM is equilateral.

Again, it has been proved that the l FKC = the l FLC,

and that the l ^ HKL, KLM are respectively double of these

angles :

.'. the ^HKL = the ^KLM.

In the same way it may be shewn that every two con-

secutive angles of the figure are equal ;

.'. the pentagon GHKLM is equiangular.

.*. the pentagon is regular, and it is circumscribed about

theÂ© ABC D. Q.E.F.

Corollary. SimilarU^ it may he proved that if tangents

are dravm at the vertices of any regular polygon inscribed in

a circle, tliey will form another regular polygon of the same

sjjecies circumscribed about the circle.

[For Exercises see p. 276.J

book iv. prop. 13. 269

Proposition 13. Problem.

To inacrihe a circle in a given regular pentagon.

Let ABCDE be the given regular pentagon :

it is required to inscribe a circle within it.

Bisect two consecutive l.^ BCD, CDE by CF and DF

which intersect at F. i. 9.

Join FB ;

and draw FH, FK perp. to BC, CD. i. 12.

Then in the AÂ« BCF, DCF,

BC = DC, ^lyi^-

Because \ and CF is common to botli ;

and the l BCF = the z. DCF ; Constr.

.-. the /.CBF=the LCDF. 1.4.

But the L CDF is half an angle of the regular pentagon :

.â– . also the z. CBF is half an angle of the regular pentagon :

that is, FB bisects the l ABC.

So it may be shewn that if FA, FE were joined, these

lines would bisect tlie z. ^ at A and E.

Again, in the A^ FCH, FCK,

( the z. FCH = the l FCK, Constr.

Because â€¢< and the ^ FHC ^-^ the z. FKC being rt. angles ;

( also FC is common ;

.-. FH = FK. I. 26.

Similarly if FG, FM, FL be drawn perp. to BA, AE, ED,

it may be shewn that the five perpendiculars drawn from F

to the sides of the pentagon are all equal.

270 EUCLID'S ELEMEXTtt.

A

From centre F, with radius FH, describe a circle;

this circle must pass through the points H, K, L, M, G ;

and it will be touched at these points by the sides of the

pentagon, for the z. ^ at H, K, L, M, G are rt. L ^ Constr.

.'. the OHKLMG is inscribed in the given pentagon, q.e.f.

Corollary. The bisectors of the angles of a regular

2)entagon meet at a point.

In the same way it may be shewn that the bisectors of the angles

of any regular polygon meet at a point. [See Ex. 1, p. 274.]

[For Exercises on Kegular Polygons see p. 276.]

MISCELLANEOUS EXERCISES.

1. Two tangents AB, AC are drawn from an external point A to

a given circle: describe a circle to touch AB, AC and the convex arc

intercepted by them on the given circle.

2. ABC is an isosceles triangle, and from the vertex A a straight

line is drawn to meet the base at D and the circumference of the cir-

cumscribed circle at E: shew that AB is a tangent to the circle

circumscribed about the triangle BDE.

3. An equilateral triangle is inscribed in a given circle : shew

that twice the square on one of its sides is equal to three times the

area of the square inscribed in the same circle.

4. ABC is an isosceles triangle in which each of the angles at B

and C is double of the angle at A: shew that the square on A B is

equal to the rectangle AB, BC with the square on BC.

book iv. prop. 14. 271

Proposition 14. Problem.

To circumscribe a circle about a given regular jyeniagon.

Let ABODE be the given regular pentagon :

it is required to circumscribe a circle about it.

Bisect the l^ BCD, CDE by CF, DF intersecting at F. I. 9.

Join FB, FA, FE.

Then in the AÂ« BCF, DCF,

I BC = DC, i^*.

Because ^ and CF is common to both ;

( and the z. BCF = the z_ DCF ; Constr.

:. the L CBF = the L CDF. i. 4.

But the ii CDF is half an angle of the regular pentagon :

.*. also the /.CBF is half an angle of the regular pentagon :

that is, FB bisects the L ABC.

So it may be shewn that FA, FE bisect the z. ^ at A and E.

Now the L^ FCD, FDC are each half an angle of tlie

given regular pentagon ;

.'. the z. FCD = the L FDC, iv. Def.

.'. FC- FD. I. 6.

Similarly it may be shewn that FA, FB, FC, FD, FE are

all equal.

From centre F, with radius FA describe a circle :

this circle must pass through the points A, B, C, D, E,

and therefore is circumscribed about the given pentagon.

Q. E. F.

In the same way a circle may be circumscribed about any regular

polygon.

ELCJ.1D8 KLEMKJSTt).

Proposition 15. Problem.

To inscribe a regular hexagon in a given circle.

Let ABDF be the given circle :

it is required to inscribe a regular hexagon in it.

Find G the centre of the ABDF ; in. 1.

and draw a diameter AG D.

From centre D, with radius DG, describe the Â©EGCH.

Join CG, EG, and produce them to cut the O** of the

given circle at F and B.

Join AB, BC, CD, DE, EF, FA.

Then ABCDEF shall be the required regular hexagon.

Now G E = G D, being radii of the ACE ;

and DG = DE, being radii of the EHC :

.". GE, ED, DG are all equal, and the AEGD is equilateral.

Hence the ;L EGD = one-third of two rt, angles, i. 32.

Similarly the z. DGC = one-third of two rt. angles.

But the z." EGD, DGC, CGB together = two rt. angles ; i. 13.

.'. the remaining l CGB = one-third of two rt. angles.

.*. the three l^ EGD, DGC, CGB are equal to one another.

And to these angles the vert. opp. l^ BGA, AGF, FGE

are respectively equal :

.-. the ^ Â» EGD, DGC, CGB, BGA, AGF, FGE are all equal ;

.'. the arcs ED, DC, CB, BA, AF, FE are all equal; ill. 26.

.â€¢. the chords ED, DC, CB, BA, AF, FE are all equal : in. 29.

.'. the hexagon is equilateral.

Again the arc FA = the arc D E : Proved.

to each of these equals add the arc ABCD ;

then the whole arc FABCD =the whole arc ABCDE :

lience the angles at the Q ^Â® which stand on these equal arcs

are equal,

ROOK IV. PROP. 15. 273

that is, the a FED = the ^AFE. in. 27.

In like manner the remaining angles of the hexagon

may be shewn to be equal.

.'. the hexagon is equiangular :

.". the hexagon is regular, and it is inscribed in the ABDF.

