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

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In like manner the process of construction may be

extended to a figure of five or more sides.

Q.E.P.

Definition. When three magnitudes are proportionals

the first is said to have to the tliird the duplicate ratio of

that which it has to the second. [Book v. Def. 13.]

BOOK VI. PROP. 19. !339

Proposition 19. Theorem.

Siinilar triangles are to one another in the duplicate ratio

of their homologoics sides.

Let ABC, DEF be similar triangles, having the /_ ABC

equal to the L DEF, and let BC and EF be homologous sides:

then shall the A ABC be to the A DEF in the duplicate

ratio of BC to EF.

To BC and EF take a third proportional BG,

so that BC : EF :: EF : BG. Yl. 11.

Join AG.

Then because the A^ ABC, DEF are similar, tlyj).

:. AB : BC :: DE : EF;

.'., alternately^ AB : DE :: BC : EF; V. 11.

but BC : EF :: EF : BG; Constr.

.'. AB : DE :: EF : BG; v. 1.

that is, the sides of the A^ ABG, DEF about the equal

angles at B and E are reciprocally proportional;

.â€¢. the A ABG = the A DEF. vi. 15.

Again, because BC : EF : : EF : BG, Constr.

.'. BC : BG in the duplicate ratio of BC to EF. Def.

But the A ABC : the A ABG :: BC : BG, vi. 1.

.'. the A ABC : the A ABG in the duplicate ratio

of BC to EF: V. 1.

I and the A ABG = the A DEF; Proved.

:. the AABC : the A DEF in the duplicate ratio

of BC : EF. Q.E.D.

22-2

340 euclid's elements.

Proposition 20. Theorem.

Similar polygons may be divided into the same number

of similar triangles^ having the same ratio each to each that

the polygons have; and the polygons are to one ariother in

the duplicate ratio of their homologous sides.

H

Let ABODE, FGHKL be similar polygons, and let AB be

the side homologous to FG ;

tlien (i) the polygons may be divided into the same number

of similar triangles;

(ii) these triangles shall have each to each the same

ratio that the polygons have;

(iii) the polygon ABODE shall be to the polygon FGHKL

in the duplicate ratio of AB to FG.

Join EB, EO, LG, LH.

(i) Then because the polygon ABODE is similar to the

polygon FGHKL, Hyp.

.-. the z.EAB-tlie z. LFG,

and EA : AB : : LF : FG ; vi. Def. 2.

.'. the A EAB is similar to the A LFG ; vi. 6.

.-. the ^ ABE = the l FGL.

But, because the polygons are similar, HyP'

:. the :L ABO -the ^ FGH, vi. Def 2.

.*. the remaining ^ E BO â€” the remaining z. LGH.

And because the A^ ABE, FGL are similar, Proved.

:. EB : BA :: LG : GF;

and because the polygons are similar, J^Iy2^-

.'. AB : BO = FG : GH; VI. Def 2.

.*., ex cequali, EB : BO :: LG : GH, V. 14.

that is, the sides about the equal l ^ EBO, LGH are

proportionals;

.*. the A EBO is similar to the A LGH. vi. 6.

BOOK VI. PROP. 20. 341

In the same way it may be proved that the A ECD is

similar to the A LHK.

.'. the polygons have been divided into the same number

of similar triangles.

(ii) Again, because the A ABE is similar to the AFGL,

.". the A ABE is to the A FGL in the duplicate ratio

of EB : LG; VI. 19.

and, in like manner,

the A EBC is to the A LGH in the duplicate ratio

of EB to LG ;

.-.the A ABE : the AFGL :: the A EBC : the A LGH. v. 1.

In like manner it can be shewn that

the A EBC : the A LGH :: the A EDC : the ALKH.

.-. the A ABE : the AFGL :: the A EBC : the A LGH

:: the A EDC : the ALKH.

But when any number of ratios are equal, as each ante-

cedent is to its consequent so is the sum of all the ante-

cedents to the sum of all the consequents; v. 12.

.-. the A ABE : the A LFG :: the fig. ABCDE : the fig. FGHKL.

(iii) Now the A EAB : the A LFG in the duplicate ratio

of AB : FG,

and the A EAB : the A LFG :: the fig. ABCDE : the fig. FGHKL;

.*. the fig. ABCDE : the fig. FGHKL in the duplicate ratio

of AB : FG. Q.E.D.

Corollary 1. Let a third proportional X be taken

to AB and FG,

then A B is to X in the duplicate ratio of AB : FG;

but the fig. ABCDE : the fig. FGHKL in the duplicate

ratio of AB : FG.

Hence, if three straight lines are proportionals, as the first

is to the third, so is any rectilineal figure described on the

first to a similar and similarly described rectilineal figure

on the second.

Corollary 2. It follows that similar rectilineal figures

are to one another as the squares on their homologous sides.

For squares are similar figures and therefore are to one

another in the duplicate ratio of their sides.

342 EUCLID'S ELEMENTS.

Proposition 21. Theorem.

Rectilineal figures which' are similar to the same recti-

lineal figure^ are also similar to each other.

Let each of the rectilineal figures A and B be similar to C :

then shall A be similar to B.

For because A is similar to C, Hyp.

.'. A is equiangular to C, and the sides about their equal

angles are proportionals- vi. Defi 2.

Again, because B is similar to C, ffyp-

.'. B is equiangular to C, and the sides about their equal

angles are proportionals. vi. Def. 2.

.'. A and B are each of them equiangular to C, and have

the sides about the equal angles proportional to the cor-

responding sides of C ;

.'. A is equiangular to B, and the sides about their equal

angles are proportionals; V. 1,

.'. A is similar to B.

Q. e. d.

BOOK VI. PROP. 22. 343

Proposition 22. Theorem.

If four straight lines he proportional and a pair of

similar rectilineal figures he similarly described on the first

and second, and also a pair on the third and fourth, these

figures shall he proportional:

Conversely, if a rectilineal figure on the first of four

straight lines he to the similar and similarly described figure

on the second as a rectilineal figure on the third is to the

similar and similarly described figure on the fourth, the four

straight lines shall be proportional.

Let AB, CD, EF, GH be proportionals,

so that AB : CD :: EF : GH;

and let similar figures KAB, LCD be similarly described on

AB, CD, and also let similar figs. MF, NH be similarly-

described on EF, GH:

then shall

the fig. KAB : the fig. LCD :: the fig. MF : the fig. NH.

To AB and CD take a third proportional X, vi. 11.

and to EF and GH take a third proportional O;

then AB : CD :: CD : X,

Constr.

and EF : GH :: GH : 0.

But AB : CD :: EF : GH;

Hyp.

.'. CD : X :: GH : 0,

V. 1.

.*., ex oiquali, AB : X :: EF : Q.

