research work in geometry

GEOMETRY TERM PAPER

BY

DR M.D.RAMAKRISHNA GUPTA MBBS FPC

17 BARNARD WYND

OAKERSIDE PARK PETERLEE Co DURHAM SR8 1LT ENGALND UK

INTRODUCTION

Geometry is a dynamic branch of mathematics that also serves as a creative tool for engineers, artists, and

Architects. Geometry is providing necessary tool for Art and Architecture. Geometry is both elegantly

simple and infinitely profound. Many professionals find the geometric shapes are very interesting. The

challenging topic when achieved will bring joy for all and new guidance to the readers I feel Geometry is

always a reliable and reassuring resource for the student throughout the school year, and again when the

student prepares for college career. I really enjoyed my school years in doing theorems and riders.

Problems after struggling for a period to solve I used to get full of joy after finding the solution to the

problems. Even now I feel joy when I find the solution to the new tasks in Geometry

In the history of the mathematics there are three problems considered tobe impossible

persisting nearly over two thousand years. The Greek problems ofantiquity were a set of geometric

problems whose solution was sought using compass and straight edge only

They are:

1. SQUARING THE CIRCLE

2. DOUBLING THE CUBE

3 THE ANGLE TRISECTION.

ALHAZEN’S BILLIARD PROBLEM

This is another Geometrical antiquity.

In a given circle, find an isosceles triangle whose legs pass through two given points inside the circle. This

can be restated as: from two points in the plane of a circle, draw lines meeting at the point of

the circumference and making equal angles with the normal vector at that point. The problem is called the

billiard problem because it corresponds to finding the point on the edgeof a circular "billiard" table at which

a cue ball at a given point must be aimed in order to carom once off theedge of the table and strike another

ball at a second given point. The problem is equivalent to the determination of the point on a spherical

mirror where a ray of light will reflect in order to pass from a given source to an observer. It is also

equivalent to the problem of finding, given two points and a circle such that the points are both inside or

outside the circle, the ellipse whose foci are the two points and which is tangent to the given circle.

The problem was first formulated by Ptolemy in 150 AD, and was named after the Arab scholar Alhazen,

who discussed it in his work on optics. The problem is insoluble using a compass and ruler construction

because the solution requires extraction of a cube root (Elkin 1965, Reide 1989, Neumann 1998). This is

the same reason that the cube duplication problem is insoluble.

SQUARING THE CIRCLE

Thanks to Dan Gaddy and Chopin who has given the following constructionin this venture The author

Claim The circle area with P as centre is equal to the square constructed on TR.

Thanks to Dan Gaddy

.

Here in this construction given by Dan Gaddy

With P as centre and PA=PC as radius a circle is drawn Then APC is a right angle . At the point A PC parallel a line is drawn. B is appoint on that line so that AC=AB Join BP and extends to meet the circumferenceof the circle at Q,BP=BR and BQ=BS, Now M is the midpoint of AS. Let TRis 90 degree to AS and TR is the side required to construct the square which is equal in area to the circle with P as centre If we consider thePA is 1 and AC=AB= Root 2

3. DOUBLING THE CUBE:

Cube duplication or doubling is also called the Delian problem, is one of the Geometric problems of

Antiquity given the length of an edge of a cube, that a second cube to be constructed having double

The volume of the first cube . The only tools allowed for the construction are the classic straight edge An

idealized mathematical object having a rigorously straight edge which can be used to draw a line.

Although geometric constructions are sometimes said to be performed with an marked ruler and compass

The term straightedge is preferable to ruler since markings on the straightedge (usually assumed to be

present on a ruler) are not allowed by the classical Greek rules.

Doubling the cube done with the marked ruler in the past . This was as follows

We can mark a ruler and we can trisect the angle and duplicate the cube, but this approach was not

permitted under ancient Greek rules. Here's a nifty construction using amarked ruler (Robin

Hartshorne, Geometry: Euclid and Beyond, Springer, 2000, p. 270; by the way, Hartshorne has a pretty

readable explanation of why this and other constructions are impossible,but it will take serious study to

understand it.)

http://mathworld.wolfram.com/Ruler.html

Let AO be the edge of the cube to be duplicated, call its length 1. Construct OB perpendicular to AB and

OC at a 30 degree angle to OB as shown. Mark off DE on a ruler, with DE = AO

Slide the ruler so it passes through A, and points D and E rest on OB and OC as shown. AD is the edge of

the doubled cube (AD = AO times cube root of two).