Q. E. F.

Corollary, Tlie side of a regular hexagon inscribed in

a circle is equal to the radius of the circle.

Proposition 16. Problem.

To inscribe a regular quindecagon in a given circle.

A

*"v

Let A BCD be the given circle :

it is required to inscribe a regular quindecagon in it.

In the 0ABCD inscribe an equilateral triangle, iv. 2.

and let AC be one of its sides.

In the same circle inscribe a regular pentagon, iv. 11.

and let AB be one of its sides.

Then of such equal parts as the whole O ^^ contains fifteen,

the arc AC, which is one-third of the O ''% contains five ;

and the arc AB, which is one-fifth of the O^*', contains three;

.". their difference, the arc BC, contains two.

Bisect the arc BC at E : in. 30.

then each of the arcs BE, EC is one-fifteenth of the O^^-

.'. if BE, EC be joined, and st. lines equal to them be

placed successively round the circle, a regular quindecagon

will be inscribed in it. q. e. f.

18

274 EUCT>TT>'

NOTE ON RKGULAR POLYGONS.

The following propositions, proved bj Euclid for a regular jienta-

gon, hold good for all regular polygons.

1. The bisectors of the angles of any regular 'polygon c,r<' eon-

current.

Let D, E, A, B, C be consecutive angular

points of a regular polygon of any number of

sides.

Bisect the z â€¢ EAB, ABC by AO, BO, which

intersect at O.

Join EO.

It is required to prove that EO bisects the l DEA.

For in the A" EAO, BAO,

{EA= BA, being sides of a regular polygon;

and AO is common;

and the z EAO = the z BAO ; Constr.

.: the zOEA = the z OBA. i. 4.

But the z OBA is half the z ABC ; Constr.

also the Z ABC = the z DEA, since the polygon is regular;

/. the z OEA is half the z DEA:

that is> EO bisects the Z DEA.

Similarly if O be joined to the remaining angular points of the

polygon, it may be proved that each joining line bisects the angle

to whose vertex it is drawn.

That is to say, the bisectors of the angles of the polygon meet at

the point O. q. e. d.

CoROLT.ARiES. Sinco the Z EAB = the zABC; Hyp.

and since the Z ^ CAB, OBA are respectively half of the Z ' EAB, ABC ;

.-. the zOAB = the z OBA.

.-. OA^OB. I. G.

Similarly OE = OA.

Hence The bisectors of the angles of a regular polygon are all equal :

and a circle described from the centre O, with radius OA, will be

circumscribed about the polygon.

Also it may be shewn, as in Proposition 13, that perpendiculars

drawn from O to the sides of the polygon are all equal ; therefore a

circle described from centre O with any one of these perpendiculars as

radius will be inscribed in the polygon.

BOOK IV. NOTE ON REGULAR POLYGONS. 275

% If a polygon inscribed in a circle is equilateral, it is also

equiangular.

Let AB, BC, CD be consecutive sides of an

equilateral polygon inscribed in the ADK;

then shall this polygon be equiangular.

Because the chord AB = the chord DC, Hyp.

.: the minor arc AB = tho minor arc DC, in. 28.

To each of these equals add the arc AKD :

then the arc BAKD^-^the arc AKDC;

, .-. the angles at the Cce^ which stand on theso

equal arcs, are equal ;

that is, the z BCD = the Z ABC. iii. 27.

Similarly the remaining angles of the polygon may be she\vn to be

equal :

.â€¢. the polygon is equiangular. q.e.d.

3. If a polygon inscribed in a circle is equiaiigular, it is also

equilateral, provided that the number of its sides is odd.

[Observe that Theorems 2 and 3 are only true of polygons inscribed

in a circle.

The accompanying figures are sufficient to shew that otherwise a

polygon may be equilateral without being equiangular, Fig. 1; or

equiangular without being equilateral, Fig. 2.]

FEg.l Fig. 2

Note. The following extensions of Euclid's constructions for

Eegular Polygons should be noticed.

By continual bisection of arcs, we are enabled to divide the

circumference of a circle,

by means of Proposition 6, into 4, 8,16,..., 2 . 2**, , . . equal parts ;

by means of Proposition 15, into 3, 6, 12, ..., 3.2'*,... equal parts;

by means of Proposition 11, into 5, 10, 20,..., 5 . 2"',... equal parts;

by means of Proposition 16, into 15, 30, 60,..., 15 . 2**,... equal parts.

Hence we can inscribe in a circle a regular polygon the number of

whose sides is included in any one of the formulae 2 . 2**, 3 . 2", 5 . 2^,

15 . 2**, 71 being any positive integer. In addition to these, it has been

shewn that a regular polygon of 2^+1 sides, provided 2"+l is a

prime number, may be inscribed in a circle.

18 2

276 Euclid's i;i

EXERCISES ON PROPOSITIONS 11 â€” 16.

1. Express in terms of a right angle the magnitude of an anp;le of

the following regular polygons :

(i) a pentagon, (ii) a hexagon, (iii) an octagon,

(iv) a decagon, (v) d. quindecagon.

2. The angle of a regular pentagon is trisected by the straight

lines which join it to the opposite vertices.

3. In a polygon of n sides the straight lines which join any

angular point to'the vertices not adjacent to it, divide the angle into

n-2 equal parts.

4. Shew how to construct on a given straight line

(i) a regular pentagon, (ii) a regular hexagon, (iii) a regular octagon.

o. An equilateral triangle and a regular hexagon are inscribed in

a given circle ; shew that

(i) the area of the triangle is half that of the hexagon ;

(ii) the square on the side of the triangle is three times the

sqitare on the side of the hexagon.

6. ABODE is a regular pentagon, and AC, BE intersect at H:

shew that

(i) ABr=CH = EH.

(ii) AB is a tangent to the circle circumscribed about the

triangle BHC.

(iii) AC and BE cut one another in medial section.

7. The straight lines which join alternate vertices of a regular

pentagon intersect so as to form another regular pentagon.

8. The straight lines which join alternate vertices of a regular

polygon of n sides, intersect so as to form another regular polygon of

n sides.

If n = 6, shew that the area of the resulting hexagon is one-third of

the given hexagon.

9. By means of iv. 16, inscribe in a circle a triangle whose

angles are as the numbers 2, 5, 8.

10. Shew that the area of a regular hexagon inscribed in a circle

is three-fourths of that of the corresponding circumscribed hexagon.