V. 14.

But AB : X :: the fig. KAB : the fig. LCD,

V]

[. 20, Co^.

and EF : O :: the fig. MF : the fig. NH;

.-. the fig. KAB : the fig. LCD :: the fig. MF : the

fig. NH.

V. 1.

344 EUCLID'S ELEMENTS.

O P R

Conversely^

let the fig. KAB : the fig. LCD :: the fig. MF : the fig. NH;

then shall AB : CD :: EF : GH.

To AB, CD, and EF take a fourth proportional PR : vi. 12.

and on PR describe the fig. SR similar and similarly situated

to either of the figs. MF, NH. vi. 18.

Then because AB : CD :: EF : PR, Constr.

.*., by the former part of the proposition,

the fig. KAB : the fig. LCD :: the fig. MF : the fig. SR.

But

the fig. KAB : the fig. LCD :: the fig. MF : the fig. NH. Hyp.

.'. the fig. MF : the fig. SR :: the fig. MF : the fig. NH, v. 1.

.-. the fig. SR = the fig. NH.

And since the figs. SR and NH are similar and similarly

situated,

.'. PR=GH*

Now AB : CD :: EF : PR; Constr.

:. AB : CD :: EF : GH.

Q. E. D.

* Euclid here assumes that if two similar and similarly situated

figures are equal, their homologous sides are equal. The proof is

easy and may be left as an exercise for the student.

Definition. "When there are any number of magnitudes

of the same kind, the first is said to have to the last the

ratio compounded of the ratios of the first to the second, of

the second to the third, and so on up to the ratio of the

last but one to the last magnitude. [Book v. Def. 12.]

book vi. prop. 23. 345

Proposition 23. Theorem.

Parallelograms which are equiangular to one another

have to one another the ratio which is compounded of the

ratios of their sides.

A D H

V

B

K L M E F

Let the par"" AC be equiangular to the par" CF, having tlie

z. BCD equal to the l ECG :

then shall the par"" AC have to the par*" CF the ratio com-

pounded of the ratios BC : CG and DC : CE.

Let the par"" be placed so that BC and CG are in a st. line;

then DC and CE are also in a st. line. i. 14.

Complete the par"* DG.

Take any st. line K,

and to BC, CG, and K find a fourth proportional L; vi. 12.

and to DC, CE, and L take a fourth proportional M ;

then BC : CG :: K : L,

and DC : CE :: L : M.

But K : M is the ratio compounded of the ratios

K : L and L : M, v. Def. 12.

that is, K : M is the ratio compounded of the ratios

BC : CG and DC : CE.

Now the parâ„¢ AC : the parâ„¢ CH :: BC : CG vi. 1.

:: K : L, ' Constr.

and the parâ„¢ CH : the parâ„¢ CF :: DC : CE vi. 1.

:: L : M, Constr.

.'., ex cequali, the parâ„¢ AC : the parâ„¢ CF :: K : M. v. 14.

But K : M is the ratio compounded of the ratios of the sides;

.". the parâ„¢ AC has to the parâ„¢ CF the ratio compounded

of the ratios of the sides. Q. e. d.

EXERCISE.

The areas of two triangles or parallelograms are to one another

in the ratio compounded of the ratios of their bases and of their

altitudes.

34f> Euclid's elements.

Proposition 24. Theorem.

Parallelograms about a diagonal of any parallelogram

are similar to the wlwle parallelograin and to one another.

A E B

Ih

Â°^

D K

Let A BCD be a par"" of which AC is a diagonal;

and let EG, HK be par"'* about AC :

then shall the par*"' EG, HK be similar to the par"* ABCD,

and to one another.

For, because DC is par' to GF,

.-. the z.ADC-the ^AGF; 1.29.

and because BC is par* to EF,

.'. the ^ ABC = the ^AEF; i. 29.

and each of the l^ BCD, EFG is equal to the opp. z. BAD,

.-. the ^ BCD = the Z.EFG; [i. 34.

.". the par"" ABCD is equiangular to the parâ„¢ AEFG.

Again in the A^ BAC, EAF,

because the l ABC = the /_ AEF, i. 29.

and the z. BAC is common;

.*. A'' BAC, EAF are equiangular to one another; i. 32.

.-. AB : BC :: AE : EF. VI. 4.

But BC = AD, and EF = AG ; I. 34.

.*. AB : AD :: AE : AG;

and DC : CB :: GF : FE,

and CD : DA :: FG : GA,

.*. the sides of the par"'' ABCD, AEFG about their equal

angles are proportional;

.'. the par"' ABCD is similar to the par"'AEFG. vi. Def. 2.

In the same way it may be proved that the parâ„¢ ABCD

is similar to the parâ„¢ FHCK,

.'. each of the parâ„¢^ EG, HK is similar to the whole parâ„¢;

.*. the parâ„¢ EG is similar to the parâ„¢ HK. vi. 21.

Q, E. D.

BOOK vr. PROP. 25. 347

Proposition 25. Problej/:.

To describe a rectilineal figure which shall he equal to

one and similar to another rectilineal figure.

I z^

Let E and S be two rectilineal figures :

it is required to describe a figure equal to the fig. E and

similar to the fig. S.

On AB a side of the fig. S describe a par"* A BCD equal to S,

and on BC describe a par"* CBGF equal to the fig. E, and

having the L CBG equal to the L DAB: i. 45.

then AB and BG are in one st. line, and also DC and CF in

one st. line.

Between AB and BG find a mean proportional HK; vi. 13.

and on H K describe the fig. P, similar and similarly situated

to the fig. S; vi. 18.

then P shall be the figure required.

Because AB : HK :: HK : BG, Constr.

.'. AB : BG :: the fig. S : the fig. P. vi. 20, Cor.

But AB : BG :: the parâ„¢ AC : the parâ„¢ BF;

.*. the fig. S : the fig. P :: the parâ„¢ AC : the par-" BF; v. i.

and the fig. S =the parâ„¢ AC; Constr.

.'. the fig. P =:the parâ„¢ BF

= the fig. E. Constr.

And since, by construction, the fig. P is similar to the fig. S,

.*. P is the rectil. figure required.

Q. E. F.

348 EUCLID'S ELEMENTS.

Proposition 26. Theorem.

If two similar parallelograms have a com/man angle, and

be similarly situated., they are about the same diagonal.

Let tlie par"" ABCD, AEFG be similar and similarly situated,

and have the common angle BAD :

then shall these par"'* be about the same diagonal.

Join AC.

Then if AC does not pass through F, let it cut FG, or FG

produced, at H.

Join AF;

and through H draw HK par' to AD or BC. i. 31.

Then the parâ„¢' BD and KG are similar, since they are about

the same diagonal AHC; vi. 24.

