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homologous, then the third angles are either equal or sup-
plementary ; and iri the former case the triangles are similar.




Let ABC, DEF be two triangles having the L ABC equal to
the z_ DEF, and the sides about the angles at A and D pro-
portional, so that

BA : AC :: ED : DF;
tlien shall the l ^ ACB, DFE be either equal or supple-
mentary, and in the former case the triangles shall be
similar.

If the L. BAC = the /. EDF,
then the ^ BCA = the _ EFD; 1.32.

and the A^ are equiangular and therefore similar, vi. 4.
But if the L BAC is not equal to the z_ EDF, one of them
must be the greater.

Let the z. EDF be greater than the L BAC.
At D in ED make the i. EDF' equal to the z. BAC. i. 23.
Then the A^ BAC, EDF' are equiangular, Constr.



:. BA


: AC :: ED : DF';


VI. 4.


but BA


: AC :: ED : DF;


Hyp.


.'. ED


: DF :: ED : DF',
. DF = DF',


V. 1.


Y. the L


DFF'=the ^DF'F.


I. 5.


But the i. « DF'F,


DF'E are supplementary.


I. 13.


.-. the L^ DFF',


DF'E are supplementary:




hat is, the /_« DFE


ACB are supplementary.


q. E. D.

21



euclid's elements.

Corollaries to Proposition 7.
A




Three cases of this theorem deserve special attention.

It has been proved that if the angles ACB, DFE are not supple-
mentanj, they are equal:

and we know that of angles which are supplementary and unequal,
one must be acute and the other obtuse.

Hence, in addition to the hypothesis of this theorem,

(i) If the angles ACB, DFE, opposite to the two homologous
sides AB, DE are both acute, both obtuse, or if one of
them is a right angle,
it follows that these angles are equal ;
and therefore the triangles are similar.

(ii) If the two given angles are right angles or obtuse angles,
it follows that the angles ACB, DFE must be both acute,
and therefore equal, by (i) :
so that the triangles are similar,
(iii) If in each triangle the side opposite the given angle is not
IBS'? than the other given side; that is, if AC and DF are
not less than AB and DE respectively, then
the angles ACB, DFE cannot be greater than the angles
ABC, DEF, respectively;

therefore the angles ACB, DFE, are both acute;
hence, as above, they are equal ;
and the triangles ABC, DEF similar.



BOOK VI. TROP. 7. 323

EXERCISES.
ON Propositions 1 to 7.

1. Shew that the diagonals of a trapezium cut one another in
the same ratio.

2. If three straight lines drawn from a point cut two parallel
straight lines in A, B, C and P, Q, R respectively, prove that

AB : BC :: PQ : QR.

3. From a point O, a tangent OP is drawn to a given circle, and
OQR is drawn cutting it in Q and R ; shew that

OQ: OP :: OP: OR.

4. If two triangles are on equal bases and between the same parallels,
any straight line parallel to their bases will cut off equal areas from the
tico triangles.

5. If two straight lines PQ, XY intersect in a point O, so that
PO : OX :: YO : OQ., prove that P, X, Q, Y are concyclic.

6. On the same base and on the same side of it two equal
triangles ACB, ADB are described; AC and BD intersect in O, and
through O lines parallel to DA and OB are drawn meeting the base
in E and F. Shew that AE= BF.

7. BD, CD are perpendicular to the sides AB, AC of a triangle
ABC, and CE is drawn perpendicular to AD, meeting AB in E : shew
that the triangles ABC, ACE are similar.

8. AC and BD are drawn perpendicular to a given straight line
CD from two given points A and B ; AD and BC intersect in E, and
EF is perpendicular to CD : shew that AF and BF make equal angles
with CD.

9. A BCD is a parallelogram; P and Q are points in a straight
line parallel to AB ; PA and QB meet at R, and PD and QC meet at
S : shew that RS is parallel to AD.

10. In the sides AB, AC of a triangle ABC two points D, E are
taken such that BD is equal to CE ; if DE, BC produced meet at F,
shew that AB : AC :: EF : DF.

11. Find a point the perpendiculars from which on the sides of a
given triangle shall be in a given ratio.



21-2



324 euclid'h elements.



Proposition S. Theorem.



In a rlyht-arbyled triangle if a perpendicular be drawn
froin tlie right angle to the hypotenuse, the triangles on earJi
side of it are similar to tlie wJiole triangle and to one anotJier.