THEOREMS AND EXAMPLES ON BOOK IV.

I. ON THE TRIANGLE AND ITS CIRCLES.

1. D, F, E are the points of contact of the inscribed circle of the

triangle ABC, and D^, Fj, Ej tJie points of contact of the escribed

circle, which touches BC and the other sides produced: a, b, c denote

the lengths of the sides BC, CA, AB; s the semi-perimeter of the

triangle, and r, r^ the radii of the inscribed and escribed circles.

A

Prove the folloicing equalities: â€”

(i)

(ii)

(iii)

(iv)

(V)

(vi)

AE = AF =

BD-BE.

CD = CF:

â– s - a,

:S-b,

:S-C.

CDi = CFi = s-6,

BDy=BE^ = s-c.

CD^BDi and BD^CDi.

EEi=FFi = a.

The area of the a ABC

3=rs = rj (s-a).

278

EUCLID*S ELEMENTS.

2. In tlie triangle ABC, I is the centre of the inscribed circle, and

I,, I.,, L the centres of the escribed circles touching respectively the

sides BC, CA, AB and the other sides produced.

Prove the following properties : â€”

(i) The points A, I, Ij are coUincar; so are B, I, \.,; and C, I, I3.

(ii) The points I.,, A, I3 are collinear ; so are I3, B, Ij; and

(iii) Tlie triangles BIjC, CLA, AI3B are equiangular to one

another.

(iv) The triangle l^lglg is equiangular to the triangle formed by

joining the points of contact of the inscribed circle.

(v) Of the four points I, Ij, 1^, I3 each is the orthocentre of the

triangle whose vertices are the other three.

(vi) The four circles, each of lohich passes through three of the

points I, Ij, Ij, Ij,, are all equal.

THEOREMS AND EXAMPLES ON BOOK IV. 279

3. With the notation of page 277, shew that in a triangle ABC.

if the angle at C is a right angle,

r=s-c; ri = s-h; r,^ = s-a; r^=s.

4. With the figure given on page 278, sItcw that if the circles

whose centres are I, Ij, lo, 1 3 touch BC at D, D^, D,, D3, then

(i) DD,z=DiD3=&. (ii) DD3=DiQ

1^2"

â– C.

(iii) Dp.^^h + c. (iv) DD^ = h~c.

5. Shew that the orthocentre and vertices of a triangle are the

centres of the inscribed and escribed circles of the pedal triangle.

[See Ex. 20, p. 225.]

6. Given the base and vertical angle of a triangle, find the locus of

tJie centre of the inscribed circle. [See Ex. 36, p. 228.]

7. Given the base and vertical angle of a triangle, find the locus of

the centre of the escribed circle which touches the base.

8. Given the base and vertical angle of a triangle^ shew that the

centre of the circumscribed circle is fixed.

9. Given the base BC, and the vertical angle A of a triangle, find

the locus of the centre of the escribed circle which touches AC.

10. Given the base, the vertical angle, and the radius of the

inscribed circle ; construct the triangle.

11. Given the base, the vertical angle, and the radius of the

escribed circle, (i) which touches the base, (ii) which touches one

of the sides containing the given angle ; construct the triangle.

12. Given the base, the vertical angle, and the point of contact

with the base of the inscribed circle ; construct the triangle.

13. Given the base, the vertical angle, and the point of contact

with the base, or base produced, of an escribed circle ; construct the

triangle.

14. From an external point A two tangents AB, AC are drawn to

a given circle ; and the angle BAC is bisected by a straight line which

meets the circumference in I and l^: shew that I is the centre of the

circle inscribed in the triangle ABC, and Ij the centre of one of the

escribed circles.

15. I is the centre of the circle inscribed in a triangle, and l^, I2, 1.

the centres of the escribed circles; sheic that llj, llg, II3 are bisected by

the circumference of the circumscribed circle,

16. ABC is a triangle, and Ig, I3 the centres of the escribed

circles which touch AC, and AB respectively: shew that the points

B, C, Ij, I3 lie upon a circle whose centre is on the circumference of

the circle circumscribed about ABC.

280 EUCLID'S ELEMENTS.

17. With three given points as centres describe three circles

touching one another two by two. How many solutions will there be?

18. Two tangents AB, AC are drawn to a given circle from an

external point A; aqd in AB, AC two points D and E are taken

so that DE is equal to the sum uf DB and EC: shew that DE touches

the circle.

19. Given the perimeter of a triangle, and one angle in magnitude

and position : shew that the opposite side always touches a fixed circle.

20. Given the centres of the three escribed circles ; construct the

triangle.

21. Given the centre of the inscribed circle, and the centres of

two escribed circles ; construct the triangle.

22. Given the vertical angle, perimeter, and the length of the

bisector of the vertical angle ; construct the triangle.

23. Given the vertical angle, perimeter, and altitude ; construct

the triangle.

24. Given the vertical angle, perimeter, and radius of the in-

scribed circle ; construct the triangle.

25. Given the vertical angle, the radius of the inscribed circle,

and the length of the perpendicular from the vertex to the base ;

construct the triangle.

26. Given the base, the difference of the sides containing the

vertical angle, and the radius of the inscribed circle ; construct the

triangle. [See Ex. 10, p. 258.]

27. Given the base and vertical angle of a triangle, find the locus

of the centre of the circle which passes through the three escribed

centres.

28. In a triangle ABC, I is the centre of the inscribed circle ; shew

that the centres of the circles circumscribed about the triangles BIC,

CIA, AIB lie on the circumference of the circle circumscribed about

the given triangle.

29. In a triangle ABC, the inscribed circle touches the base BC at

D ; and r, r^ are the radii of the inscribed circle and of the escribed

circle which touches BC : shew that r .i\=BD . DC.

30. ABC is a triangle, D, E, F the points of contact of its inscribed

circle; and D'E'F' is the pedal triangle of the triangle DEF: shew

that the sides of the triangle D'E'F' are parallel to those of ABC,

31. In a triangle ABC the inscribed circle touches BC at D.

Shew that the circle& inscribed in the triangles ABD, ACD touch one

another.

THEOKKM8 AND EXAMPLES ON BOOK IV.

L\S1

On the Nink-Points Chicle.

32. In any triangle the middle points of the sides, the feet of the

perpendiculars drawn from the vertices to the opposite sides, and the

middle points of the lines joining the orthocentre to the vertices are

concyclic.