.-. DA : AB :: GA : AK.

But because the par"'* BD and EG are similar; HyjJ.

:. DA : AB :: GA : AE; vi. Be/ 2.

.-. GA : AK :: GA : AE;

.'. AK = AE, which is impossible;

.'. AC must pass tlirough F;

tliat is, the par""* BD, EG are about the same diagonal.

Q.E. D.

BOOK VI. PROP. 30. 349

Ohs. Propositions 27, 28, 29 being cumbrous in form and of little

value as geometrical results are now very generally omitted.

Definitiox. a straight line is said to be divided in

extreme and mean ratio, when the whole is to the greater

segment as the greater segment is to the less.

[Book VI. Def. 4.]

Proposition 30. Problem.

2^0 divide a given straight line in extreme and mean ratio.

'3~B

Let AB be the given st. line:

it is required to divide it in extreme and mean ratio.

Divide AB in C so that the rect. AB, BC may be equal to

the sq. on AC. ii. 11.

Then because tlie rect. AB, BC = the sq. on AC,

.*. AB : AC :: AC : BC. vi. 17.

Q. E. P.

EXERCISES.

1. ABCDE is a regular pentagon; if the lines BE and AD inter-

sect in O, shew that each of them is divided in extreme and mean

ratio.

2, If the radius of a circle is cut in extreme and mean ratio, the

greater segment is equal to the side of a regular decagon inscribed in

the circle.

350

EUCLID'S ELEMENTS.

Proposition 31. Theorem.

In a right-angled triangle, any rectilineal figure described

on the hypotenuse is equal to the sum of the two sirnilar and

similarly described figures on the sides containing the right

angle.

Let ABC be a right-angled triangle of which BC is the

hypotenuse; and let P, Q, R be similar and similarly described

figures on BC, CA, AB respectively:

then shall the tig. P be equal to the sum of the tigs. Q and R.

Draw AD perp. to BC.

Then the A^ CBA, ABD are similar;

BA : BD;

the tig. P

VI. 8.

/. CB : BA ::

.-. CB : BD ::

.'., inversely, BD : BC ::

In like manner DC : BC : :

'. the sum of BD, DC : BC

the tig. R

the tig. Q

the sum of figs.

the tig. R, VI. 20, Con

the tig. P.

the tig. P;

R, Q

V. :i.

tig. P;

V. 15.

but BC = the sum of BD, DC;

the tig. P = the sum of the figs. R and Q.

Q.E.D.

Note. This proposition is a generalization of the 47th Prop, of

Book I. It will be a useful exercise for the student to deduce the

general theorem from the particular case with the aid of Prop. 20,

Cor. 2.

EXERCISES ON PROP. 31. 351

EXERCISES.

1. In a right-angled triangle if a perpendicular be drawn from the

right angle to the opposite side, the segments of the hypotenuse are in

the duplicate ratio of the sides containing the right angle.

2. If, in Proposition 31, the figure on the hypotenuse is equal

to the given triangle, the figures on the other two sides are each equal

to one of the parts into which the triangle is divided by the perpen-

dicular from the right angle to the hypotenuse.

3. AX and BY are medians of the triangle ABC which meet in

G : if XY be joined, compare the areas of the triangles AGB, XGY.

4. Shew that similar triangles are to one another in the duplicate

ratio of (i) corresponding medians, (ii) the radii of their inscribed

circles, (iii) the radii of their circumscribed circles.

5. DEF is the pedal triangle of the triangle ABC ; prove that the

triangle ABC is to the triangle DBF in the duplicate ratio of AB to

BD. Hence shew that

the fig. AFDC : the a BFD :: AD2 : BD^.

6. The base BC of a triangle ABC is produced to a point D such

that BD : DC in the duplicate ratio of BA : AC. Shew that AD is a

mean proportional between BD and DC.

7. Bisect a triangle by a line drawn parallel to one of its sides.

8. Shew how to draw a line parallel to the base of a triangle so

as to form with the other two sides produced a triangle double of the

given triangle.

9. If through any point within a triangle lines be drawn from

the angles to cut the opposite sides, the segments of any one side will

have to each other the ratio compounded of the ratios of the segments

of the other sides.

10. Draw a straight line parallel to the base of an isosceles tri-

angle so as to cut off a triangle which has to the whole triangle the

ratio of the base to a side.

11. Through a given point, between two straight lines containing

a given angle, draw a line which shall cut off a triangle equal to a

given rectilineal figure.

Obs. The 32nd Proposition as given by Euclid is de-

fective, and as it is never applied, we have omitted it.

â€¢So2

EUCLID'S ELEMKNTb.

Proposition 33. Theorem.

In equal circles, angles, wliether at the centres or the cir-

cumferences, have the same ratio as the arcs on which they

stand: so also have the sectors.

Let ABC and DEF be equal circles, and let BGC, EHF be

angles at the centres, and BAG and EDF angles at the O*^*^^;

tlien shall

(i) the L BGG : the l EHF :: the arc BG : the arc EF,

(ii) the L. BAG : the l EDF :: the arc BG : the arc EF,

(iii) the sector BGG : the sector EHF :: the arc BG : the

arc EF.

Along the O'*'' of the Â©ABG take any number of ares

OK, KL each equal to BG; and along the O^Â® of the 0DEF

take any number of arcs FM, MN, NR each equal to EF.

Join GK, GL, HM, HN, HR.

(i) Then the l^ BGC, GGK, KGL are all equal,

for they stand on the equal arcs BG, GK, KL: in. 27..

.". the z. BGL is the same multiple of the l BGG that the

arc BL is of the arc BG.

Similarly the z. EHR is the same multiple of the z_ EHF

that the arc ER is of the arc EF.

And if the arc BL = the arc ER,

the z. BGL^the ^EHR; III. 27.

and if the arc BL is greater than the arc ER,

the Â£. BGL is greater than the lEHR;

and if the arc BL is less than the arc ER,

the z. BGL is less than the l EHR.

BOOK VI. PROP. 33.

353

Kow since there are four magnitudes, namely the

L^ BGC, EHF and the arcs BC, EF; and of the antecedents

any equimultiples have been taken, namely the lBGL

and the arc BL; and of the consequents any equimultiples

have been taken, namely the z. EHR and the arc ER :

and it has been proved that the z. BGL is greater than,

equal to, or less than the z_ EHR according as BL is greater

than, equal to, or less than ER;

.'. the four magnitudes are proportionals; v. Be/. 4.

that is, the :. BGC : the z. EHF :: the arc BC : the arc EF.

(ii) And since the z. BGC = twice the l BAC, hi. 20.

and the Â£. EHF = twice the t^EDF;

.". the L BAC : the z. EDF : : the arc BC : the arc EF.. v. 8.