Proof:Using similarity, we have x/1 = (x+1)/(m+1) or xm + x = x + 1 or xm = 1

Also, h/1 = sqrt(3)m/(m+1) Since xm = 1, h = sqrt(3)(1/x)/(1/x+1) = sqrt(3)/(1+x) By the Pythagorean Theorem, we have x2 = 1 + h2

x2 = 1 + h2 = 1 + (sqrt(3)/(1+x))2 = 1 + 3/(1+x)2

Clearing fractions, we get x2 (1+x)2 = (1 + x)2 + 3 Expanding, we have x2 + 2x3 + x4 = 1 + 2x + x2 + 3 Thus 2x3 + x4 = 4 + 2x And x3(2 + x) = 2(2 + x) Thus x3 = 2

This is the proof available in the existing literature done by Robin Hartshome mathematician.

Hartshorne, Geometry: Euclid and Beyond, Springer, 2000, p. 270;

My Father has attempted and along with his other construction derived similar picture in a different way

which you can see in further pages But I am not able to derive the proof for that construction yet.

My beloved father did an ISOSCLES TRAPEZIUM THEOREM and along with thatconstruction he did

the following additional construction which gives the information about doubling the cube and we have not

developed the proof for this construction yet.

The angle GMH is bisected. And 60 degree and 90 degree lines are drawn on the point M . The line DM is extended X so that XM=MA=AR=RB=BD

Copy the angle ZMG 30 dgree+half the angle of GMH 5degree at the point X That is The angle ZXM =35

degree. Copying angle can be done by the compass easily. That 35 degreeline cuts the perpendicular line

at Y and meet the 60 degree line at Z. My father believe the cube on XY=2 cube on MX

Also note that MX =YZ. That means this doubling the cube also done. The angle ZXM=angle ZMG+half

Of angle GMH . He told me in 1980 when he came to UK . We were not ableto develop the proof yet.

This could be a hypothesis at this stage. Needs verification and proof later if it is correct .

MY SELECTED TOPIC

TRISECTION OF AN UNKNOWN GIVEN ANGLE WITH COMPASS AND STRAIGHT EDGE

When I was mathematics student in Sahyadri college Shimoga I came to know these impossible task in

Geometry. I did discuss these impossible tasks with my beloved father who has taken this task very

Seriously and he is of the opinion it is possible to do the trisection within the restricted guide lines using

the straight edge and a compass. He started working on this task from 1960 on wards.

HISTORY OF TRISECTION

As per the Author Yates Robert C. The Trisection problem was publishedin 1942 by the Franklin Press

Inc of Buton Rouge , Lousiana when the author was teaching at LousianaState Univeristy. In 1947 a

photo-lithoprint reproduction was issued by Edwards Brothers, Inc., Lithoprinters. But no solutions found.

Earlier attempts by several mathematicians:

Hippias of Elis who lived in the fifth century B.C. was one of the first to attempt to solve the Trisection

problem. He devised a curve called the Quadratrix by means of which he was able to give an exact

solution to the problem .But he was not able to do with compass and straight edge

His followers several great mathematicians Archimedes, Nicomedes, Pappus, Leonardo da vinchi , Durer,

Descartes, Ceva, Pascal , Huygens , Leibniz, Newton, Maclaurin, Mascheroni, Gauss , Steiner, Chasles,

Sylvester, Kempe, Klein, Dickson-ali and hundreds more tackled the problem worked for several years and

made advances in the mathematical science subject.

The angle trisection is the division of the arbitrary angle into three equal parts. It was one of the three

grometric problems of antiquity for which solutions using only compass and straight edge were sought. The

problem was algebraically proved impossible by Wantzel (1836).

Although trisection is not possible for a general angle using a Greek construction, there are some specific

angles, such as and radians ( and , respectively), which can be trisected. Furthermore,

some angles are geometrically trisectable, but cannot be constructed in the first place, such as

(Honsberger 1991). In addition, trisection of an arbitrary angle can beaccomplished using a marked

ruler (aNeuises construction)) as illustrated above (Courant and Robbins 1996).

An angle can also be divided into three equal parts using the quadratrixof Hippias or Trisectrix

.

Algebraic Formulation.Given the angle AOB 30, let us suppose one of the trisecting lines to be OT, Fig. 1,

so that TOB=Select an arbitrary length on OA as the unit distance and draw the parallel AC to OT, meeting

OB extended at C. Then angle DCOo.

