Geometry

Vladimir V. Kisil

Chapter 1
General Information

This is an online manual designed for students. The manual is available at the moment in HTML with frames (for easier navigation), HTML without frames and PDF formats. Each from these formats has its own advantages. Please select one better suit your needs.

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• Calculus II.

• Geometry.

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1.2  Warnings and Disclaimers

• These Web pages are designed in order to help students as a source of additional information. They are NOT an obligatory part of the course.

• The main material introduced during lectures and is contained in Textbook. This interactive manual is NOT a substitution for any part of those primary sources of information.

• It is NOT required to be familiar with these pages in order to pass the examination.

• The entire contents of these pages is continuously improved and updated. Even for material of lectures took place weeks or months ago changes are made.

Course Outline

Chapter 0
Seven Top Reasons to Enjoy Geometry

There are many reasons to enjoy Geometry. No, ``I need to pass the exam'' is not among them. These reasons are of a much pleasant nature:

• Geometry is elementary. To understand even most advanced results you need only to know simple notions like lines, circles, triangles, etc.

• Geometry is beautiful. The inner harmony of geometrical constructions is explicit.

• Geometry is real. It describes the property of the world around us.

• Geometry is principal. It was the first field of mathematical knowledge which set up a model for all other branches of mathematics.

• Geometry is reach. You may meet all variety of mathematical tools employed in geometry.

• Geometry is modern. Geometric results are used in all contemporary fields of mathematics and are source of inspiration for many new theories.

• Geometry is surprising. One is greatly impressed by the unexpected deep and beauty of geometrical results.

Exercise 1 Illustrate each of the above statements by at least one geometrical construction or theorem.

Exercise* 2 Give at least one more good reason to enjoy geometry.

Chapter 1
Points and Lines Connected with a Triangle

1.1  The extended Law of Sines

Theorem 1 [Law of Sines] For a triangle ABC with circumradius R

a sinA =  b sinB =  c sinC = 2R

PROOF. The proof is based on the Theorem A.0.1 from the Useful Theorem Chapter . ^[¯]

Exercise 2 For any triangle ABC, even if B and C is an obtuse angle, a=bcosC+c cosB. Use the Law of Sines to deduce the additional formula

sin(B+C)=sinB cosC + cosB sinC.

Exercise 3 In any triangle ABC,

a(sinB −sinC) + b(sinC − sinA) + c (sinA −sinB) = 0.

Exercise 4 In any triangle ABC, (ABC)=abc/4R¹ .

1.2  Ceva's theorem

Theorem 1 [Ceva's theorem (1678)] If three cevians AX, BY, CZ, one through each vertex of a triangle ABC, are concurrent, then

BX XC  CY YA  AZ ZB =1

PROOF. Three cevians are concurrent so they pass through one point, say P. The proof follows from a consideration of areas of triangles ABP, BPC, CPA. The key point is Lemma on the area of two triangle with a common altitude . ^[¯]

Exercise 2 If X, Y, Z are midpoints of the sides, the three cevians are concurrent.

Exercise 3 Let XB/XC=p, YC/YA=q, and AX, BY, CZ are concurrent. Find AZ/ZB.

Exercise 4 Cevians perpendicular to the opposite sides are concurrent.

Exercise 5 Let ABC and A′B′C′ be two non-congruent triangles whose sides are respectively parallel. Then the three lines AA′, BB′, and CC′ (extended) are concurrent.

1.3  Points of intersect

The most important points and lines of intersect in a triangle are:

1. Circumcenter -the center of the circle circumscribed about a triangle, denoted by O. The circle is called circumcircle. Its radius ( circumradius) denoted by R.

2. The cevians that joint the vertices of a triangle to the midpoints of the opposite sides are called medians. They are concurrent (see Exercise 1.2.2) and the common point is centroid G, which is ``center of gravity'' of the triangle.

   Theorem 1 A triangle is dissected by its medians into six smaller triangles of equal area.