(iii) Join BC, CK; and in the arcs BC, CK take any

points X, O.

Join BX, XC, CO, OK.

Then in the AÂ« BGC, CGK,

[ BG=:CG,

Because - GC = GK,

[and the z. BGC = the ^ CGK;

.'. BC = CK;

and the A BGC -= the A CGK.

And because the arc BC = the arc CK,

.*. the remaining arc BAC =the remaining arc CAK:

.'. the z.BXC=the Z.COK; m. 27.

.*. the segment BXC is similar to the segment COK; iii. Be/.

and they stand on equal chords BC, CK;

.'. the segment BXC = the segment COK. iii. 24.

And the A BGC = the A CGK;

.'. the sector BGC ^ the sector CGK.

III. 27.

I. 4.

Constr.

H. E.

23

354

EUCLID'S ELEMENTS.

Similarly it may be shewn that the sectors BGC, CGK,

KGL are all equal;

and likewise the sectors EHF, FHM, MHN, NHR are all equal.

.', the sector BGL is the same multiple of the sector BGC

that the arc BL is of the arc BC;.

and the sector EHR is the same multiple of the sector EHF

that the arc ER is of the arc EF:

And if the arc BL = the arc ER,

the sector BGL = the sector EHR : Proved.

and if the arc BL is greater than the arc ER,

the sector BGL is greater than the sector EHR:

and if the arc BL is less than the arc ER,

the sector BGL is less than the sector EHR.

Now since there are four magnitudes, namely, the sec-

tors BGC, EHF and the arcs BC, EF; and of the antecedents

any equimultiples have been taken, namely the sector BGL

and the arc BL; and of the consequents any equimultiples

have been taken, namely the sector EHR and the arc ER :

and it has been shewn that the sector BGL is greater than,

equal to, or less than the sector EHR according as the

arc BL is greater than, equal to, or less than the arc ER;

.'. the four magnitudes are proportionals; v. Def. 4.

that is,

the sector BGC : the sector EHF : : tlie arc BC : the arc EF.

Q. E. D.

book vi. prop. b. 355

Proposition B. Theorem.

If the vertical angle of a triangle he bisected by a straight

line which cuts the base, the rectangle contained by the sides

of the triangle shall be equal to the rectangle contained by

the segments of the base, together ivith the square on the

st7'aight line which bisects the angle.

Let ABC be a triangle having the l BAG bisected by AD:

then shall

the rect. BA, AC = the rect. BD, DC, with the sq. on AD.

Describe a circle about the A ABC, iv. 5.

and produce AD to meet the O^^ in E.

Join EC.

Then in the A^ BAD, EAC,

because the Â£. BAD = the a. EAC, ^J^yp-

and the L ABD = the /_ AEC in the same segment; iii. 21.

.'. the A BAD is equiangular to the A EAC. i. 32.

.-. BA : AD :: EA : AC; A^l. 4.

.'. the rect. BA, AC = the rect. EA, AD, VI. 16.

= the rect. ED, DA, with the sq. on AD.

II. 3.

But the rect. ED, DA = the rect. BD, DC; iii. 35.

.'. the rect. BA, AC = the rect. BD, DC, with the sq. on AD.

Q. E. D.

EXERCISE.

If the vertical angle BAC be externally bisected by a straight line

which meets the base in D, shew that the rectangle contained by BA,

AC together with the square on AD is equal to the rectangle contained

by the segments of the base.

23-2

356 EUCLID'S ELEMENTS.

Proposition C. Theorem.

If from the vertical angle of a triangle a straight line he

drawn perjyendicular to the ba,se, the rectarigle contained hy

the sides of the triangle shall he equal to the rectangle con-

tained hy the perpendicular and the diameter of the circlf.

described about the triangle.

Let ABC be a triangle, and let AD be the perp. from A

to BC:

then the rect. BA, AC shall be equal to the rect. contained

by AD and the diameter of the circle circumscribed about

o the A ABC.

Describe a circle about the A ABC; iv. 5.

draw the diameter AE, and join EC.

Then in the AÂ« BAD, EAC,

the rt. angle BDA == the rt. angle ACE, in the semicircle ACE,

and the L ABD = the l. AEC, in the same segment; in. 21.

.". the A BAD is equiangular to the A EAC; i. 32.

.â€¢. BA : AD :: EA : AC; vi. 4.

.". the rect. BA, AC - the rect, EA, AD. vi. 16.

Q.E.D.

book vi. prop. d. 357

Proposition D. Theorem.

The rectangle contained hy the diagonals of a quadri-

lateral inscribed in a circle is equal to the sum of the two

rectangles contained by its opposite sides.

Let A BCD be a quadrilateral inscribed in a circle, and let

AC, BD be its diagonals:

then the rect. AC, BD shall be equal to the sum of the rect-

angles AB, CD and BC, AD.

Make the l DAE equal to the l BAC; i. 23.

to each add the z_ EAC,

then the L DAC = the Â£. BAE.

Then in the AÂ« EAB, DAC,

the L EAB = the L DAC,

and the l ABE = the l ACD in the same segment; ill. 21.

.â– . the triangles are equiangular to one another; i. 32.

.-. AB : BE :: AC : CD; VI. 4.

,'. the rect. AB, CD = the rect. AC, EB. vi 16.

Again in the A* DAE, CAB,

the z. DAE = the zL CAB, Constr.

and the L ADE ==the L ACB, in the same segment, iii. 21.

.'. the triangles are equiangular to one another ; i. 32.

.â€¢. AD : DE :: AC : CB; VI. 4.

.*. the rect. BC, AD = the rect. AC, DE. vi. 16.

But the rect. AB, CD = the rect. AC, EB. Proved.

.'. the sum of the rects. BC, AD and AB, CD = the sum of

the rects. AC, DE and AC, EB ;

that is, the sum of the rects. BC, AD and AB, CD

= the rect. AC, BD. ii. 1.

358 boclid's elements.

Note. Propositions B, C, and D do not occur in Euclid, but were

added by Kobert Simson.

Prop. D is usually known as Ptolemy's theorem, and it is the par-

ticular case of the following more general theorem :

The rectangle contained by the diagonals of a quadrilateral is leas

than the sum of tlie recta/ngles contained by its opposite sides, unless a

circle can be circumscribed about the quadrilateral, in which case it is

equal to that sum.

EXERCISES.

1. ABC is an isosceles triangle, and on the base, or base pro-

duced, any point X is taken : shew that the circumscribed circles of

the triangles ABX, ACX are equal.

2. From the extremities B, C of the base of an isosceles triangle

ABC, straight lines are drawn perpendicular to AB, AC respectively,

and intersecting at D : shew that the rectangle BC, AD is double of

the rectangle AB, DB.