In the A ABC, let X, Y, Z be the

middle points of the sides BC, CA,

AB ; let D, E, F be the feet of the

perps drawn to these sides from A,

B, C ; let O be the orthocentre, and

a, j8, 7 the middle points of OA,

OB, OC:

then shall the nine points X, Y, Z,

D, E, F, a, /3, 7 be concyclic.

Join XY, XZ, Xa, Ya, Za.

Nowfromthe a ABO,sinceAZ = ZB,

andAa = aO, Hyp.

.-. Za is pari to BO. Ex. 2, p. 96.

And from the a ABC, since BZ = Z A,

andBX = XC, Hyp.

.: ZX is par' to AC.

But BO makes a rt. angle with AC :

.'. the Z XZa is a rt. angle.

Similarly, the Z XYa is a rt. angle. i. 29.

.â€¢. the points X, Z, a, Y are concyclic :

that is, a lies on the 0"^Â° of the circle, which passes through X, Y, Z ;

and Xa is a diameter of this circle.

Similarly it may be shewn that /3 and y lie on the C"^ of the circle

which passes through X, Y, Z.

_ Again, since aDX is a rt. angle, Hyp.

.'. the circle on Xa as diameter passes through D.

Similarly it may be shewn that E and F lie on the circumference

of the same circle.

.'. the points X, Y, Z, D, E, F, a, /3, y are concyclic. q.e.d.

Hyp.

From this property the circle which passes through the middle

points of the sides of a triangle is called the Nine-Points Circle ; many

of its properties may be derived from the fact of its being the circle

circumscribed about the pedal triangle.

282

EUCLID'S ELEMENTS.

33. To prove that

(i) the centre of the nine-points circle is the middle point of

the straight line tohich joins the orthocentre to the circumscribed centre:

(ii) the radius of the nine-points circle is half the radius of the

circumscribed circle :

(iii) the centroid is collinear with the circumscribed centre^ the

nine-points centre, and the orthocentre.

In the A ABC, let X, Y, Z be the

middle points of the sides; D, E, F

the feet of the perp"; O the ortho-

centre; S and N the centres of the

circumscribed and nine-points circles

respectively.

(i) To prove that N is the

middle point of SO.

It may be shewn that the perp.

to XD from its middle point bisects

SO; Ex. 14, p. 98.

Similarly the perp. to EY at its

middle point bisects SO :

that is, these perp^ intersect at the middle point of SO :

And since XD and EY are chords of the nine-points circle,

.â€¢. the intersection of the lines which bisect XD and EY at rt. angles

is its centre: iii. 1.

.â€¢. the centre N is the middle point of SO.

(ii) To prove that the radius of the nine-points circle is half

the radius of the circumscribed circle.

By the last Proposition, Xa is a diameter of the nine-points circle.

.â€¢. the middle point of Xa is its centre:

but the middle point of SO is also the centre of the nine-points circle.

(Proved.)

Hence Xa and SO bisect one another at N.

Then from the a' SNX, ONa

j SN = ON,

Because < and NX = Na,

(and the z SNX = the zONa;

.-. SX = Oa

= Aa.

And SX is also par' to Aa,

.-. SA = Xa.

But SA is a radius of the circumscribed circle;

and Xa is a diameter of the nine-points circle;

.'. the radius of the nine-points circle is half the radius of the circum-

scribed circle.

1.15.

1.4.

I. 33.

THEOKEMS AND EXAMPLES ON BOOK IV. 283

(iii) To prove that the centroid is colHnear with points S, N , O.

Join AX and draw ag par^ to SO,

Let AX meet SO at G.

Then from the a AGO, since Aa = aO and ag is par' to OG,

.: Ag = gG. Ex. 13, p. 98.

And from the a Xag, since aN =^ NX, and NG is par> to ag,

.: gG^GX. Ex. 13, p. 98.

.-. AG=f of AX;

.-. G is the centroid of the triangle ABC.

That is, the centroid is colHnear with the points S, N, O. q.e.d,

34. Given the base and vertical angle of a triangle, find the locus

of the centre of the nine-points circle.

35. The nine-points circle of any triangle ABC, whose centre is

O, is also the nine-points circle of each of the triangles AOB, BOC,

COA.

36. If I, Ij, ly, I3 are the centres of the inscribed and escribed

circles of a triangle ABC, then the circle circumscribed about ABC is

the nine-points circle of each of the four triangles formed by joining

three of the points I, 1^, I2, I3.

37. All triangles which have the same orthocentre and the same

circumscribed circle, have also the same nine-points circle.

38. If S, I are the centres, and R, r the radii of the circumscribed

and inscribed circles of a triangle, and if N is the centre of the nine-

Ijoints circle; prove that

(i) S|2=R2-2Rr,

(ii) Nl =iR-r.

And establish corresponding properties for the escribed circles.

39. Employ the preceding theorem to shew that the nine-points

circle touches the inscribed and escribed circles of a triangle.

II. MISCELLANEOUS EXAMPLES.

1. If four circles are described to touch every three sides of a

quadrilateral, shew that their centres are concyclic.

2. If the straight lines which bisect the angles of a rectilineal

figure are concurrent, a circle may be inscribed in the figure.

3. Within a given circle describe three equal circles touching one

another and the given circle.

4. The perpendiculars drawn from the centres of the three

escribed circles of a triangle to the sides which they touch, are con-

current.

284 EUCLID'S ELEMENTS.

5. Given an angle and the radii of the inscribed and circumscribed

circles; construct the triangle.

6. Given the base, an angle at the base, and the distance between

the centre of the inscribed circle and the centre of the escribed circle

which touches the base ; construct the triangle.

7. In a given circle inscribe a triangle such that two of its sides

may pass through two given points, and the third side be of given

length.

8. In any triangle ABC, I, Ij, 1^, 1^ are the centres of the in-

scribed and escribed circles, and S^, S._,, S3 are the centres of the

circles circumscribed about the triangles BIC, CIA, AIB: shew that

the triangle S^S.^Sghas its sides parallel to those of the triangle Ijl.jly,

and is one-fourth of it in area: also that the triangles ABC and

S^SoSg have the same circumscribed circle.

9. O is the orthocentre of a triangle ABC : shew that

AOHBC^=B02-)-CA2^CO'-^-i-AB2 = d^

where d is the diameter of the circumscribed circle.