3. The medians of a triangle divide one another in the ratio 2:1; in other words, the medians of a triangle ``trisect'' one another.

4. Cevians perpendicular to corresponding lines are altitudes. It follows from Exercise 1.2.4 that altitudes are concurrent, they intersect in orthocenter H. Feets of altitudes form the orthic triangle.

5. Bisectors are cevians which divide angles to two equal parts.

   Theorem 2 Each angle bisector of a triangle divides the opposite side into segments proportional in length to the adjacent sides.

   PROOF. There at least two ways to make a proof:

   1. Applying Law of Sines to two resulting triangles.

   2. Considering the ratio of areas of those two triangles.

   ^[¯]

   Theorem 3 The internal bisectors of the three angles of a triangle are concurrent.

   PROOF. The proof follows from observation that points of bisectors are equidistant from the sides of the triangle. The point of concurrence I is incenter, that is the center of inscribed circle, which has all three sides for tangents. Its radius is inradius. ^[¯]

Exercise 4 The circumcenter and orthocenter of an obtuse-angled triangle lie outside the triangle.

Exercise 5 Find the ratio of the area of a given triangle to that of triangle whose sides have the same lengths as medians of the original triangle.

Exercise 6 Any triangle having two equal medians is isosceles.

Exercise 7 Any triangle having two equal altitudes is isosceles.

Exercise 8 Use Cevas Theorem to obtain another proof of Theorem 1.3.3.

Exercise 9 The product of two sides of a triangle is equal to the product of the circumdiameter and the altitude on the third side.

1.4  The incircle and excircles

Theorem 1 x=s−a, y=s−b, z=s−c.

PROOF. It follows from the Theorem on two tangents . ^[¯]

Theorem 2 (ABC)=sr.

PROOF. It follows from the Theorem on areas . ^[¯]

Theorem 3 The external bisectors of any two angles of a triangle are concurrent with the internal bisector of the third angle.

The circles with with above centers ( excenters) are excircles escribed to the triangle. Their radii ( exradii) denoted r_a, r_b, r_c. Incircle together with excircles are four tritangent circles.

Exercise 4 The cevians AX, BY, CZ are concurrent. Their common point is the Gergonne point.

Exercise 5 ABC is the orthic triangle of the triangle formed by excenters.

Exercise 6 (ABC)=(s−a)r_a=(s−b)r_b=(s−c)r_c

Exercise 7

1 ra +  1 rb +  1 rc =  1 r .

1.6  The orthic triangle

Theorem 1 The orthocenter H of an acute-angled triangle is the incenter of its orthic triangle.

PROOF. Let AD, CF, and BE are altitudes of ABC, O be the circumcenter and α = 90^° −A. Then the following angles are α: OBC, OCB, ABE, ACF. Because CDFA is inscribe to a circle then FDA=FCA=α and by the same reason ADE=ABE=α. Thus DA bisect FDE. ^[¯]

We also see that OB⊥FD, OC⊥DE, OA⊥FE.

Exercise 2 AEF ∼ DBF  ∼ DEC  ∼ ABC.

Exercise 3 HAO=| B−C |.

1.7  The medial triangle and Euler line

The triangle A′B′C′ formed by joining the midpoints A′, B′, C′ of the sides of a given triangle ABC will be called the medial triangle. We have the following set of conclusions:

1. The medial triangle A′B′C′ is similar to the given ABC with the ration 1:2. In fact we have four equal triangles!

2. ABC and A′B′C′ have the same centroid G.

3. The circumcenter O of ABC is the orthocenter of A′B′C′.

4. ABC and A′B′C′ are homothetic with a center G and ratio 2:1.

Theorem 1 The orthocenter, centroid and circumcenter of any triangle are collinear. The centroid divides the distance from the orthocenter to the circumcenter in the ratio 2:1.

The line on which these three points lie is called the Euler line of the triangle.

