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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:
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.
Theorem 1 [Law of Sines]
For a triangle ABC with circumradius R
a
=
b
=
c
= 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/4R^{1} .
Theorem 1 [Ceva's theorem (1678)]
If three cevians AX, BY, CZ, one through each vertex of a
triangle ABC, are concurrent, then
BX
CY
AZ
=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 noncongruent triangles whose sides are respectively parallel. Then the three lines AA′, BB′, and CC′ (extended) are concurrent.
The most important points and lines of intersect in a triangle are:
Theorem 1 A triangle is dissected by its medians into six smaller triangles of equal area.
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:
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 obtuseangled 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.
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
+
1
+
1
=
1
.
Theorem 1 The orthocenter H of an acuteangled 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 .
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:
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 OH^{2}=9R^{2}−a^{2}−b^{2}−c^{2}.
Exercise 4 DA′= b^{2}−c^{2} /2a.
Exercise 5 If ABC has the special property that its Euler line is parallel to its side BC, then tanB tanC = 3.
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′ = PT^{2} 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′ = R^{2} −d^{2} and if it is outside then PA ×PA′ = d^{2} − R^{2}.
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
d^{2}=R^{2}−2rR.
PROOF.
Let bisector AL meet circumcircle at L, ML be diameter of
circumcircle, then LM ⊥BC. BLI is isosceles, thus
BL=IL. Then
This is an example of
synthetic proof, compare with proof of
Radial Axis Theorem .
^{[¯]}
R^{2}−d^{2}
=
LI ×IA = BL ×IA
=
LM
LB/LM
IY = LM
sinA/2
IY
=
LM ×IY = 2rR.
If we adopt the Newton convention:
 (2.1) 

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 PT^{2} − PU^{2}=QT^{2}.
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 nonconcentric 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.

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.
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:
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 A_{1}B_{1}C_{1} be the pedal triangle of the isosceles triangle ABC for an arbitrary point P on the circle. Then PA_{1}^{2}=PB_{1}× PC_{1}.
It is easy to see that AB_{1}C_{1} 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
,
by
,
cz
. (2.2)
The Simpson line is degenerate case of the
pedal triangle, nevertheless the above
formulas are true and moreover A_{1}B_{1}+B_{1}C_{1}=A_{1}C_{1} we deduce
c CP + a AP = b BP, thus

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 A_{1}B_{1} +B_{1}C_{1} > A_{1}C_{1}.
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).
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:
We agree to count the area of triangle
positive
or
negative
if its vertices are named
in counterclockwise or clockwise order. For example
 (3.1) 

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).
Theorem 1 [Brahmagupta]
If a cyclic quadrangle has sides a, b, c, d and
semiperimeter s, its
area
K
is given by
K^{2} = (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).
Theorem 1 [Menelaus]
If points X, Y, Z on sides BC, CA, AB (suitable
extended) of ∆ABC are collinear, then
Conversely, if this equation holds for points X, Y, Z on the
three sides, then these three points are collinear.
BX
CY
AZ
=1.
For the directed segments it could
be rewritten as follows:
BX
CY
AZ
= −1.
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
on the sides of this triangle UVW. Then the product of first three
identities divided by the last two ones gives
LDE, AMF, BCN, ACE, BDF
Thus by Menelaus's Theorem L, M, N are
collinear.
^{[¯]}
VL
WM
UN
= −1.
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.
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.
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.
For Eucleadean geometry the important transformations are isometries. There are several of them: translations , rotations (particularly halfturn ), reflections .
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!).
The characteristic property of rotations among isometries is: each ray come to ray rotated by the α.
Theorem 1 Composition of two halfturn is a traslation by the vector 2→O_{1}O_{2}.
Using halfturns we could easily prove that if digonals of a quadrangle bisect each other then it is a parallelogramm.
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 halfturn are partucular cases of dilations with ratio 1 and −1 correspondingly.
Theorem 1 If squares, with centers O_{1}, O_{2}, O_{3}, are erected externally on the sides of ∆ABC, then line segments O_{1}O_{2} and CO_{3}. 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.
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
Halfturn
Theorem 1 If four points A, B, C, D do not all lie on the circle or line, there exist two nonintercecting 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
with the equals sign only then AC//BD.
AB ×CD + BC ×AD ≥ 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. ^{[¯]}

Theorem 1
The cross ratios of four distinct points A, B, C, D satisfy
iff AC // BD.
{AD, BC} + {AB, DC} = 1
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.
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

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.
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′=
k^{2} AB
.
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.
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 selfconjugate point, and any line through A (on ω) is conjugate to the tangent a, which is a selfconjugate line.
Reciprocation allows us to introduce a vocabulary for projective duality
point  line 
lie on  pass through 
line joining two points  intersection of two lines 
concurrent  collinear 
quadrangle  quadrilateral 
pole  polar 
locus  envelope 
tangent  point of contact 
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.
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.
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.
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 .
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:
If the unknown element is a line segment or an angletry 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.
If you meet a problemtry to investigate a particular case. If you study an angle inscribed in a circleconsider 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.




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