10. If from any point within a regular polygon of n sides perpen-

Again, it has been proved that the l FKC = the l FLC,

and that the l ^ HKL, KLM are respectively double of these

angles :

.'. the ^HKL = the ^KLM.

In the same way it may be shewn that every two con-

secutive angles of the figure are equal ;

.'. the pentagon GHKLM is equiangular.

.*. the pentagon is regular, and it is circumscribed about

theÂ© ABC D. Q.E.F.

Corollary. SimilarU^ it may he proved that if tangents

are dravm at the vertices of any regular polygon inscribed in

a circle, tliey will form another regular polygon of the same

sjjecies circumscribed about the circle.

[For Exercises see p. 276.J

book iv. prop. 13. 269

Proposition 13. Problem.

To inacrihe a circle in a given regular pentagon.

Let ABCDE be the given regular pentagon :

it is required to inscribe a circle within it.

Bisect two consecutive l.^ BCD, CDE by CF and DF

which intersect at F. i. 9.

Join FB ;

and draw FH, FK perp. to BC, CD. i. 12.

Then in the AÂ« BCF, DCF,

BC = DC, ^lyi^-

Because \ and CF is common to botli ;

and the l BCF = the z. DCF ; Constr.

.-. the /.CBF=the LCDF. 1.4.

But the L CDF is half an angle of the regular pentagon :

.â– . also the z. CBF is half an angle of the regular pentagon :

that is, FB bisects the l ABC.

So it may be shewn that if FA, FE were joined, these

lines would bisect tlie z. ^ at A and E.

Again, in the A^ FCH, FCK,

( the z. FCH = the l FCK, Constr.

Because â€¢< and the ^ FHC ^-^ the z. FKC being rt. angles ;

( also FC is common ;

.-. FH = FK. I. 26.

Similarly if FG, FM, FL be drawn perp. to BA, AE, ED,

it may be shewn that the five perpendiculars drawn from F

to the sides of the pentagon are all equal.

270 EUCLID'S ELEMEXTtt.

A

From centre F, with radius FH, describe a circle;

this circle must pass through the points H, K, L, M, G ;

and it will be touched at these points by the sides of the

pentagon, for the z. ^ at H, K, L, M, G are rt. L ^ Constr.

.'. the OHKLMG is inscribed in the given pentagon, q.e.f.

Corollary. The bisectors of the angles of a regular

2)entagon meet at a point.

In the same way it may be shewn that the bisectors of the angles

of any regular polygon meet at a point. [See Ex. 1, p. 274.]

[For Exercises on Kegular Polygons see p. 276.]

MISCELLANEOUS EXERCISES.

1. Two tangents AB, AC are drawn from an external point A to

a given circle: describe a circle to touch AB, AC and the convex arc

intercepted by them on the given circle.

2. ABC is an isosceles triangle, and from the vertex A a straight

line is drawn to meet the base at D and the circumference of the cir-

cumscribed circle at E: shew that AB is a tangent to the circle

circumscribed about the triangle BDE.

3. An equilateral triangle is inscribed in a given circle : shew

that twice the square on one of its sides is equal to three times the

area of the square inscribed in the same circle.

4. ABC is an isosceles triangle in which each of the angles at B

and C is double of the angle at A: shew that the square on A B is

equal to the rectangle AB, BC with the square on BC.

book iv. prop. 14. 271

Proposition 14. Problem.

To circumscribe a circle about a given regular jyeniagon.

Let ABODE be the given regular pentagon :

it is required to circumscribe a circle about it.

Bisect the l^ BCD, CDE by CF, DF intersecting at F. I. 9.

Join FB, FA, FE.

Then in the AÂ« BCF, DCF,

I BC = DC, i^*.

Because ^ and CF is common to both ;

( and the z. BCF = the z_ DCF ; Constr.

:. the L CBF = the L CDF. i. 4.

But the ii CDF is half an angle of the regular pentagon :

.*. also the /.CBF is half an angle of the regular pentagon :

that is, FB bisects the L ABC.

So it may be shewn that FA, FE bisect the z. ^ at A and E.

Now the L^ FCD, FDC are each half an angle of tlie

given regular pentagon ;

.'. the z. FCD = the L FDC, iv. Def.

.'. FC- FD. I. 6.

Similarly it may be shewn that FA, FB, FC, FD, FE are

all equal.

From centre F, with radius FA describe a circle :

this circle must pass through the points A, B, C, D, E,

and therefore is circumscribed about the given pentagon.

Q. E. F.

In the same way a circle may be circumscribed about any regular

polygon.

ELCJ.1D8 KLEMKJSTt).

Proposition 15. Problem.

To inscribe a regular hexagon in a given circle.

Let ABDF be the given circle :

it is required to inscribe a regular hexagon in it.

Find G the centre of the ABDF ; in. 1.

and draw a diameter AG D.

From centre D, with radius DG, describe the Â©EGCH.

Join CG, EG, and produce them to cut the O** of the

given circle at F and B.

Join AB, BC, CD, DE, EF, FA.

Then ABCDEF shall be the required regular hexagon.

Now G E = G D, being radii of the ACE ;

and DG = DE, being radii of the EHC :

.". GE, ED, DG are all equal, and the AEGD is equilateral.

Hence the ;L EGD = one-third of two rt, angles, i. 32.

Similarly the z. DGC = one-third of two rt. angles.

But the z." EGD, DGC, CGB together = two rt. angles ; i. 13.

.'. the remaining l CGB = one-third of two rt. angles.

.*. the three l^ EGD, DGC, CGB are equal to one another.

And to these angles the vert. opp. l^ BGA, AGF, FGE

are respectively equal :

.-. the ^ Â» EGD, DGC, CGB, BGA, AGF, FGE are all equal ;

.'. the arcs ED, DC, CB, BA, AF, FE are all equal; ill. 26.

.â€¢. the chords ED, DC, CB, BA, AF, FE are all equal : in. 29.

.'. the hexagon is equilateral.

Again the arc FA = the arc D E : Proved.

to each of these equals add the arc ABCD ;

then the whole arc FABCD =the whole arc ABCDE :

lience the angles at the Q ^Â® which stand on these equal arcs

are equal,

ROOK IV. PROP. 15. 273

that is, the a FED = the ^AFE. in. 27.

In like manner the remaining angles of the hexagon

may be shewn to be equal.

.'. the hexagon is equiangular :

.". the hexagon is regular, and it is inscribed in the ABDF.