3. If the diagonals of a quadrilateral inscribed in a circle are at

extended to a figure of five or more sides.

Q.E.P.

Definition. When three magnitudes are proportionals

the first is said to have to the tliird the duplicate ratio of

that which it has to the second. [Book v. Def. 13.]

BOOK VI. PROP. 19. !339

Proposition 19. Theorem.

Siinilar triangles are to one another in the duplicate ratio

of their homologoics sides.

Let ABC, DEF be similar triangles, having the /_ ABC

equal to the L DEF, and let BC and EF be homologous sides:

then shall the A ABC be to the A DEF in the duplicate

ratio of BC to EF.

To BC and EF take a third proportional BG,

so that BC : EF :: EF : BG. Yl. 11.

Join AG.

Then because the A^ ABC, DEF are similar, tlyj).

:. AB : BC :: DE : EF;

.'., alternately^ AB : DE :: BC : EF; V. 11.

but BC : EF :: EF : BG; Constr.

.'. AB : DE :: EF : BG; v. 1.

that is, the sides of the A^ ABG, DEF about the equal

angles at B and E are reciprocally proportional;

.â€¢. the A ABG = the A DEF. vi. 15.

Again, because BC : EF : : EF : BG, Constr.

.'. BC : BG in the duplicate ratio of BC to EF. Def.

But the A ABC : the A ABG :: BC : BG, vi. 1.

.'. the A ABC : the A ABG in the duplicate ratio

of BC to EF: V. 1.

I and the A ABG = the A DEF; Proved.

:. the AABC : the A DEF in the duplicate ratio

of BC : EF. Q.E.D.

22-2

340 euclid's elements.

Proposition 20. Theorem.

Similar polygons may be divided into the same number

of similar triangles^ having the same ratio each to each that

the polygons have; and the polygons are to one ariother in

the duplicate ratio of their homologous sides.

H

Let ABODE, FGHKL be similar polygons, and let AB be

the side homologous to FG ;

tlien (i) the polygons may be divided into the same number

of similar triangles;

(ii) these triangles shall have each to each the same

ratio that the polygons have;

(iii) the polygon ABODE shall be to the polygon FGHKL

in the duplicate ratio of AB to FG.

Join EB, EO, LG, LH.

(i) Then because the polygon ABODE is similar to the

polygon FGHKL, Hyp.

.-. the z.EAB-tlie z. LFG,

and EA : AB : : LF : FG ; vi. Def. 2.

.'. the A EAB is similar to the A LFG ; vi. 6.

.-. the ^ ABE = the l FGL.

But, because the polygons are similar, HyP'

:. the :L ABO -the ^ FGH, vi. Def 2.

.*. the remaining ^ E BO â€” the remaining z. LGH.

And because the A^ ABE, FGL are similar, Proved.

:. EB : BA :: LG : GF;

and because the polygons are similar, J^Iy2^-

.'. AB : BO = FG : GH; VI. Def 2.

.*., ex cequali, EB : BO :: LG : GH, V. 14.

that is, the sides about the equal l ^ EBO, LGH are

proportionals;

.*. the A EBO is similar to the A LGH. vi. 6.

BOOK VI. PROP. 20. 341

In the same way it may be proved that the A ECD is

similar to the A LHK.

.'. the polygons have been divided into the same number

of similar triangles.

(ii) Again, because the A ABE is similar to the AFGL,

.". the A ABE is to the A FGL in the duplicate ratio

of EB : LG; VI. 19.

and, in like manner,

the A EBC is to the A LGH in the duplicate ratio

of EB to LG ;

.-.the A ABE : the AFGL :: the A EBC : the A LGH. v. 1.

In like manner it can be shewn that

the A EBC : the A LGH :: the A EDC : the ALKH.

.-. the A ABE : the AFGL :: the A EBC : the A LGH

:: the A EDC : the ALKH.

But when any number of ratios are equal, as each ante-

cedent is to its consequent so is the sum of all the ante-

cedents to the sum of all the consequents; v. 12.

.-. the A ABE : the A LFG :: the fig. ABCDE : the fig. FGHKL.

(iii) Now the A EAB : the A LFG in the duplicate ratio

of AB : FG,

and the A EAB : the A LFG :: the fig. ABCDE : the fig. FGHKL;

.*. the fig. ABCDE : the fig. FGHKL in the duplicate ratio

of AB : FG. Q.E.D.

Corollary 1. Let a third proportional X be taken

to AB and FG,

then A B is to X in the duplicate ratio of AB : FG;

but the fig. ABCDE : the fig. FGHKL in the duplicate

ratio of AB : FG.

Hence, if three straight lines are proportionals, as the first

is to the third, so is any rectilineal figure described on the

first to a similar and similarly described rectilineal figure

on the second.

Corollary 2. It follows that similar rectilineal figures

are to one another as the squares on their homologous sides.

For squares are similar figures and therefore are to one

another in the duplicate ratio of their sides.

342 EUCLID'S ELEMENTS.

Proposition 21. Theorem.

Rectilineal figures which' are similar to the same recti-

lineal figure^ are also similar to each other.

Let each of the rectilineal figures A and B be similar to C :

then shall A be similar to B.

For because A is similar to C, Hyp.

.'. A is equiangular to C, and the sides about their equal

angles are proportionals- vi. Defi 2.

Again, because B is similar to C, ffyp-

.'. B is equiangular to C, and the sides about their equal

angles are proportionals. vi. Def. 2.

.'. A and B are each of them equiangular to C, and have

the sides about the equal angles proportional to the cor-

responding sides of C ;

.'. A is equiangular to B, and the sides about their equal

angles are proportionals; V. 1,

.'. A is similar to B.

Q. e. d.

BOOK VI. PROP. 22. 343

Proposition 22. Theorem.

If four straight lines he proportional and a pair of

similar rectilineal figures he similarly described on the first

and second, and also a pair on the third and fourth, these

figures shall he proportional:

Conversely, if a rectilineal figure on the first of four

straight lines he to the similar and similarly described figure

on the second as a rectilineal figure on the third is to the

similar and similarly described figure on the fourth, the four

straight lines shall be proportional.

Let AB, CD, EF, GH be proportionals,

so that AB : CD :: EF : GH;

and let similar figures KAB, LCD be similarly described on

AB, CD, and also let similar figs. MF, NH be similarly-

described on EF, GH:

then shall

the fig. KAB : the fig. LCD :: the fig. MF : the fig. NH.

To AB and CD take a third proportional X, vi. 11.

and to EF and GH take a third proportional O;

then AB : CD :: CD : X,

Constr.

and EF : GH :: GH : 0.

But AB : CD :: EF : GH;

Hyp.

.'. CD : X :: GH : 0,

V. 1.

.*., ex oiquali, AB : X :: EF : Q.