Now draw OD equal to the unit length so that triangle AOD is isosceles with base angles 20, angles DAO

and AOT being equal since they are alternate interior angles. It is evident, since angle ADO is the sum of

the opposite interior angles of triangles DCO and angle DCO 0,that angle DOC-0.Therefor,-.triangle DCO

is isosceles and DC = DO =1.Let x denote the distance OC, 2y the distance AD, and a the projection of OA

upon the side OB. From similar triangles CMD, CNA, and CLO, all right triangles with equal angles at C

alternate interior angles.It is evident, since angle ADO is the sum of the opposite interior angles of triangles

DCO and angle DCO 0,that angle DOC-0.Therefor,-.triangle DCO is isosceles andDC = DO =1.Let x

denote the distance OC, 2y the distance AD,and a the projection of OA upon the side OB. From similar

trianglesCMD, CNA, and CLO, all right triangles with equal angles at C,we find:

x/2= (x+ a)/(1+2y)=(1+y)/x,

which givex2=2±2y and 1±2y =2(x+a)/x.

From these we eliminate y by substitution to obtain:

X2-1 =-2(x±a)/x or X3-3x- 2a=0

This relation, as will be seen shortly, is fundamental to the problem and is called the Trisection Equation.It

is a cubic equation with the term in X2 missing.

Plato and Archimedes were trisecting the angle by means of compass andgraduated ruler as follows but

they are not accepted. The others came forward with some sort of instruments or apparatus to trisect the

angle. The following are the pictures of those apparatus.

Three bar apparatus

Cev’s panlograph

These instruments trisection were not acceptable as the solutions required to use straight edge and a compass only.

My father’s constructions:-

Here in this construction MA=AR=RB=BD and With the same as radius with A,R nd B as centre 3 circles

drawn and they intersect at G and H It is interesting to note that DG=MH=2/3 of MD=DK=MLalso

MK=KL=LD=1/3MD

1. To find out the trisecting points 2/3 arc DK and ML radius will help to trisect other angles or

not? This is a straight angle MRD 180 degree. Trisection of the semicircle is cut the arc with the

same radius get 60 degree easily and that point coincided with this2/3 arc. Hence the next step is to

change the straight angle MRD 180 degree into different angles andto see whether that is going to

help us to trisect other angles or not needs to be studied.

2. Now we have created an angle MRD .With centre points A and B of MRand RD two semi circles

Constructed and the DK 2/3 of MD as radius an arc is drawn to cut the semicircle at G

GR supposed to trisect the angle MRD and we are not able to justify the construction with the

proper proof in this construction so far.

3. By looking at my father’s above construction the following theorem has been derived .

THEOREM No. 17

DASAPPAGUPTA’S ISOSCLES TRAPEZIUM THEOREM

In the isosceles trapezium if the parallel lines are in 1:4 ratio the diagonals are equal and also they

trisect the base angles of the isosceles trapezium

Data; GMDH is an isosceles trapezium with GH II MD and GH:MD is 1:4

Construction: MD line is divided into 4 equal parts MA=AR=RB=BD, MK=KL=LD =1/3 of the MD With

A,R and B as centres and MA=AR=RB=BD as radius 3 circles are drawn. TheA circle and R circle

intersect in G , and R circle and B circle intersect at H, Join lines GH,MG,DH,GD and

HM . GD and MH are two diagonals .

It is required to prove: HM=GD and they trisect the base angles of the isosceles triangle respectively.

Proof

In this construction Triangles AGR and GRH and HRB all are equal equilateral triangles and

They are all equal in all respects. Angles HGR=GRA=60degrees Therefore GH is parallel to AR and MD

The triangle HRD angle HRD=60 degree Angle RHD=90 degree and hence the remaining angle must be 30 degree

Therefore both base angles of isosceles trapezium are equal to 30 degree.

In triangles HRM and GRD angles GRD=HRM=120 degrees by construction, GR=RH= radius of the same circle also MR=RD

Therefore triangles HMR and GDR are equal in all respects. Therefore GD=HM, Therefore the diagonals are equal

GRD is an 120 degree obtuse angle triangle with 2GR=RD i.e 1:2 ratio The remaining 60 degrees has to be shared in

1:2 ratio Angle RDG+angle RGD= 60 degrees one should be 40 degree and the remaining must be 20 degree

as per our Obtuse angle triangle theorem

Since GH is parallel to MD angles HGD=GDR=20 degrees But we know angle HDR is 30 degree and angle RGH is 60

Degrees . Therefore GD trisect the angle HDR into 10 and 20 degree and angle RGH into 40 and 20 degrees.