Q. E. F.

Corollary, Tlie side of a regular hexagon inscribed in

a circle is equal to the radius of the circle.

Proposition 16. Problem.

To inscribe a regular quindecagon in a given circle.

A

*"v

Let A BCD be the given circle :

it is required to inscribe a regular quindecagon in it.

In the 0ABCD inscribe an equilateral triangle, iv. 2.

and let AC be one of its sides.

In the same circle inscribe a regular pentagon, iv. 11.

and let AB be one of its sides.

Then of such equal parts as the whole O ^^ contains fifteen,

the arc AC, which is one-third of the O ''% contains five ;

and the arc AB, which is one-fifth of the O^*', contains three;

.". their difference, the arc BC, contains two.

Bisect the arc BC at E : in. 30.

then each of the arcs BE, EC is one-fifteenth of the O^^-

.'. if BE, EC be joined, and st. lines equal to them be

placed successively round the circle, a regular quindecagon

will be inscribed in it. q. e. f.

18

274 EUCT>TT>'

NOTE ON RKGULAR POLYGONS.

The following propositions, proved bj Euclid for a regular jienta-

gon, hold good for all regular polygons.

1. The bisectors of the angles of any regular 'polygon c,r<' eon-

current.

Let D, E, A, B, C be consecutive angular

points of a regular polygon of any number of

sides.

Bisect the z â€¢ EAB, ABC by AO, BO, which

intersect at O.

Join EO.

It is required to prove that EO bisects the l DEA.

For in the A" EAO, BAO,

{EA= BA, being sides of a regular polygon;

and AO is common;

and the z EAO = the z BAO ; Constr.

.: the zOEA = the z OBA. i. 4.

But the z OBA is half the z ABC ; Constr.

also the Z ABC = the z DEA, since the polygon is regular;

/. the z OEA is half the z DEA:

that is> EO bisects the Z DEA.

Similarly if O be joined to the remaining angular points of the

polygon, it may be proved that each joining line bisects the angle

to whose vertex it is drawn.

That is to say, the bisectors of the angles of the polygon meet at

the point O. q. e. d.

CoROLT.ARiES. Sinco the Z EAB = the zABC; Hyp.

and since the Z ^ CAB, OBA are respectively half of the Z ' EAB, ABC ;

.-. the zOAB = the z OBA.

.-. OA^OB. I. G.

Similarly OE = OA.

Hence The bisectors of the angles of a regular polygon are all equal :

and a circle described from the centre O, with radius OA, will be

circumscribed about the polygon.

Also it may be shewn, as in Proposition 13, that perpendiculars

drawn from O to the sides of the polygon are all equal ; therefore a

circle described from centre O with any one of these perpendiculars as

radius will be inscribed in the polygon.

BOOK IV. NOTE ON REGULAR POLYGONS. 275

% If a polygon inscribed in a circle is equilateral, it is also

equiangular.

Let AB, BC, CD be consecutive sides of an

equilateral polygon inscribed in the ADK;

then shall this polygon be equiangular.

Because the chord AB = the chord DC, Hyp.

.: the minor arc AB = tho minor arc DC, in. 28.

To each of these equals add the arc AKD :

then the arc BAKD^-^the arc AKDC;

, .-. the angles at the Cce^ which stand on theso

equal arcs, are equal ;

that is, the z BCD = the Z ABC. iii. 27.

Similarly the remaining angles of the polygon may be she\vn to be

equal :

.â€¢. the polygon is equiangular. q.e.d.

3. If a polygon inscribed in a circle is equiaiigular, it is also

equilateral, provided that the number of its sides is odd.

[Observe that Theorems 2 and 3 are only true of polygons inscribed

in a circle.

The accompanying figures are sufficient to shew that otherwise a

polygon may be equilateral without being equiangular, Fig. 1; or

equiangular without being equilateral, Fig. 2.]

FEg.l Fig. 2

Note. The following extensions of Euclid's constructions for

Eegular Polygons should be noticed.

By continual bisection of arcs, we are enabled to divide the

circumference of a circle,

by means of Proposition 6, into 4, 8,16,..., 2 . 2**, , . . equal parts ;

by means of Proposition 15, into 3, 6, 12, ..., 3.2'*,... equal parts;

by means of Proposition 11, into 5, 10, 20,..., 5 . 2"',... equal parts;

by means of Proposition 16, into 15, 30, 60,..., 15 . 2**,... equal parts.

Hence we can inscribe in a circle a regular polygon the number of

whose sides is included in any one of the formulae 2 . 2**, 3 . 2", 5 . 2^,

15 . 2**, 71 being any positive integer. In addition to these, it has been

shewn that a regular polygon of 2^+1 sides, provided 2"+l is a

prime number, may be inscribed in a circle.

18 2

276 Euclid's i;i

EXERCISES ON PROPOSITIONS 11 â€” 16.

1. Express in terms of a right angle the magnitude of an anp;le of

the following regular polygons :

(i) a pentagon, (ii) a hexagon, (iii) an octagon,

(iv) a decagon, (v) d. quindecagon.

2. The angle of a regular pentagon is trisected by the straight

lines which join it to the opposite vertices.

3. In a polygon of n sides the straight lines which join any

angular point to'the vertices not adjacent to it, divide the angle into

n-2 equal parts.

4. Shew how to construct on a given straight line

(i) a regular pentagon, (ii) a regular hexagon, (iii) a regular octagon.

o. An equilateral triangle and a regular hexagon are inscribed in

a given circle ; shew that

(i) the area of the triangle is half that of the hexagon ;

(ii) the square on the side of the triangle is three times the

sqitare on the side of the hexagon.

6. ABODE is a regular pentagon, and AC, BE intersect at H:

shew that

(i) ABr=CH = EH.

(ii) AB is a tangent to the circle circumscribed about the

triangle BHC.

(iii) AC and BE cut one another in medial section.

7. The straight lines which join alternate vertices of a regular

pentagon intersect so as to form another regular pentagon.

8. The straight lines which join alternate vertices of a regular

polygon of n sides, intersect so as to form another regular polygon of

n sides.

If n = 6, shew that the area of the resulting hexagon is one-third of

the given hexagon.

9. By means of iv. 16, inscribe in a circle a triangle whose

angles are as the numbers 2, 5, 8.

10. Shew that the area of a regular hexagon inscribed in a circle

is three-fourths of that of the corresponding circumscribed hexagon.