A




Let ABC be a triangle riglit-anglecl at A, and let AD l)e
perp. to BC:

then shall the A^ DBA, DAC be similar to the A ABC and
to one another.

In the A*' DBA, ABC,
the z. BDA = the z. BAC, being rt. angles,
and the l. ABC is common to both;
.*. the remaining z. BAD = the remaining i. BCA, i. 32.
that is, the A^ DBA, ABC are equiangular;

.'. they are similar. vi. 4.

In the same way it may be proved that the A^ DAC,
ABC are similar.

Hence the A" DBA, DAC, being equiangular to the same
A ABC, are equiangular to one another;

.'. they are similar. vi. 4.

Q. E. D.

Corollary. Because the A^ BDA, ADC are similar,

.'. BD : DA :: DA : DC;
and because the A* CBA, ABD are similar,

/. CB : BA :: BA : BD;
and because the A^ BCA, ACD are similar,

.-. BC : CA :: CA : CD.

EXERCISES.

1. Prove that the hypotenuse is to one side as the second side is
to the perpendicular.

2. Shew that the radius of a circle is a vieau proportiomd heticeen
the sepments of any tangent between its point of contact and a pair
of parallel tangents.



BOOK VI. PROP. 9. 325

Definition. A less magnitude is said to be a sub-
multiple of a greater, when the less is contained an exact
number of times in the greater. [Book v. Def. 2.]



Proposition 9. Problem.

Froini a given straight line to cut off any required suh-
multiple.



G.




A F B

Let AB be the given straight line.
It is required to cut off a certain submultiple from AB.

From A draw a straight line AG of indefinite length making
any angle with AB.

In AG take any point D; and, by cutting off successive

parts each equal to AD, make AE to contain AD as many

times as AB contains the required submultiple.

Join EB.

Through D draw DF par^ to EB, meeting AB in F.

Then shall AF be the required submultiple.

\ Because DF is par^ to EB, a side of the AAEB,

.-. BF : FA :: ED : DA; VI. 2.

, componendo, BA : AF :: EA : AD. v. 13.

But AE contains AD the required number of times; Constr.

I.'. AB contains AF the required number of times;
that is, AF is the required submultiple. Q.e.f.

EXERCISES.

1. Divide a straight line into five equal parts.

2. Give a geometrical construction for cutting off two-sevenths of
a given straight line.



326 Euclid's elements.

Proposition 10. Problem.

To divide a straight line similarly to a given divided
straight line.

A

f/Np

7K




B K



Let AB be the given straight line to be divided, and AC
the given straight line divided at the points D and E.
It is required to divide AB similarly to AC.

Let AB, AC be placed so as to form any angle.

Join CB.

Through D draw DF par' to CB, i. 31.

and through E draw EG par' to CB,

and through D draw DHK par' to AB.

Then AB shall be divided at F and G similarly to AC.

For by construction each of the figs. FH, HB is a par'";

.'. DH = FG, and HK = GB. I. 34.

Now since HE is par' to KC, a side of the A DKC,

.'. KH : HD :: CE : ED. VI. 2.

But KH = BG, and HD = GF;

.*. BG : GF :: CE : ED. V. 1.

Again, because FD is par' to GE, a side of the A AGE,

.-. GF : FA :: ED : DA, vi. 2.

and it has been shewn that

BG : GF :: CE : ED,
,'., ex cequali, BG : FA :: CE : DA : v. 14.

.*. AB is divided similarly to AC. Q. e. f.

EXERCISE.

Divide a straight line internally and externally in a given ratio. Is
this ahcays possible ?



book vi. prop. 11. 327

Proposition 11. Problem.
To find a third j^roportional to two given straight lines.

K




BA D



Let A, B be two given straight lines.
It is required to find a third proportional to A and B.

Take two st. lines DL, DK of indefinite length, containing
any angle:

from DL cut off DG equal to A, and GE equal to B;

and from DK cut off DH equal to B. i. 3.

Join GH.
Through E draw EF par' to GH, meeting DK in F. i. 31.
Then shall H F be a third proportional to A and B.

Because GH is par' to EF, a side of the A DEF;

.-. DG : GE :: DH : HF. vi. 2.

But DG =^ A; and GE, DH each = B; Constr.