V. 14.

But AB : X :: the fig. KAB : the fig. LCD,

V]

[. 20, Co^.

and EF : O :: the fig. MF : the fig. NH;

.-. the fig. KAB : the fig. LCD :: the fig. MF : the

fig. NH.

V. 1.

344 EUCLID'S ELEMENTS.

O P R

Conversely^

let the fig. KAB : the fig. LCD :: the fig. MF : the fig. NH;

then shall AB : CD :: EF : GH.

To AB, CD, and EF take a fourth proportional PR : vi. 12.

and on PR describe the fig. SR similar and similarly situated

to either of the figs. MF, NH. vi. 18.

Then because AB : CD :: EF : PR, Constr.

.*., by the former part of the proposition,

the fig. KAB : the fig. LCD :: the fig. MF : the fig. SR.

But

the fig. KAB : the fig. LCD :: the fig. MF : the fig. NH. Hyp.

.'. the fig. MF : the fig. SR :: the fig. MF : the fig. NH, v. 1.

.-. the fig. SR = the fig. NH.

And since the figs. SR and NH are similar and similarly

situated,

.'. PR=GH*

Now AB : CD :: EF : PR; Constr.

:. AB : CD :: EF : GH.

Q. E. D.

* Euclid here assumes that if two similar and similarly situated

figures are equal, their homologous sides are equal. The proof is

easy and may be left as an exercise for the student.

Definition. "When there are any number of magnitudes

of the same kind, the first is said to have to the last the

ratio compounded of the ratios of the first to the second, of

the second to the third, and so on up to the ratio of the

last but one to the last magnitude. [Book v. Def. 12.]

book vi. prop. 23. 345

Proposition 23. Theorem.

Parallelograms which are equiangular to one another

have to one another the ratio which is compounded of the

ratios of their sides.

A D H

V

B

K L M E F

Let the par"" AC be equiangular to the par" CF, having tlie

z. BCD equal to the l ECG :

then shall the par"" AC have to the par*" CF the ratio com-

pounded of the ratios BC : CG and DC : CE.

Let the par"" be placed so that BC and CG are in a st. line;

then DC and CE are also in a st. line. i. 14.

Complete the par"* DG.

Take any st. line K,

and to BC, CG, and K find a fourth proportional L; vi. 12.

and to DC, CE, and L take a fourth proportional M ;

then BC : CG :: K : L,

and DC : CE :: L : M.

But K : M is the ratio compounded of the ratios

K : L and L : M, v. Def. 12.

that is, K : M is the ratio compounded of the ratios

BC : CG and DC : CE.

Now the parâ„¢ AC : the parâ„¢ CH :: BC : CG vi. 1.

:: K : L, ' Constr.

and the parâ„¢ CH : the parâ„¢ CF :: DC : CE vi. 1.

:: L : M, Constr.

.'., ex cequali, the parâ„¢ AC : the parâ„¢ CF :: K : M. v. 14.

But K : M is the ratio compounded of the ratios of the sides;

.". the parâ„¢ AC has to the parâ„¢ CF the ratio compounded

of the ratios of the sides. Q. e. d.

EXERCISE.

The areas of two triangles or parallelograms are to one another

in the ratio compounded of the ratios of their bases and of their

altitudes.

34f> Euclid's elements.

Proposition 24. Theorem.

Parallelograms about a diagonal of any parallelogram

are similar to the wlwle parallelograin and to one another.

A E B

Ih

Â°^

D K

Let A BCD be a par"" of which AC is a diagonal;

and let EG, HK be par"'* about AC :

then shall the par*"' EG, HK be similar to the par"* ABCD,

and to one another.

For, because DC is par' to GF,

.-. the z.ADC-the ^AGF; 1.29.

and because BC is par* to EF,

.'. the ^ ABC = the ^AEF; i. 29.

and each of the l^ BCD, EFG is equal to the opp. z. BAD,

.-. the ^ BCD = the Z.EFG; [i. 34.

.". the par"" ABCD is equiangular to the parâ„¢ AEFG.

Again in the A^ BAC, EAF,

because the l ABC = the /_ AEF, i. 29.

and the z. BAC is common;

.*. A'' BAC, EAF are equiangular to one another; i. 32.

.-. AB : BC :: AE : EF. VI. 4.

But BC = AD, and EF = AG ; I. 34.

.*. AB : AD :: AE : AG;

and DC : CB :: GF : FE,

and CD : DA :: FG : GA,

.*. the sides of the par"'' ABCD, AEFG about their equal

angles are proportional;

.'. the par"' ABCD is similar to the par"'AEFG. vi. Def. 2.

In the same way it may be proved that the parâ„¢ ABCD

is similar to the parâ„¢ FHCK,

.'. each of the parâ„¢^ EG, HK is similar to the whole parâ„¢;

.*. the parâ„¢ EG is similar to the parâ„¢ HK. vi. 21.

Q, E. D.

BOOK vr. PROP. 25. 347

Proposition 25. Problej/:.

To describe a rectilineal figure which shall he equal to

one and similar to another rectilineal figure.

I z^

Let E and S be two rectilineal figures :

it is required to describe a figure equal to the fig. E and

similar to the fig. S.

On AB a side of the fig. S describe a par"* A BCD equal to S,

and on BC describe a par"* CBGF equal to the fig. E, and

having the L CBG equal to the L DAB: i. 45.

then AB and BG are in one st. line, and also DC and CF in

one st. line.

Between AB and BG find a mean proportional HK; vi. 13.

and on H K describe the fig. P, similar and similarly situated

to the fig. S; vi. 18.

then P shall be the figure required.

Because AB : HK :: HK : BG, Constr.

.'. AB : BG :: the fig. S : the fig. P. vi. 20, Cor.

But AB : BG :: the parâ„¢ AC : the parâ„¢ BF;

.*. the fig. S : the fig. P :: the parâ„¢ AC : the par-" BF; v. i.

and the fig. S =the parâ„¢ AC; Constr.

.'. the fig. P =:the parâ„¢ BF

= the fig. E. Constr.

And since, by construction, the fig. P is similar to the fig. S,

.*. P is the rectil. figure required.

Q. E. F.

348 EUCLID'S ELEMENTS.

Proposition 26. Theorem.

If two similar parallelograms have a com/man angle, and

be similarly situated., they are about the same diagonal.

Let tlie par"" ABCD, AEFG be similar and similarly situated,

and have the common angle BAD :

then shall these par"'* be about the same diagonal.

Join AC.

Then if AC does not pass through F, let it cut FG, or FG

produced, at H.

Join AF;

and through H draw HK par' to AD or BC. i. 31.

Then the parâ„¢' BD and KG are similar, since they are about

the same diagonal AHC; vi. 24.

.-. DA : AB :: GA : AK.