There fore the diagonals GD and HR trisects the angles on the both ends of the lines.

Now in triangles GMD and HDM , GM= HD MD is common and angles GMD=HDM=30 degrees

Therefore both triangles are equal in all respects Therefore the lines GD and MH are also equal

Therefore the diagonals GD and MH of the Isosceles trapezium GMDH are equal.

Hence my father’s new theorem came to existence from 1980 when he is onholiday in UK in our house.

My Father’s few constructions are added here and his correspondence with professor also added.

My father did some correspondences with Professor Johnson of New CastleUniversity in 1979 when he came to UK and professor Johnson says the proof is not correct and he later changed the proof also he changed the construction also.

All his later constructions and the details are here with enclosed

SRI M.R.DASAPPAGUPTA, LANDLORD ,SRINGERI KARNATAKA INDIA

My father Sri M.R.DasappaGupta and my self also attempted several constructions and started trying to

develop a geometrical proof since 1960 on wards . While attempting to solve this Trisection we derived

several new theorems and several new constructions

1

This is one of the construction to increase the given angle into one and a half times and three times

Here the trisection noticed using the isosceles triangle and circumcentre method

2. This is the construction to derive the proof for obtuse angle triangle to prove if the sides are in 1:2 ratio

the other angles also in 1:2 ratio and also to prove the angle formed atthe circum centre also trisected

Here P is the circumcentre of BAC and Q is the circum centre of XAC. Since XAC is 120 degree angle in this construction QAX and QAC=60 degree, Since QX=QA=QC as it is circum radius. Both QXA and QCA are two equal equilateral triangles and QXAC is a rhombus, Angle QXC=Angle QCX=30 degree=angle AXC=ACX. Angle CXA=1/2 of angle AQC as the angle at

the centre and angle at the circumference. Here in this picture Let BC and PA intersect at the point S Then the triangles CSA and PAC are twosimilar triangles. PAC is an isosceles triangle and one of the base angle is the same angle PAC and CAS. Angle APC=SCA=40 degree The remaining angle ASC angle PCA must be equal to 70 degree

As per our theorem

Angles APC =angle BCA=40degree, Angle BPA at the centre and angle BCA atthe circumference

Hence angle BPA=80 degree when the angle BCA is 40 degree. Since the angle APC =40 degree and BPA

is 80 degree the Line AC trisects the angle BPC

3

In this construction verifies the angles and the circumcentre of the both triangles on each side of the

semicircle

4

Trisection inside the obtuse angle by practical measurements noticed.

5

A method of trisection tried with smaller angle about 90 degree the obtuse angle came with 135 degree and

the remaining angle came 15 and 30 degree verified

6 Trisecting the angle in the similar method three different sizes verified.

By the properties of the circumcentre and the isosceles triangle it is possible to increase the given angle to

one and a half times and also to 3 times.

Data:-Let Angle BAC is given angle, Let ABC is an isosceles triangle withAB=AC. Let O be the circum

centre of the ABC triangle. So OA=OB=OC radius, Join BO and extendedto meet AC in X . Let the

perpendicular bisector of BX meet AC on D

To Prove that Angle BDC is equal to 3times the angle BAC

Proof: Since the O is a circum centre of an isosceles triangle ABCwith angle BAC as the apex angle

OA bisects the angle BAC and OA=OB=OC radius of the circum circle.

Hence Angles OAB=OAC =OBA=OCA. In the triangle BXA angle BXC= AnglesXBA+XAB The

angles OBA+OAB+OAC= angles 1/2BAC+ BAC= Angle BXC The angles BAC=BAX same angle

Angle BXC= 1+1/2 Angle BAC

Hence the given angle increased to one and half times.

In triangle BDX angle BDC= Angles DBX+DXB and they are equal as DB=DXas D is on the

perpendicular bisector of BX

Angles DXB and BXC are same angle=angle DBX

Hence Angle BDC=Angles DBX+ angle DXB

Therefore Angle BDC= 3 times the angle BAC

Hence the given angle increased to 3 times

This construction attempted in 1968 with a view to find out the wayof finding the reverse of the

construction

The one final construction my father and myself together did it as follows. We derived the construction into

a theorem and called TRISECTION THEOREM

Theorem No. 11 TRISECTION THEOREM

In concentric arcs or in circles the 1/4th of the 4th arc is equal to 1/3of the third arc where the arc radius

are equidistant ( in 1,2,3 and 4 value). This will trisect the subtended given angle also.