THEOREMS AND EXAMPLES ON BOOK IV.

I. ON THE TRIANGLE AND ITS CIRCLES.

1. D, F, E are the points of contact of the inscribed circle of the

triangle ABC, and D^, Fj, Ej tJie points of contact of the escribed

circle, which touches BC and the other sides produced: a, b, c denote

the lengths of the sides BC, CA, AB; s the semi-perimeter of the

triangle, and r, r^ the radii of the inscribed and escribed circles.

A

Prove the folloicing equalities: â€”

(i)

(ii)

(iii)

(iv)

(V)

(vi)

AE = AF =

BD-BE.

CD = CF:

â– s - a,

:S-b,

:S-C.

CDi = CFi = s-6,

BDy=BE^ = s-c.

CD^BDi and BD^CDi.

EEi=FFi = a.

The area of the a ABC

3=rs = rj (s-a).

278

EUCLID*S ELEMENTS.

2. In tlie triangle ABC, I is the centre of the inscribed circle, and

I,, I.,, L the centres of the escribed circles touching respectively the

sides BC, CA, AB and the other sides produced.

Prove the following properties : â€”

(i) The points A, I, Ij are coUincar; so are B, I, \.,; and C, I, I3.

(ii) The points I.,, A, I3 are collinear ; so are I3, B, Ij; and

(iii) Tlie triangles BIjC, CLA, AI3B are equiangular to one

another.

(iv) The triangle l^lglg is equiangular to the triangle formed by

joining the points of contact of the inscribed circle.

(v) Of the four points I, Ij, 1^, I3 each is the orthocentre of the

triangle whose vertices are the other three.

(vi) The four circles, each of lohich passes through three of the

points I, Ij, Ij, Ij,, are all equal.

THEOREMS AND EXAMPLES ON BOOK IV. 279

3. With the notation of page 277, shew that in a triangle ABC.

if the angle at C is a right angle,

r=s-c; ri = s-h; r,^ = s-a; r^=s.

4. With the figure given on page 278, sItcw that if the circles

whose centres are I, Ij, lo, 1 3 touch BC at D, D^, D,, D3, then

(i) DD,z=DiD3=&. (ii) DD3=DiQ

1^2"

â– C.

(iii) Dp.^^h + c. (iv) DD^ = h~c.

5. Shew that the orthocentre and vertices of a triangle are the

centres of the inscribed and escribed circles of the pedal triangle.

[See Ex. 20, p. 225.]

6. Given the base and vertical angle of a triangle, find the locus of

tJie centre of the inscribed circle. [See Ex. 36, p. 228.]

7. Given the base and vertical angle of a triangle, find the locus of

the centre of the escribed circle which touches the base.

8. Given the base and vertical angle of a triangle^ shew that the

centre of the circumscribed circle is fixed.

9. Given the base BC, and the vertical angle A of a triangle, find

the locus of the centre of the escribed circle which touches AC.

10. Given the base, the vertical angle, and the radius of the

inscribed circle ; construct the triangle.

11. Given the base, the vertical angle, and the radius of the

escribed circle, (i) which touches the base, (ii) which touches one

of the sides containing the given angle ; construct the triangle.

12. Given the base, the vertical angle, and the point of contact

with the base of the inscribed circle ; construct the triangle.

13. Given the base, the vertical angle, and the point of contact

with the base, or base produced, of an escribed circle ; construct the

triangle.

14. From an external point A two tangents AB, AC are drawn to

a given circle ; and the angle BAC is bisected by a straight line which

meets the circumference in I and l^: shew that I is the centre of the

circle inscribed in the triangle ABC, and Ij the centre of one of the

escribed circles.

15. I is the centre of the circle inscribed in a triangle, and l^, I2, 1.

the centres of the escribed circles; sheic that llj, llg, II3 are bisected by

the circumference of the circumscribed circle,

16. ABC is a triangle, and Ig, I3 the centres of the escribed

circles which touch AC, and AB respectively: shew that the points

B, C, Ij, I3 lie upon a circle whose centre is on the circumference of

the circle circumscribed about ABC.

280 EUCLID'S ELEMENTS.

17. With three given points as centres describe three circles

touching one another two by two. How many solutions will there be?

18. Two tangents AB, AC are drawn to a given circle from an

external point A; aqd in AB, AC two points D and E are taken

so that DE is equal to the sum uf DB and EC: shew that DE touches

the circle.

19. Given the perimeter of a triangle, and one angle in magnitude

and position : shew that the opposite side always touches a fixed circle.

20. Given the centres of the three escribed circles ; construct the

triangle.

21. Given the centre of the inscribed circle, and the centres of

two escribed circles ; construct the triangle.

22. Given the vertical angle, perimeter, and the length of the

bisector of the vertical angle ; construct the triangle.

23. Given the vertical angle, perimeter, and altitude ; construct

the triangle.

24. Given the vertical angle, perimeter, and radius of the in-

scribed circle ; construct the triangle.

25. Given the vertical angle, the radius of the inscribed circle,

and the length of the perpendicular from the vertex to the base ;

construct the triangle.

26. Given the base, the difference of the sides containing the

vertical angle, and the radius of the inscribed circle ; construct the

triangle. [See Ex. 10, p. 258.]

27. Given the base and vertical angle of a triangle, find the locus

of the centre of the circle which passes through the three escribed

centres.

28. In a triangle ABC, I is the centre of the inscribed circle ; shew

that the centres of the circles circumscribed about the triangles BIC,

CIA, AIB lie on the circumference of the circle circumscribed about

the given triangle.

29. In a triangle ABC, the inscribed circle touches the base BC at

D ; and r, r^ are the radii of the inscribed circle and of the escribed

circle which touches BC : shew that r .i\=BD . DC.

30. ABC is a triangle, D, E, F the points of contact of its inscribed

circle; and D'E'F' is the pedal triangle of the triangle DEF: shew

that the sides of the triangle D'E'F' are parallel to those of ABC,

31. In a triangle ABC the inscribed circle touches BC at D.

Shew that the circle& inscribed in the triangles ABD, ACD touch one

another.

THEOKKM8 AND EXAMPLES ON BOOK IV.

L\S1

On the Nink-Points Chicle.

32. In any triangle the middle points of the sides, the feet of the

perpendiculars drawn from the vertices to the opposite sides, and the

middle points of the lines joining the orthocentre to the vertices are

concyclic.