But because the par"'* BD and EG are similar; HyjJ.

:. DA : AB :: GA : AE; vi. Be/ 2.

.-. GA : AK :: GA : AE;

.'. AK = AE, which is impossible;

.'. AC must pass tlirough F;

tliat is, the par""* BD, EG are about the same diagonal.

Q.E. D.

BOOK VI. PROP. 30. 349

Ohs. Propositions 27, 28, 29 being cumbrous in form and of little

value as geometrical results are now very generally omitted.

Definitiox. a straight line is said to be divided in

extreme and mean ratio, when the whole is to the greater

segment as the greater segment is to the less.

[Book VI. Def. 4.]

Proposition 30. Problem.

2^0 divide a given straight line in extreme and mean ratio.

'3~B

Let AB be the given st. line:

it is required to divide it in extreme and mean ratio.

Divide AB in C so that the rect. AB, BC may be equal to

the sq. on AC. ii. 11.

Then because tlie rect. AB, BC = the sq. on AC,

.*. AB : AC :: AC : BC. vi. 17.

Q. E. P.

EXERCISES.

1. ABCDE is a regular pentagon; if the lines BE and AD inter-

sect in O, shew that each of them is divided in extreme and mean

ratio.

2, If the radius of a circle is cut in extreme and mean ratio, the

greater segment is equal to the side of a regular decagon inscribed in

the circle.

350

EUCLID'S ELEMENTS.

Proposition 31. Theorem.

In a right-angled triangle, any rectilineal figure described

on the hypotenuse is equal to the sum of the two sirnilar and

similarly described figures on the sides containing the right

angle.

Let ABC be a right-angled triangle of which BC is the

hypotenuse; and let P, Q, R be similar and similarly described

figures on BC, CA, AB respectively:

then shall the tig. P be equal to the sum of the tigs. Q and R.

Draw AD perp. to BC.

Then the A^ CBA, ABD are similar;

BA : BD;

the tig. P

VI. 8.

/. CB : BA ::

.-. CB : BD ::

.'., inversely, BD : BC ::

In like manner DC : BC : :

'. the sum of BD, DC : BC

the tig. R

the tig. Q

the sum of figs.

the tig. R, VI. 20, Con

the tig. P.

the tig. P;

R, Q

V. :i.

tig. P;

V. 15.

but BC = the sum of BD, DC;

the tig. P = the sum of the figs. R and Q.

Q.E.D.

Note. This proposition is a generalization of the 47th Prop, of

Book I. It will be a useful exercise for the student to deduce the

general theorem from the particular case with the aid of Prop. 20,

Cor. 2.

EXERCISES ON PROP. 31. 351

EXERCISES.

1. In a right-angled triangle if a perpendicular be drawn from the

right angle to the opposite side, the segments of the hypotenuse are in

the duplicate ratio of the sides containing the right angle.

2. If, in Proposition 31, the figure on the hypotenuse is equal

to the given triangle, the figures on the other two sides are each equal

to one of the parts into which the triangle is divided by the perpen-

dicular from the right angle to the hypotenuse.

3. AX and BY are medians of the triangle ABC which meet in

G : if XY be joined, compare the areas of the triangles AGB, XGY.

4. Shew that similar triangles are to one another in the duplicate

ratio of (i) corresponding medians, (ii) the radii of their inscribed

circles, (iii) the radii of their circumscribed circles.

5. DEF is the pedal triangle of the triangle ABC ; prove that the

triangle ABC is to the triangle DBF in the duplicate ratio of AB to

BD. Hence shew that

the fig. AFDC : the a BFD :: AD2 : BD^.

6. The base BC of a triangle ABC is produced to a point D such

that BD : DC in the duplicate ratio of BA : AC. Shew that AD is a

mean proportional between BD and DC.

7. Bisect a triangle by a line drawn parallel to one of its sides.

8. Shew how to draw a line parallel to the base of a triangle so

as to form with the other two sides produced a triangle double of the

given triangle.

9. If through any point within a triangle lines be drawn from

the angles to cut the opposite sides, the segments of any one side will

have to each other the ratio compounded of the ratios of the segments

of the other sides.

10. Draw a straight line parallel to the base of an isosceles tri-

angle so as to cut off a triangle which has to the whole triangle the

ratio of the base to a side.

11. Through a given point, between two straight lines containing

a given angle, draw a line which shall cut off a triangle equal to a

given rectilineal figure.

Obs. The 32nd Proposition as given by Euclid is de-

fective, and as it is never applied, we have omitted it.

â€¢So2

EUCLID'S ELEMKNTb.

Proposition 33. Theorem.

In equal circles, angles, wliether at the centres or the cir-

cumferences, have the same ratio as the arcs on which they

stand: so also have the sectors.

Let ABC and DEF be equal circles, and let BGC, EHF be

angles at the centres, and BAG and EDF angles at the O*^*^^;

tlien shall

(i) the L BGG : the l EHF :: the arc BG : the arc EF,

(ii) the L. BAG : the l EDF :: the arc BG : the arc EF,

(iii) the sector BGG : the sector EHF :: the arc BG : the

arc EF.

Along the O'*'' of the Â©ABG take any number of ares

OK, KL each equal to BG; and along the O^Â® of the 0DEF

take any number of arcs FM, MN, NR each equal to EF.

Join GK, GL, HM, HN, HR.

(i) Then the l^ BGC, GGK, KGL are all equal,

for they stand on the equal arcs BG, GK, KL: in. 27..

.". the z. BGL is the same multiple of the l BGG that the

arc BL is of the arc BG.

Similarly the z. EHR is the same multiple of the z_ EHF

that the arc ER is of the arc EF.

And if the arc BL = the arc ER,

the z. BGL^the ^EHR; III. 27.

and if the arc BL is greater than the arc ER,

the Â£. BGL is greater than the lEHR;

and if the arc BL is less than the arc ER,

the z. BGL is less than the l EHR.

BOOK VI. PROP. 33.

353

Kow since there are four magnitudes, namely the

L^ BGC, EHF and the arcs BC, EF; and of the antecedents

any equimultiples have been taken, namely the lBGL

and the arc BL; and of the consequents any equimultiples

have been taken, namely the z. EHR and the arc ER :

and it has been proved that the z. BGL is greater than,

equal to, or less than the z_ EHR according as BL is greater

than, equal to, or less than ER;

.'. the four magnitudes are proportionals; v. Be/. 4.

that is, the :. BGC : the z. EHF :: the arc BC : the arc EF.

(ii) And since the z. BGC = twice the l BAC, hi. 20.

and the Â£. EHF = twice the t^EDF;

.". the L BAC : the z. EDF : : the arc BC : the arc EF.. v. 8.

(iii) Join BC, CK; and in the arcs BC, CK take any

points X, O.