Here we have used only the compass and straight edge with out any measurement on the straight edge..

While trying to derive this hypothesis at initial stages we did not got the proof and later we got the proof

also as explained in the next few lines. We found this will trisect anyangle.

DATA

MRD is the given angle and with R as centre four concentric circles constructed with 1cm , 2cm,

3 cm, and 4 cm radius. RP 3cms radius and RM=RD as 4cms radius. Let Angle MRD is θ

CONSTRUCTION

Angle MRD is bisected ER and angle ERM is bisected by BR. With MB as radius and P as centre on the

third circle an arc is drawn to cut the third arc on the point C Join CR,

TO PROOVE THAT: CR is going to trisect the angle MRD

PROOF: Arc MD= 2 π4θ /360 Because Radius is 4 cms

Hence ¼ of Arc MD= ¼ x2 π4θ /360 = 2 πθ /360

Arc PX= 2 π3θ /360 Because Radius is 3 cms

Hence 1/3 of ARC PX= 1/3X2 π3θ /360=2 πθ /360

Therefore 1/4th of the 4th arc= 1/3rd of the 3rd arc =2 πθ /360

Arc MB=Arc PC by construction Therefore ARC , PX is trisected at the point C

Arc PC=Arc CY=Arc YX Hence CR trisect the angle MRD and also the arc PX

This is the correct trisection of an unknown given angle by using straight edge and compassonly by

myself and my beloved father Sri M .R. Dasappa Gupta in 1970 in Sringeri TownChickmagalore

district South India before me coming to UK

Another hypothesis I produced is :

In an obtuse angle of more than 105 degree triangle if the two sides arein 1;2 ratio the angles also are in 1;2 ratio.

To check the hypothesis practically angle measured on the drawings found very interesting and supported

my hypothesis.

Obtuse angle theorem and its applications

And verifications by direct measurements by case by case

In an obtuse angle with an angle 105 degree or more if the sides are in 1:2 ratio the remaining two angles

are also in 1:2 ratio

Data:

BAC is an obtuse angle triangle with two sides AB=2AC are in 1:2 ratio

To prove that: Angle ABC: Angle ACB is also in 1:2 ratio

Construction: A triangle is constructed with AB=2AC and angle BAC is 111 degree. With an obtuse angle

of 111 degree and sides are in 1:2 ratio as per the data.

Proof by direct measurement

Angles measured; Angle ABC is found 23 degree and Angle ACB is found46 degree Angle BAC is 111 degree And Side AB= 2AC

That means the angle measured is also 1:2 ratio by direct measure

Proof by Exhaustion (Case by Case)

DATA: BAC is the obtuse angle triangle more than 120 degree

AC: AB is 1:2 That is to say AB =2AC

By measurement we notice angle ABC is 16 degree and angle ACB is 32 degree

The angles also are in the same ratio ABC: ACB is also 1:2 ratio

Example case number 1


In this example Angle ABC is 8 degree and ANGLE ACB is 16 degree


In this example we have increased the obtuse angle further and the angles are 6 degree and 12 degree and the sides are in 1:2 ratio

Example case number 4 same as 1 with angles 16 and 32 degree and two sides also 1:2


Here in this example we have two sides are in 1:2 ratio and angle measured as 23 and 46

ANGLE ABC IS 23 degree and Angle ACB is 46 degree Angle BAC should be 111 degree

Side AB= 2AC


The angles are ABC is 19 degree and ACB is 38 degree found proportionate to the sides 1:2

The obtuse angle BAC is 123 degree

.


Obtuse angle 150 degree ratio of angles 10:20 and ratio of sides also 1:2


Obtuse angle increased to 162 degree

Angles measured came to 6 degree and 12 degree keeping the sides in 1:2 ratio


We have increased the obtuse angle to 171 degree nearer to the maximum

We have noticed 3degree on one side and 6 degree on the other angle still maintaining the 1:2

Ratio keeping the sides also 1:2 ratio


Obtuse angle reduced to 120 keeping the sides in1:2 ratio

Angles came to 20 and 40 degree

By looking at the above examples we can appreciate the ratio of the angle to the sides in an obtuse angle triangle over 105 degree is proportional to 1:2.

This is a kind of proof by examples and also proof by exhaustion. Because we tried big obtuse angles also

and we have tried just 120 degree obtuse angles also. We came to know by the trial and errors the valuation

of 1:2 ratio may not hold good for the angles below 105 degree obtuse angles as we have not checked.