:. A : B :: B : HF;
that is, HF is a third proportional to A and B.

Q. E. P.
I

exercises.

1. AB is a diameter of a circle, and through A any straight Hne
is drawn to cut the circumference in C and the tangent at B in D :
shew that AC is a third proportional to AD and AB.

2. ABC is an isosceles triangle having each of the angles at the
base double of the vertical angle BAC ; the bisector of the angle BCA
meets AB at D. Shew that AB, BC, BD are three proportionals.

3. Two circles intersect at A and B ; and at A tangents are
drawn, one to each circle, to meet the circumferences at C and D :
shew that if CB, BD are joined, BD is a third proportional to CB,
BA.



328 Euclid's elements.

Proposition 12. Problem.
Tojlnd a fourth proportional to three given straight lines.




ABC

Let A, B, C be the three given straight lines. ,
It is required to find a fourth proportional to A, B, C.

Take two straight lines DL, DK containing any angle:
from DL cut off DG equal to A, GE equal to B;

and from DK cut off DH equal to C. i. 3.

Join GH.
Through E draw EF par' to GH. L 31.

Then shall HF be a fourth proportional to A, B, C.

Because GH is par' to EF, a side of the A DEF;

.-. DG : GE :: DH : HF. vi. 2.

But DG = A, GE = B, and DH - C; Consir.

.'. A : B :: : HF;
that is, HF is a fourth proportional to A, B, C.

Q. E. F.



EXERCISES.

1. If from D, one of the angular points of a parallelogram
A BCD, a straight line is drawn meeting AB at E and CB at F ; shew
that CF is a fourth proportional to EA, AD, and AB,

2. In a triangle ABC the bisector of the vertical angle BAC
meets the base at D and the circumference of the circumscribed circle
at E : shew that BA, AD, EA, AC are four proportionals.

3. Prom a point P tangents PQ, PR are drawn to a circle whose
centre is C, and QT is drawn perpendicular to RC produced : shew
that QT is a fourth proportional to PR, RC, and RT,



BOOK VI. PROP. 13. ,329

PPtOPOsiTiox 13. Problem.

To find a mecui lyroportional between two given straight
lines.




Let AB, BC be the two given straight lines.
It is required to find a mean proportional between them.

Place AB, BC in a straight line, and on AC describe the
semicircle ADC.

From B draw BD at rt. angles to AC. i. 11.

Then shall BD be a mean proportional between AB and BC.
Join AD, DC.

Now the L ADC being in a semicircle is a rt. angle; iii. 31.
and because in the right-angled A ADC, DB is drawn from
the rt. angle perp. to the hypotenuse,

.*. the A^ ABD, DBC are similar; vi. 8.

.•. AB : BD :: BD : BC;
that is, BD is a mean proportional between AB and BC.

Q. E. F.

I

EXERCISES.

1. If from one angle A of a parallelogram a straight line be
drawn cutting the diagonal in E and the sides in P, Q, shew that AE
is a mean proportional between PE and EQ.

2. A, B, C are three points in order in a straight line : find a
point P in the straight line so that PB may be a mean proportional
between PA and PC.

3. The diameter AB of a semicircle is divided at any point C,
and CD is drawn at right angles to AB meeting the circumference in
D ; DO is drawn to the centre, and CE is perpendicular to CD : shew
that DE is a third proportional to AG and DC.



330 Euclid's elements.

4. AC is the diameter of a semicircle on which a point B is taken
so that BC is equal to the radius : shew that AB is a mean propor-
tional between BC and the sum of BC, CA.

5. A is any point in a semicircle on BC as diameter; from D any
point in BC a perpendicular is drawn meeting AB, AC, and the cir-
cumference in E, G, F respectively; shew that DG is a third propor-
tional to DE and DF.

G. Two circles touch externally, and a common tangent touches
them at A and B : prove that AB is a mean proportional between the
diameters of the circles. [See Ex. 21, p. 219.]

7. If a straight line be divided in two given points, determine
a third point such that its distances from the extremities may be
proportional to its distances from the given points.

8. AB is a straight line divided at C and D so that AB, AC, AD
are in continued proportion; from A a line AE is drawn in any direc-
tion and equal to AC ; shew that BC and CD subtend equal angles at E.

9. In a given triangle draw a straight line parallel to one of the
sides, so that it may be a mean proportional between the segments of
the base.