Theorem 2 The circumcenter of the medial triangle lies at the midpoint of segment HO of the Euler line of the parent triangle. The circumradius of the medial triangle equals half the circumradius of the parent triangle.

Exercise 3 OH²=9R²−a²−b²−c².

Exercise 4 DA′=| b²−c² |/2a.

Exercise 5 If ABC has the special property that its Euler line is parallel to its side BC, then tanB tanC = 3.

Chapter 2
Some properties of Circles

2.1  The power of point with respect to a circle

Theorem 1 If two lines through a point P meet a circle at points A, A′ (possibly coincident) and B, B′ (possibly coincident), respectively, then PA ×PA′ = PB ×PB′.

PROOF. The proof follows from the similarity of triangles PAB′ and PBA′ in both cases if P inside or outside of the circle. Notably in the second case PA ×PA′ = PT² where PT is tangent to circle and T belong to it. ^[¯]

Let R is the radius of the circle and d is the distance to its center. If P is inside then PA ×PA′ = R² −d² and if it is outside then PA ×PA′ = d² − R².

Theorem 2 Let O and I be the circumcenter and incenter, respectively, of a triangle with circumradius R and r; let d be the distance OI. Then

d2=R2−2rR.

PROOF. Let bisector AL meet circumcircle at L, ML be diameter of circumcircle, then LM ⊥BC. BLI is isosceles, thus BL=IL. Then

R2−d2 = LI ×IA = BL ×IA = LM  LB/LM IY/IA IY = LM  sinA/2 sinA/2 IY = LM ×IY = 2rR.

This is an example of synthetic proof, compare with proof of Radial Axis Theorem . ^[¯]

If we adopt the Newton convention:

PA = −AP

(2.1)

then identity

d2−R2=PA ×PA′

became universally true for any secant or chord. Its value is power of P with respect to the circle .

Exercise 3 What is smallest possible value of the power of a point with respect to a circle of radius R? Which point has this critical power?

Exercise 4 What is the locus of points of constant power?

Exercise 5 If PT and PU are tangents from P to two concentric circles, with T on the smaller, and if the segment PT meets the larger circle at Q, then PT² − PU²=QT².

2.2  The radical axis of two circles

Theorem 1 The locus of all points whose powers with respect to two nonconcentric circles are equal is a line perpendicular to the line of centers of the two circles.

PROOF. The proof could be done by means of analytic geometry, or analytic proof. Namely we express the problem by means of equations and solve them afterwards. The key ingredient is the equation of circle known from the Calculus I course. ^[¯]

The locus of points of equal power with respect to two non-concentric circles is called their radical axis.

Exercise 2 Give a simple indication of radical axis when two circles intersect or are tangent.

Exercise 3 Let PAB, AQB, ABR, P′AB, AQ′B, ABR′ be six similar triangles all on the same side of their common side AB. Then points P, Q, R, P′, Q′, R′ all lie on one circle.

2.3  Coaxal circles

1. a any if c < 0.

2. a ∈ [−√c, √c] if c > 0.

Theorem 1 If the centers of three circles form a triangle, there is just one point whose powers with respect to the three circles are equal. Its name is radical center.

Exercise 2 Two circles are in contact internally at a point T. Let the chord AB of the largest circle be tangent to the smaller circle at point P. Then the line TP bisect ATB.

2.5  Simpson line

Theorem 1 The feet of the perpendiculars from a point to the sides of a triangle are collinear iff the point lies on the circumcircle.

PROOF. Let A′, B′, C′ be feets of perpediculars from point P on the circumcircle. Observations:

1. Quadragles PB′A′C and C′PB′A are inscribed to a circles.

2. From the above: A′PC=A′B′C and AB′C′=APC′.

3. C′PA′=APC;

4. From the above C′PA=A′PC.

5. From 2 and 4: AB′C′=CB′A′, thus A′, B′, and C′ are collinear.

Exercise 2 What point on the circumcirle has CA as its Simpson line?