Hence we said this hold good only with obtuse angle triangles with over 105 degree angles only. In

addition to the above examples we have attempted to write the following triangles and found it is true. In the

following examples also Keeping the BA=2BC in the BAC triangle we noticed the following anglesWe

noticed the angle ACB=2ABC

By increasing the angle ABC by one degree we noticed the angle ACB raises by 2 degree and obtuse angle

increases by 3 degree in the following table keeping BA=2BC

Angle BAC obtuse angle Angle ABC Angle ACB

105 25 50

111 23 46

114 22 44

117 21 42

120 20 40

123 19 38

126 18 36

129 17 34

132 16 32

135 15 30

138 14 28

141 13 26

144 12 24

147 11 22

150 10 20

153 9 18

156 8 16

159 7 14

162 6 12

165 5 10

168 4 8

171 3 6

174 2 4

Example case number 8 repeated picture shows the same results. 6 degreeand 12 degree only

I tried to construct an obtuse angle triangle with 105 degree and keep the ratio of the sides 1:2

I surprisingly got the measurements of the angles to 25 and 50 degrees.

Since we have several examples we conclude the proof by exhaustion in obtuse angle triangles with over

105 degree onwards only if the sides of the triangle are in 1:2 ratio the angles are also in 1:2 ratio.

After seeing these examples and several trials I believe my hypothesis appears to be correct

This make me to think it is possible in the obtuse angle triangle of 105Degree also when we keep the ratio

of the sides 1:2

I tried with the different measurement for the sides keeping 1:2 ratios of the sides and keep the obtuse angle

120 degree and we got the measurements 20 degree and 40 degree angles with 1:2 ratios also.

But this is not tried on the obtuse angle below 105 degree so far. We know it is not possible in the right

angle triangle and other triangles.. The theorem is only for the obtuse angle above 105 degree and only if

the sides are in 1:2 ratio on

Here in this example the obtuse angle is 159 degree and other angles are 7 degree and 14 degree by

measurement. Keeping the ratio of the sides 1:2 ratio. Sum of the two angles 7+14=21 degree external

angle noted

In modern times, more than 2000 years after they were formulated, stillthe three ancient problems proved

Impossible to solve using only COMPASS AND STRAIGHT EDGE.

But our constructions are with compass and unmarked straight edge only.

Hippias of Elis, who lived in the Fifth century B. C.,was one ofthe first to attempt to solve the

Trisection Problem. The very same obstacles presented themselves to thwart his efforts but,

freeing him-self from the Platonian rules, he devised a curve called theQuadratrix,to be

discussed later, by means of which he was able to given exact solution to the problem. But,

we must understand, it was not achieved by straightedge and compasses alone.

A partial list of his followers will show you what a powerful pull ithad upon the attention of

the great and the near-great.Archimedes, Nicomedes, Pappus, Leonardo daVinci,

Darer,Descartes, Ceva, Pascal, Huygens, Leibniz, Newton,Maclaurin,Mascheroni, Gauss,

Steiner, Chas les, Sylvester, Kempe, Klein, Dick-sonali of these, andhundreds more, attacked

the problem directly or created the mathematics by which substantialadvances could be made

toward a full understanding of the situation. Archimedes did the firsttrisection and he used

the straight edge to find out the points on the line and the circlesince he has used the straight

edge to measure and find out the points this was not followed with theconditions imposed.

Now draw OD equal to the unit length so that triangle AOD is isosceles with base angles 20, angles DAO

and AOT being equal since they are base angles

FIG. Ialternate interior angles.It is evident, since angle ADO is the sum of the opposite interior angles of triangles DCO and angle DCO 0,that angleDOC-0.Therefor,-.triangle DCO isisosceles and DC = DO =1.Let x denote the distance OC, 2y the distance AD,and a the projection of OA upon the side OB. From similar trianglesCMD, CNA, and CLO, all right triangles with equal angles at C,we find:

of Constructibility.If we are giventwo line segments, a and b, these segments can be added, subtracted,

multiphed, and divided geometrically, using only straightedge and

10THE TRISECTION PROBLEMcompasses.These operations on the given quantities are calledrational.The first two need no explanation and are evident from themeanings of the words sum and difference.The multiplication of aand bis effected by drawing the line PQ, Fig. 2A, at any angle withPR and constructing the parallel line to produce similar triangles asof;I

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