10. On the radius OA of a quadrant OAB, a semicircle ODA is
described, and at A a tangent AE is drawn ; from O any line ODFE is
drawn meeting the circumferences in D and F and the tangent in E :
if DG is drawn perpendicular to OA, shew that OE, OF, CD, and OG
are in continued proportion.

11. From any point A, in the circumference of the circle ABE, as
centre, and with any radius, a circle BDC is described cutting the
former circle in B and C ; from A any line AFE is drawn meeting the
chord BC in F, and the circumferences BDC, ABE in D, E respec-
tively: shew that AD is a mean proportional between AF and AE.



Definition. Two figures are said to have their sides
about two of their angles reciprocally proportional, when
a side of the first is to a side of the second as the remaining
side of the second is to the remaining side of the first.

[Book VI. Def. 3.]



BOOK VI. PROP. 14.



331



Proposition 14. Theorem.

Parallelograms which are equal in area, and which have
one angle of the one equal to one angle of the other, have
their sides about the equal angles reciprocally proportional :

Conversely, parallelograms which have one angle of the
one equal to one angle of the other, and the sides about these
angles reciprocally proportional, are equal in area.

A F

/




Let the par""' AB, BC be of equal area, and have the
L DBF equal to the z. GBE:

then shall the sides about these equal angles be reciprocally
proportional,

that is, DB : BE :: GB : BF.
Place the par*"' so that DB, BE may be in the same straight
line;

.'. FB, BG are also in one straight line.

Complete the par" FE.
Then because the par'" AB = the par"" BC,
and FE is another par"",
.*. the par™ AB : the par"" FE :: the par™ BC : the par
but the par™ AB : the par™ FE :: DB : BE,
and the par™ BC : the par™ FE :: GB : BF,
.•. DB : BE :: GB : BF.
Conversely, let the l DBF be equal to the z. GBE,
and let DB : BE :: GB : BF.
Then shall the par™ AB be equal in area to the par™ BC.
For, with the same construction as before,
by hypothesis DB : BE :: GB : BF;
but DB : BE :: the par™ AB : the par™ FE, vi. 1,

and GB : BF :: the par™ BC : the par™ FE,
.". the par™ AB : the par™ FE : : the par™ BC : the par'" FE; v. 1 ,
.*. the par™ AB =the par™ BC.

Q. E. D,



I. 14.

Hyp.

•™ FE;
VI. 1.

V. 1.



^^'^ OF THE

^rrtSPTTr'W'RIsTTV^



^32 Euclid's elements.

Proposition 15. Theorem.

Trlamjles which are equal in area, aiul which have one
angle of the one equal to one aiigle of the other ^ liave their
sides about the equal angles reciprocally jyroportional :

Conversely^ triangles which have one angle of the one
equal to one angle of the other, and the sides about these
angles reciprocally prroportional, are equal in area.




Let the A** ABC, ADE be of equal area, and liave tlie
L CAB equal to the z_ EAD :

then shall the sides of the triangles about these angles be
reciprocally proportional,

that is, CA : AD :: EA : AB.
Place the A^ so that CA and AD may be in the same st. line;
.'. BA, AE are also in one st. line. i. 14.

Join BD.
Then because the A CAB = the AEAD, Hyp.

and ABD is another triangle;
.'. the A CAB : the A ABD :: the AEAD : the A ABD;
but the A CAB : the A ABD : : CA : AD, VI. ].

and the AEAD : the A ABD : : EA : AB,

.-. CA : AD :: EA : AB. v. 1.

Conversely, let the l. CAB be equal to the /_ EAD,
and let CA : AD :: EA : AB.
Then shall the A CAB = A EAD.
For, with tlie same construction as before,
by hypothesis CA : AD :: EA : AB;

but CA : AD :: the A CAB : the A ABD, w. 1.
and EA : AB :: the AEAD : the A ABD,
.'. the A CAB : the A ABD :: the AEAD : the A ABD; v. 1.
.". the A CAB -^ the AEAD. q. e.b.



BOOK VI, PROP. 15. .333

KXERCISES.

ON Propositions 14 and 15.

1. Parallelofjravis tvhich are equal in area and ivhich have their
sides reciprocally proportional, have their angles respectively equal.