In the A ABC, let X, Y, Z be the

middle points of the sides BC, CA,

AB ; let D, E, F be the feet of the

perps drawn to these sides from A,

B, C ; let O be the orthocentre, and

a, j8, 7 the middle points of OA,

OB, OC:

then shall the nine points X, Y, Z,

D, E, F, a, /3, 7 be concyclic.

Join XY, XZ, Xa, Ya, Za.

Nowfromthe a ABO,sinceAZ = ZB,

andAa = aO, Hyp.

.-. Za is pari to BO. Ex. 2, p. 96.

And from the a ABC, since BZ = Z A,

andBX = XC, Hyp.

.: ZX is par' to AC.

But BO makes a rt. angle with AC :

.'. the Z XZa is a rt. angle.

Similarly, the Z XYa is a rt. angle. i. 29.

.â€¢. the points X, Z, a, Y are concyclic :

that is, a lies on the 0"^Â° of the circle, which passes through X, Y, Z ;

and Xa is a diameter of this circle.

Similarly it may be shewn that /3 and y lie on the C"^ of the circle

which passes through X, Y, Z.

_ Again, since aDX is a rt. angle, Hyp.

.'. the circle on Xa as diameter passes through D.

Similarly it may be shewn that E and F lie on the circumference

of the same circle.

.'. the points X, Y, Z, D, E, F, a, /3, y are concyclic. q.e.d.

Hyp.

From this property the circle which passes through the middle

points of the sides of a triangle is called the Nine-Points Circle ; many

of its properties may be derived from the fact of its being the circle

circumscribed about the pedal triangle.

282

EUCLID'S ELEMENTS.

33. To prove that

(i) the centre of the nine-points circle is the middle point of

the straight line tohich joins the orthocentre to the circumscribed centre:

(ii) the radius of the nine-points circle is half the radius of the

circumscribed circle :

(iii) the centroid is collinear with the circumscribed centre^ the

nine-points centre, and the orthocentre.

In the A ABC, let X, Y, Z be the

middle points of the sides; D, E, F

the feet of the perp"; O the ortho-

centre; S and N the centres of the

circumscribed and nine-points circles

respectively.

(i) To prove that N is the

middle point of SO.

It may be shewn that the perp.

to XD from its middle point bisects

SO; Ex. 14, p. 98.

Similarly the perp. to EY at its

middle point bisects SO :

that is, these perp^ intersect at the middle point of SO :

And since XD and EY are chords of the nine-points circle,

.â€¢. the intersection of the lines which bisect XD and EY at rt. angles

is its centre: iii. 1.

.â€¢. the centre N is the middle point of SO.

(ii) To prove that the radius of the nine-points circle is half

the radius of the circumscribed circle.

By the last Proposition, Xa is a diameter of the nine-points circle.

.â€¢. the middle point of Xa is its centre:

but the middle point of SO is also the centre of the nine-points circle.

(Proved.)

Hence Xa and SO bisect one another at N.

Then from the a' SNX, ONa

j SN = ON,

Because < and NX = Na,

(and the z SNX = the zONa;

.-. SX = Oa

= Aa.

And SX is also par' to Aa,

.-. SA = Xa.

But SA is a radius of the circumscribed circle;

and Xa is a diameter of the nine-points circle;

.'. the radius of the nine-points circle is half the radius of the circum-

scribed circle.

1.15.

1.4.

I. 33.

THEOKEMS AND EXAMPLES ON BOOK IV. 283

(iii) To prove that the centroid is colHnear with points S, N , O.

Join AX and draw ag par^ to SO,

Let AX meet SO at G.

Then from the a AGO, since Aa = aO and ag is par' to OG,

.: Ag = gG. Ex. 13, p. 98.

And from the a Xag, since aN =^ NX, and NG is par> to ag,

.: gG^GX. Ex. 13, p. 98.

.-. AG=f of AX;

.-. G is the centroid of the triangle ABC.

That is, the centroid is colHnear with the points S, N, O. q.e.d,

34. Given the base and vertical angle of a triangle, find the locus

of the centre of the nine-points circle.

35. The nine-points circle of any triangle ABC, whose centre is

O, is also the nine-points circle of each of the triangles AOB, BOC,

COA.

36. If I, Ij, ly, I3 are the centres of the inscribed and escribed

circles of a triangle ABC, then the circle circumscribed about ABC is

the nine-points circle of each of the four triangles formed by joining

three of the points I, 1^, I2, I3.

37. All triangles which have the same orthocentre and the same

circumscribed circle, have also the same nine-points circle.

38. If S, I are the centres, and R, r the radii of the circumscribed

and inscribed circles of a triangle, and if N is the centre of the nine-

Ijoints circle; prove that

(i) S|2=R2-2Rr,

(ii) Nl =iR-r.

And establish corresponding properties for the escribed circles.

39. Employ the preceding theorem to shew that the nine-points

circle touches the inscribed and escribed circles of a triangle.

II. MISCELLANEOUS EXAMPLES.

1. If four circles are described to touch every three sides of a

quadrilateral, shew that their centres are concyclic.

2. If the straight lines which bisect the angles of a rectilineal

figure are concurrent, a circle may be inscribed in the figure.

3. Within a given circle describe three equal circles touching one

another and the given circle.

4. The perpendiculars drawn from the centres of the three

escribed circles of a triangle to the sides which they touch, are con-

current.

284 EUCLID'S ELEMENTS.

5. Given an angle and the radii of the inscribed and circumscribed

circles; construct the triangle.

6. Given the base, an angle at the base, and the distance between

the centre of the inscribed circle and the centre of the escribed circle

which touches the base ; construct the triangle.

7. In a given circle inscribe a triangle such that two of its sides

may pass through two given points, and the third side be of given

length.

8. In any triangle ABC, I, Ij, 1^, 1^ are the centres of the in-

scribed and escribed circles, and S^, S._,, S3 are the centres of the

circles circumscribed about the triangles BIC, CIA, AIB: shew that

the triangle S^S.^Sghas its sides parallel to those of the triangle Ijl.jly,

and is one-fourth of it in area: also that the triangles ABC and

S^SoSg have the same circumscribed circle.

9. O is the orthocentre of a triangle ABC : shew that

AOHBC^=B02-)-CA2^CO'-^-i-AB2 = d^

where d is the diameter of the circumscribed circle.

10. If from any point within a regular polygon of n sides perpen-

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