Exercise 3 The tangent at two points B and C on a circle meet at A. Let A₁B₁C₁ be the pedal triangle of the isosceles triangle ABC for an arbitrary point P on the circle. Then PA₁²=PB₁× PC₁.

2.6  Ptolemy's theorem and its extension

It is easy to see that AB₁C₁ inscribed to a circle with the diameter AP. Then from the Theorem of sines follows that:

Theorem 1 If the pedal point is distant x, y, z from the vertices of ABC, the pedal triangle has sides

ax 2R ,         by 2R ,         cz 2R .

(2.2)

The Simpson line is degenerate case of the pedal triangle, nevertheless the above formulas are true and moreover A₁B₁+B₁C₁=A₁C₁ we deduce c CP + a AP = b BP, thus

AB×CP + BC ×AP = AC ×BP.

Theorem 2 If a quadralaterial is inscribed in a circle, the sum of the product of the two pairs of opposite sides is equal to the product of the diagonals.

The inverse theorem could be modified accordingly to the triangle inequality A₁B₁ +B₁C₁ > A₁C₁.

Theorem 3 If ABC is a triangle and P is not on the arc CA of the circumcircle, then

AB×CP + BC ×AP > AC ×BP.

Exercise 4 If a point P lies on the arc CD of the circumcircle of a square ABCD, then PA(PA+PC)=PB(PB+PD).

Chapter 3
Collinearity and Concurrence

3.1  Quadrangles; Varignon's theorem

A polygon is a cyclically ordered set of points in a plane, with no three successive points collinear, together with the line segments joining consecutive pairs of the points. First few names are triangle, quadrangle, pentagon, hexagon, and so on.

Two sides of a quadrangle are said to be adjacent or opposite according as they do or do not have a vertex in common. The lines joining pairs of opposite vertices are called diagonals.

There three different types of the quadrangles:

1. convex -both diagonals are inside;

2. re-entrant -one diagonal is in, another is out;

3. crossed -both diagonals are outside.

We agree to count the area of triangle positive or negative if its vertices are named in counterclockwise or clockwise order. For example

(ABC)=−(BAC).

(3.1)

For all convex and re-entrant quadrangles area is:

(ABCD)=(ABC)+(CDA).

Remark 1 Combined the idea of signed area with directed segments could extend the proof of the Ceva's theorem to the case, then points divides sides externally.

Theorem 2 [Varignon 1731] The figure formed when the midpoints of the sides of a quadrangle are joined in order is a parallelogram, and its area is half that of the quadrangle.

Theorem 3 The segments joining the midpoints of pairs of the opposite sides of the a quadrangle and segment joining the midpoints of the diagonals are concurrent and bisect one another.

Theorem 4 If one diagonal divides a quadrangle into two triangles of equal area, it bisect the other diagonal. Conversely, if one diagonal bisect the other, it bisect the area of the quadrangle.

Theorem 5 If a quadrangle ABCD has its opposite sides AD and BC (extended) meeting at W, while X and Y are the midpoints of the diagonals AC and BD, then (WXY)=1/4(ABCD).

3.2  Cyclic quadrangles; Brahmagupta's formula

Theorem 1 [Brahmagupta] If a cyclic quadrangle has sides a, b, c, d and semiperimeter s, its area K is given by

K2 = (s−a)(s−b)(s−c)(s−d).

Corollary 2 [Heron] Area of a triangle is given by

(ABC)2=s(s−a)(s−b)(s−c).

3.4  Menelaus's theorem

Theorem 1 [Menelaus] If points X, Y, Z on sides BC, CA, AB (suitable extended) of ∆ABC are collinear, then

BX CX  CY AY  AZ BZ =1.

Conversely, if this equation holds for points X, Y, Z on the three sides, then these three points are collinear.

For the directed segments it could be rewritten as follows:

BX XC  CY YA  AZ ZB = −1.

3.5  Pappus's theorem

The following theorem is the first belonging to projective geometry. It is formulated entirely in terms of collinearity.