2. Triangles which are equal in area, and which have the sides
about a pair of angles reciprocally proportional, have those angles equal
or supplemental^.

3. AC, BD are the diagonals of a trapezium which intersect in
O ; if the side AB is parallel to CD, use Prop. 15 to prove that the
triangle AOD is equal to the triangle BOC.

4. Prom the extremities A, B of the hypotenuse of a right-
angled triangle ABC lines AE, BD are drawn perpendicular to AB,
and meeting BC and AC produced in E and D respectively: employ
Prop. 15 to shew that the triangles ABC, ECD are equal in area.

5. On AB, AC, two sides of any triangle, squares are described
externally to the triangle. If the squares are ABDE, ACFG, shew
that the triangles DAG, FAE are equal in area.

6. ABCD is a parallelogram; from A and C any two parallel
straight lines are drawn meeting DC and AB in E and F respectively;
EG, which is parallel to the diagonal AC, meets AD in G : shew that
the triangles DAF, GAB are equal in area.

7. Describe an isosceles triangle equal in area to a given triangle
and having its vertical angle equal to one of the angles of the given
triangle.

8. Prove that the equilateral triangle described on the hypotenuse
of a right-angled triangle is equal to the sum of the equilateral
triangles described on the sides containing the right angle.

[Let ABC be the triangle right-angled at C ; and let BXC, CYA,
AZB be the equilateral triangles. Draw CD perpendicular to AB ;
and join DZ. Then shew by Prop. 15 that the A AYC = the A DAZ ;
and similarly that the ABXC = the aBDZ.]



33



KUCIJD'S KLKiMKNTK.



Proposition 16. Theorem.

If four straight lines are p^'oportiorial, the rectangle
contained by the extremes is equal to the rectangle contained
by the means:

Conversely, if the rectangle contained by the extremes is
equal to the rectangle contained by the means, the four
straight lines are proportional.



B C



D E G



Let the St. lines AB, CD, EF, GH be proportional, so that
AB : CD :: ^F : GH.
Then shall the rect. AB, GH =the rect. CD, EF.
From A draw AK perp. to AB, and equal to GH. i. 11, 3,
From C draw CL perp. to CD, and equal to EF.
Complete the par""' KB, LD.



Then because AB : CD :: EF : GH



Hyp.
Constr.



KB, LD are



and EF = CL, and GH = AK;
.*. AB : CD :: CL : AK;
that is, the sides about equal angles of par
reciprocally proportional;

.•.KB = LD. VI. 14.

But KB is the rect. AB, GH, for AK = GH, Constr.
and LD is the rect. CD, EF, for CL- EF;
.'. the rect. AB, GH= - the rect. CD, EF.

Conversely, let the rect. AB, GH = the rect. CD, EF:

then shall AB : CD :: EF : GH.
For, with the same construction as before,
because the rect. AB, GH =:the rect. CD, EF
and the rect. AB, GH = KB, for GH = AK,
and the rect. CD, EF = LD, for EF = CL :
,•. KB = LD;



Hyp.

Constr.



BOOK YI. PROP. 17.



:«o



that is, the par"'' KB, LD, which have the angle at A equal
to the angle at C, are equal in area;
.'. the sides about the equal angles are reciprocally
proportional :

that is, AB : CD :: CL : AK;
.-. AB : CD :: EF : GH.

Q. E. D.



Proposition 1'



Theorem.



If three straight lines are proportional the rectangle con-
tained hy the extremes is equal to the square on the mean:

Conversely, if the rectangle contaiiied hy the extremes is
equal to the square on the mean, the three straight lines are
proportional.



B



Let the three st. lines A, B, C be proportional, so that
A : B :: B : C.
Then shall the rect. A, C be equal to the sq. On B.
Take D equal to B.
Then because A : B : : B : C, and D = B ;
.•. A : B :: D : C;
.*. the rect. A, C = the rect. B, D;
but the rect. B, D = the sq. on B, for
.', the rect. A, C =^ the sq. on B.
Conversely, let the rect. A, C = the sq. on B :
then shall A : B : : B : C.
For, with the same construction as before,
because the rect. A, C = the sq. on B,
and the sq. on B — the rect. B, D
.'. the rect. A, C = the rect. B, D
.". A : B :: D : C,
that is, A : B :: B : C.



16.



for D



Hyp.
YI. 16.



CJ. E. D.