Join BX, XC, CO, OK.

Then in the AÂ« BGC, CGK,

[ BG=:CG,

Because - GC = GK,

[and the z. BGC = the ^ CGK;

.'. BC = CK;

and the A BGC -= the A CGK.

And because the arc BC = the arc CK,

.*. the remaining arc BAC =the remaining arc CAK:

.'. the z.BXC=the Z.COK; m. 27.

.*. the segment BXC is similar to the segment COK; iii. Be/.

and they stand on equal chords BC, CK;

.'. the segment BXC = the segment COK. iii. 24.

And the A BGC = the A CGK;

.'. the sector BGC ^ the sector CGK.

III. 27.

I. 4.

Constr.

H. E.

23

354

EUCLID'S ELEMENTS.

Similarly it may be shewn that the sectors BGC, CGK,

KGL are all equal;

and likewise the sectors EHF, FHM, MHN, NHR are all equal.

.', the sector BGL is the same multiple of the sector BGC

that the arc BL is of the arc BC;.

and the sector EHR is the same multiple of the sector EHF

that the arc ER is of the arc EF:

And if the arc BL = the arc ER,

the sector BGL = the sector EHR : Proved.

and if the arc BL is greater than the arc ER,

the sector BGL is greater than the sector EHR:

and if the arc BL is less than the arc ER,

the sector BGL is less than the sector EHR.

Now since there are four magnitudes, namely, the sec-

tors BGC, EHF and the arcs BC, EF; and of the antecedents

any equimultiples have been taken, namely the sector BGL

and the arc BL; and of the consequents any equimultiples

have been taken, namely the sector EHR and the arc ER :

and it has been shewn that the sector BGL is greater than,

equal to, or less than the sector EHR according as the

arc BL is greater than, equal to, or less than the arc ER;

.'. the four magnitudes are proportionals; v. Def. 4.

that is,

the sector BGC : the sector EHF : : tlie arc BC : the arc EF.

Q. E. D.

book vi. prop. b. 355

Proposition B. Theorem.

If the vertical angle of a triangle he bisected by a straight

line which cuts the base, the rectangle contained by the sides

of the triangle shall be equal to the rectangle contained by

the segments of the base, together ivith the square on the

st7'aight line which bisects the angle.

Let ABC be a triangle having the l BAG bisected by AD:

then shall

the rect. BA, AC = the rect. BD, DC, with the sq. on AD.

Describe a circle about the A ABC, iv. 5.

and produce AD to meet the O^^ in E.

Join EC.

Then in the A^ BAD, EAC,

because the Â£. BAD = the a. EAC, ^J^yp-

and the L ABD = the /_ AEC in the same segment; iii. 21.

.'. the A BAD is equiangular to the A EAC. i. 32.

.-. BA : AD :: EA : AC; A^l. 4.

.'. the rect. BA, AC = the rect. EA, AD, VI. 16.

= the rect. ED, DA, with the sq. on AD.

II. 3.

But the rect. ED, DA = the rect. BD, DC; iii. 35.

.'. the rect. BA, AC = the rect. BD, DC, with the sq. on AD.

Q. E. D.

EXERCISE.

If the vertical angle BAC be externally bisected by a straight line

which meets the base in D, shew that the rectangle contained by BA,

AC together with the square on AD is equal to the rectangle contained

by the segments of the base.

23-2

356 EUCLID'S ELEMENTS.

Proposition C. Theorem.

If from the vertical angle of a triangle a straight line he

drawn perjyendicular to the ba,se, the rectarigle contained hy

the sides of the triangle shall he equal to the rectangle con-

tained hy the perpendicular and the diameter of the circlf.

described about the triangle.

Let ABC be a triangle, and let AD be the perp. from A

to BC:

then the rect. BA, AC shall be equal to the rect. contained

by AD and the diameter of the circle circumscribed about

o the A ABC.

Describe a circle about the A ABC; iv. 5.

draw the diameter AE, and join EC.

Then in the AÂ« BAD, EAC,

the rt. angle BDA == the rt. angle ACE, in the semicircle ACE,

and the L ABD = the l. AEC, in the same segment; in. 21.

.". the A BAD is equiangular to the A EAC; i. 32.

.â€¢. BA : AD :: EA : AC; vi. 4.

.". the rect. BA, AC - the rect, EA, AD. vi. 16.

Q.E.D.

book vi. prop. d. 357

Proposition D. Theorem.

The rectangle contained hy the diagonals of a quadri-

lateral inscribed in a circle is equal to the sum of the two

rectangles contained by its opposite sides.

Let A BCD be a quadrilateral inscribed in a circle, and let

AC, BD be its diagonals:

then the rect. AC, BD shall be equal to the sum of the rect-

angles AB, CD and BC, AD.

Make the l DAE equal to the l BAC; i. 23.

to each add the z_ EAC,

then the L DAC = the Â£. BAE.

Then in the AÂ« EAB, DAC,

the L EAB = the L DAC,

and the l ABE = the l ACD in the same segment; ill. 21.

.â– . the triangles are equiangular to one another; i. 32.

.-. AB : BE :: AC : CD; VI. 4.

,'. the rect. AB, CD = the rect. AC, EB. vi 16.

Again in the A* DAE, CAB,

the z. DAE = the zL CAB, Constr.

and the L ADE ==the L ACB, in the same segment, iii. 21.

.'. the triangles are equiangular to one another ; i. 32.

.â€¢. AD : DE :: AC : CB; VI. 4.

.*. the rect. BC, AD = the rect. AC, DE. vi. 16.

But the rect. AB, CD = the rect. AC, EB. Proved.

.'. the sum of the rects. BC, AD and AB, CD = the sum of

the rects. AC, DE and AC, EB ;

that is, the sum of the rects. BC, AD and AB, CD

= the rect. AC, BD. ii. 1.

358 boclid's elements.

Note. Propositions B, C, and D do not occur in Euclid, but were

added by Kobert Simson.

Prop. D is usually known as Ptolemy's theorem, and it is the par-

ticular case of the following more general theorem :

The rectangle contained by the diagonals of a quadrilateral is leas

than the sum of tlie recta/ngles contained by its opposite sides, unless a

circle can be circumscribed about the quadrilateral, in which case it is

equal to that sum.

EXERCISES.

1. ABC is an isosceles triangle, and on the base, or base pro-

duced, any point X is taken : shew that the circumscribed circles of

the triangles ABX, ACX are equal.

2. From the extremities B, C of the base of an isosceles triangle

ABC, straight lines are drawn perpendicular to AB, AC respectively,

and intersecting at D : shew that the rectangle BC, AD is double of

the rectangle AB, DB.

3. If the diagonals of a quadrilateral inscribed in a circle are at

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