Theorem 1 [Pappus, 300 A.D.] If A, C, E are three points on one line, B, D, F on another, and if the three lines AB, CD, EF meet DE, FA, BC, respectively, then three points of intersection L, M, N are collinear.

PROOF. Let lines AB, CD, EF form triangle UVW. Apply the Menelaus's Theorem to the five triads of points

LDE,    AMF,    BCN,    ACE,     BDF

on the sides of this triangle UVW. Then the product of first three identities divided by the last two ones gives

VL LM  WM MU  UN NV = −1.

Thus by Menelaus's Theorem L, M, N are collinear. ^[¯]

3.6  Perspective triangles; Desargues's theorem

If two specimens of a figure, composed of points and lines, can be put into correspondence in such a way that pairs of corresponding points are joined by concurrent lines, we say that two specimens are perspective from a point. If the correspondence is such that pairs of corresponding lines meet at collinear points, we say that two specimens are perspective from a line.

Theorem 1 [Desargues, 1650] If two triangles are perspective from a point then they are perspective from a line.

In other words, If two triangles are perspective from a point, and if their pairs of corresponding sides meet, then three points of intersection are collinear.

PROOF. Let PQR and P′Q′R′ are the triangles perspective from point O, and let D=RQ·R′Q′, E=PR·P′R′, F=PQ·P′Q′. Apply Menelaus's Theorem to triads DR′Q′, EP′R′, FQ′P′ and triangles OQR, ORP, OPQ. ^[¯]

The converse theorem is also true.

Theorem 2 If two triangles are perspective from a line, they are perspective from a point.

If two triangles are perspective from a line, and if two pairs of corresponding vertices are joined by intersecting lines, the triangles are perspective from the point of intersection of these lines.

3.7  Hexagons

Two vertices of a hexagon are said to be adjacent , alternate , opposite according as they are separated by one sides, two sides, or three sides. The join of two opposite vertices is called a diagonal .

Exercise 1 Count the number of ways a given hexagon could be labelled as ABCDEF (Answer: 12).

Exercise 2 Count number of different hexagons defined by given 6 point, no three collinear. (Answer: 60).

In term of hexagon we could reformulate Pappus's Theorem as follows:

If each set of three alternate vertices of a hexagon is a set of three collinear points, and the three pairs of opposite sides intersect, then the three points of intersection are collinear.

3.8  Pascal's theorem

Theorem 1 [Pascal's Theorem] If all six vertices of a hexagon lie on a circle and the three pairs of opposite sides intersect, then the three points of intersection are collinear.

PROOF. The proof consists of application Menelaus's Theorem four times. ^[¯]

This theorem of a projective nature and hexagon could be in fact inscribed in any conic . Under such a formulation it has an inverse:

Theorem 2 If the three pairs of opposite sides of a hexagon meet at three collinear points, then the six vertices lie on a conic.

Some degenerated cases of the Pascal's Theorem are of interest

Corollary 3 Let ABDE be a cyclic crossed quadrangle. Tangents to the circle in points B and E meet in a point N which is collinear with points L=AB·DE and M=BD·EA.

Chapter 4
Transformations

For Eucleadean geometry the important transformations are isometries. There are several of them: translations , rotations (particularly half-turn ), reflections .

4.1  Translations

Exercise 1 Inscribe in a given circle a rectangle with two opposite sides equal and parallel to a given line segment a.

The characteristic property of translation among isometries is: each ray come to a parllel ray (prove it!).

4.2  Rotations

The characteristic property of rotations among isometries is: each ray come to ray rotated by the α.

4.3  Half-turn

Theorem 1 Composition of two half-turn is a traslation by the vector 2→O₁O₂.

Using half-turns we could easily prove that if digonals of a quadrangle bisect each other then it is a parallelogramm.

4.4  Reflections

4.7  Dilation

Exercise* 1 Prove that suchtransformations preserve collinearity and angles.