336 EUCLID'S ELEMENTS.

EXERCISES.
ON PrOPOSITIOKS 16 AND 17.

1. Apply Proposition 16 to prove that if two chords of a circle
intersect, the rectangle contained by the segments of the one is equal
to the rectangle contained by the segments of the other.

2. Prove that the rectangle contained by the sides of a right-
angled triangle is equal to the rectangle contained by the hypotenuse
and the perpendicular on it from the right angle.

3. On a given straight line construct a rectangle equal to a given
rectangle.

4. ABCD is a parallelogram; from B any straight line is drawn
cutting the diagonal AC at F, the side DC at G, and the side AD pro-
duced at E : shew that the rectangle EF, FG is equal to the square
on BF.

5. On a given straight line as base describe an isosceles triangle
equal to a given triangle.

0. AB is a diameter of a circle, and any line ACD cuts the circle
in C and the tangent at B in D ; shew by Prop. 17 that the rectangle
AC, AD is constant.

7. The exterior angle A of a triangle ABC is bisected by a straight
line which meets the base in D and the circumscribed circle in E :
shew that the rectangle BA, AC is equal to the rectangle EA, AD.

8. If two chords AB, AC drawn from any point A in the cir-
cumference of the circle ABC be produced to meet the tangent at the
other extremity of the diameter through A in D and E, shew that the
triangle AED is similar to the triangle ABC.

9. At the extremities of a diameter of a circle tangents are drawn ;
these meet the tangent at a point P in Q and R : shew that the rect-
angle QP, PR is constant for all positions of P.

10. A is the vertex of an isosceles triangle ABC inscribed in a
circle, and ADE is a straight line which cuts the base in D and the
circle in E ; shew that the rectangle EA, AD is equal to the square on
AB.

11. Two circles touch one another externally in A; a straight line
touches the circles at B and C, and is produced to meet the straight
line joining their centres at S : shew that the rectangle SB, SC is
equal to the square on SA.

12. Divide a tiiangle into two equal parts by a straight line at
right angles to one of the sides.



BOOK VI. PROP. 18. 337

Definition. Two similar rectilineal figures are said to
be similarly situated with respect to two of their sides
when these sides are homologous. [Book vi. Def. 5.]



Proposition 18. Problem.

On a given straight line to describe a rectili^ieal Jigure
similar and shnilarly situated to a given rectilineal figure.




'^



Let AB be the given st. line, and CDEF the given rectil.
figure: first suppose CDEF to be a quadrilateral.

It is required to describe on the st. line AB, a rectil.
figure similar and similarly situated to CDEF.

Join DF.
At A in BA make the l. BAG equal to the L DCF, i. 23.
iind at B in AB make the l. ABG equal to the l CDF;

.*. the remaining z. AG B = the remaining z.CFDji. 32.
and the AAGB is equiangular to the ACFD.

Again at B in GB make the z. GBH equal to the z. FDE,

and at G in BG make the ^ BGH equal to the l DFE; l. 23.

.'. the remaining z. BHG = the remaining z. DEF ; i. 32.

and the A BHG is equiangular to the A DEF.

Then shall ABHG be the required figure.

(i) To prove that the quadrilaterals are equiangular.
IHL Because the z. AGB = the z. CFD,

^B and the ^ BGH = the z. DFE; Constr.

^^K .". the whole ^ AG H = the whole z_ CFE. Ax. 2.

lip Similarly the ^ ABH -^ the ^ CDE ;

and the angles at A and H are respectively equal to the
angles at C and E ; Constr.

the fig. ABHG is equiangular to the fig. CDEF.



"O'



H. E. 22



338 EUCLID'S ELEMENTS.




'd



(ii) To prove that the quadrilaterals have the sides about
their equal angles proportional.

Because the A" BAG, DCF are equiangular;

.*. AG : GB :: CF : FD. VL 4.

And because the A^ BGH, DFE are equiangular;
.•. BG : GH :: DF : FE,
.'., ex cequali, AG : GH :: CF : FE. v. 14.

Similarly it may be shewn that
AB : BH :: CD : DE.
Also BA : AG :: DC : CF, VL 4.

and GH : HB :: FE : ED;
.'. tlie figs. ABHG, CDEF have their sides about the equal
angles proportional ;

.'. ABHG is similar to CDEF. Def. 2.


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