A simplest kind is dilation, which transforms each line into a parallel line. If a dilation is not a translation then its central dilation . Translations and half-turn are partucular cases of dilations with ratio 1 and −1 correspondingly.

4.8  Spiral symmetry

Theorem 1 If squares, with centers O₁, O₂, O₃, are erected externally on the sides of ∆ABC, then line segments O₁O₂ and CO₃. are equal and perpendicular.

PROOF. It is follows from consideration of A(√2,45^°) and C(√2,−45^°). ^[¯]

It is interesting that there are no other direct similarities besides spiral ones:

Theorem 2 Any two directly similar figures are related either by a translation or by a spiral similarity.

Corollary 3 If ABC and A′B′C′ are two directly similar triangles, while AA′A′′, BB′B′′, CC′C′′ are three directly similar triangles, then ∆A′′B′′C′′ and ∆ABC are directly similar.

4.9  A genealogy of transformations

We could put the following transformation in a genealogical tree:

Transformation

Continuous transormation

Linear transformation

Similarity Procrustean stretch

Isometry Dilation Spiral similarity

Reflection Translation Rotation Central dilation

Half-turn

Chapter 5
An Introduction to Inversive Geometry

5.1  Separation

Theorem 1 If four points A, B, C, D do not all lie on the circle or line, there exist two non-intercecting circles, one through A and C, the other through B and D.

Two distinct point pairs, AC and BD are said to sepatrate each other if A, B, C, D lie on a circle (or a line) in such an order that either of the arcs AC contains one but not both of the remaining points B and D. It is denoted by AC//BD. Another characterizations are

Theorem 2 Two distinct point pairs, AC and BD are said to sepatrate each other if every circle through A and C intersects (or coinsides with) every circle through B and D.

Alternatively

Theorem 3 The mutual distances of four distinct points A, B, C, D satisfy

AB ×CD + BC ×AD ≥ AC×BD,

with the equals sign only then AC//BD.

PROOF. It is follows directly from consideration of directed line segments if the points are collinear and is a consequence of the Ptolemy's theorem if points lie on a circle or are not collinear. ^[¯]

5.2  Cross Ratio

Theorem 1 The cross ratios of four distinct points A, B, C, D satisfy

{AD, BC} + {AB, DC} = 1

iff AC // BD.

Now instead of defining separation in the term of circles we could define circles in the term of separation:

Definition 2 The circle determined by three points A, B, C is set of points consisiting of the three points themselves along with all the points X such that

BC //AX or CA //BX or AB //CX.

5.3  Inversion

For a given circle ω with the center O and radius k we define a point P′=i(P) being inverse to P if P′ ∈ OP and

OP ×OP′=k2.

It is obvious from this conditions that P=i(i(P)) for any point P (different from O). The inverse for O is not defined. There is a simple geometrical constraction.

Theorem 1 The inverse of any line a, not through 0, is a circle through O, and the diametre through O of the circle is perpendicular to a.

The inverse of any circle through O is a perpendicular to the diametr through O.

We could construct inverse points using Peaucellier's cell.

Considering images under inversion of three points we could observe

Theorem 2 For a suitable circel inversion, any three distinct points A, B, C can be inverted into the vertices of a triangle A′B′C′ congruent to a given triangle.

5.4  The inversive plane

Theorem 1 If a circle with center O and radius k invert point pair AB into A′B′, the distance are related by the equation

A′B′=  k2 AB OA×OB .

Theorem 2 If A, B, C, D invert into A′, B′, C′, D′, then

{A′B′,C′D′}={AB,CD}.

Theorem 3 If A, B, C, D invert into A′, B′, C′, D′ and Ac//BD then A′C′//B′D′.

If we will think on lines as circles with infinite radius then we could state

Theorem 4 The inverse of any circle is a circle.

To define inversion for all points we may add to a plane a one special point: point at infinity p_∞. Then inverse of O is p_∞ and vise verse. A plane together with p_∞q form the inversive plane.

Chapter 6
An Introduction to Projective Geometry

6.1  Reciprocation

Let ω be a circle with center O and radius k. Each point P (different from O) determine a corresponding line p, called the polar of P; it is the line perpendicular to OP through the inverse of P. Conversely, each line p determine a point P, the pole of p; it is the inverse of the foot of the perpendicular from O to p.

Theorem 1 If B lies on a, then b passes through A.

We say that A and B are conjugate points; a and b are conjugate line. Any point on a tangent a is conjugate to the point of contact A, which is self-conjugate point, and any line through A (on ω) is conjugate to the tangent a, which is a self-conjugate line.

Reciprocation allows us to introduce a vocabulary for projective duality

Theorem 2 The pole of any secant AB (except a diameter) is the common point of the tangents at A and B. The polar of any exterior point is the line joining the points of contact of two tangents from this point. The pole of any line p (except a diameter) is the common point of the polars of two exterior points on p. The polar of any point P (except the center) is the line joining the poles of two secants through P.

6.3  Conics

Definition 1 A conic is the reciprocal of a circle with a center A and radius r. Let ε = OA/r be the eccentricity of the conic.

1. If ε < 1 then it is ellipce, particularly ε = 0 is the circle.

2. If ε = 1 then it is parabola.

3. If ε > 1 then it is hyperbola.

6.5  The projective plane

Theorem 1 If P is not on the conic, its polar joins the points of intersection AB ·DE and AE·BD, where AD and BE are any two secant through P.

Theorem 2 With respect to any conic except a circle, a directrix is the polar of the corresponding focus.

6.7  Stereographic and gnomonic projection

If a plane is tangent to the sphere of inversion at A then its image is the sphere σ with a diameter OA. And the image of any point P in the plane is just another point of intersection of line OP with σ. Sphere σ is a model for inversive plane . The mapping between plane and sphere is stereographic projection. Its preserve angles between directions in any points.

If we take a sphere Σ and construct the map from a tangent plane to pairs of antipodal points as intersections of line OP and Σ then we obtain gnomonic map. It maps big circles (shortest distances on the sphere) to straight lines (shortest distances on the plane). Identifying pairs antipodal points on the sphere we obtain a model of projective plane .

Appendix A
Some Useful Theorems

Theorem 1 An angle inscribed in an arc of a circle has a measure which is a half of angular measure of the complementary arc.

Corollary 2 An angle inscribed in semicircle is is a right angle.

Theorem 3 Two tangemts to a circle from any external point are equal.

Theorem 4 The following geometric objects have indicated areas S:

1. Rectangle with sides a and b: S=ab.

2. Parallelogram with a base a and altitude h: S=ah.

3. Triangle with a side a and corresponding altitude h_a: S=ah_a/2.

Corollary 5

1. Two triangles with a common altitude h have areas proportional to their sides: [(S₁)/(S₂)]=[(a₁)/(a₂)].

2. Two triangles with a common side a have areas proportional to their altitudes: [(S₁)/(S₂)]=[(h₁)/(h₂)].

Appendix B
Some Useful Tricks

B.1  Look for a triangle-the golden rule of geometry

If the unknown element is a line segment or an angle-try to find a triangle, which contains this element and such that other parameters of the triangle are given or could be found from the given conditions.

If you are questioned about two elements (like a ratio of two line segments, for example) try to find two triangles with some common elements and each containing one of the unknown line segments.

B.2  Investigate a particular case

If you meet a problem-try to investigate a particular case. If you study an angle inscribed in a circle-consider first a case when the angle goes through the center of circle. If this particular investigation was successful try to use this particular case for solution the general one.

Bibliography

[1]

H.S.M. Coxeter and S.L. Greitzer. Geometry Revisited. This famous book was reprinted many times and translated to many foreign languages.

Index (showing section)

Footnotes:

¹We alway denote area of a figure by its name enclosed in parentheses.