The circle in mathematics is one of the most important and important figures. It is necessary for many calculations. Knowledge of the properties of this figure from school curriculum will certainly come in handy in life. The circumference is required when calculating many materials with a circular cross-section. Working on drawings, building a fence near a flowerbed - this will require knowledge of a geometric figure and its properties.
The concept of a circle and its main elements
A figure on a plane consisting of numerous points located on equal distance from the central one is called a circle. A segment extending from the center and connecting it to one of the points forming the circle is called a radius. A chord is a segment that connects a pair of points located along the perimeter of a circle to each other. If it is positioned so that it passes through the central point, then it is also a diameter.
The length of the radius of a circle is equal to the length of the diameter, halved. A pair of divergent points located on a circle divides it into two arcs. If a segment with ends at these points passes through the central point (thereby being a diameter), then the arcs formed will be semicircles.
Circumference
The calculation of the perimeter of a circle is determined in several ways: through the diameter or through the radius. In practice, it was discovered that the circumference of a circle (l) when divided by its diameter (d) always gives one number. This is the number π, which is equal to 3.141692666... The calculation is made using the formula: π= l/ d. Transforming it, we get the circumference. The formula is: l=πd.
To find the radius we use the following formula: d=2r. This became possible thanks to division. After all, the radius is half the diameter. Once we have obtained the above values, we can calculate what the circumference is equal to, using the formula the following type: l=2πr.
Basic properties
The area of a circle is always greater when compared with the areas of other closed curves. A tangent is a line that touches the circle at only one point. If a line intersects it in two places, then it is a secant. The point at which 2 different circles touch each other is always on a line passing through their center points. Circles that intersect on a plane are those that have 2 common points. The angle between them is calculated as the angle formed by the tangents to the points of contact.
If through a point that is not a point on a circle, two straight lines secant to it are drawn, then the angle formed by them will be equal to the difference in the lengths of the arcs, halved. This rule also applies in the opposite case, when we're talking about about two chords. Two intersecting chords form an angle equal to the sum arc lengths reduced by half. Arcs in such a situation are chosen in given corner and the corner opposite. Optical property circle says the following: rays of light reflected from mirrors placed around the perimeter of the circle are collected back to its center. IN in this case the light source should be installed at the center point of the circle.
\[(\Large(\text(Central and inscribed angles)))\]
Definitions
A central angle is an angle whose vertex lies at the center of the circle.
An inscribed angle is an angle whose vertex lies on a circle.
The degree measure of an arc of a circle is the degree measure central angle which rests on it.
Theorem
The degree measure of an inscribed angle is equal to half the degree measure of the arc on which it rests.
Proof
We will carry out the proof in two stages: first, we will prove the validity of the statement for the case when one of the sides of the inscribed angle contains a diameter. Let point \(B\) be the vertex of the inscribed angle \(ABC\) and \(BC\) be the diameter of the circle:
Triangle \(AOB\) is isosceles, \(AO = OB\) , \(\angle AOC\) is external, then \(\angle AOC = \angle OAB + \angle ABO = 2\angle ABC\), where \(\angle ABC = 0.5\cdot\angle AOC = 0.5\cdot\buildrel\smile\over(AC)\).
Now consider an arbitrary inscribed angle \(ABC\) . Let us draw the diameter of the circle \(BD\) from the vertex of the inscribed angle. There are two possible cases:
1) the diameter cuts the angle into two angles \(\angle ABD, \angle CBD\) (for each of which the theorem is true as proven above, therefore it is also true for the original angle, which is the sum of these two and therefore equal to half the sum of the arcs to which they rely, that is equal to half arc on which it rests). Rice. 1.
2) the diameter did not cut the angle into two angles, then we have two more new inscribed angles \(\angle ABD, \angle CBD\), whose side contains the diameter, therefore, the theorem is true for them, then it is also true for the original angle (which is equal to the difference of these two angles, which means it is equal to the half-difference of the arcs on which they rest, that is, equal to half the arc on which it rests). Rice. 2.
Consequences
1. Inscribed angles subtending the same arc are equal.
2. An inscribed angle subtended by a semicircle is a right angle.
3. An inscribed angle is equal to half the central angle subtended by the same arc.
\[(\Large(\text(Tangent to the circle)))\]
Definitions
There are three types relative position straight line and circle:
1) straight line \(a\) intersects the circle at two points. Such a line is called a secant line. In this case, the distance \(d\) from the center of the circle to the straight line is less than the radius \(R\) of the circle (Fig. 3).
2) straight line \(b\) intersects the circle at one point. Such a line is called a tangent, and their common point \(B\) is called the point of tangency. In this case \(d=R\) (Fig. 4).
Theorem
1. A tangent to a circle is perpendicular to the radius drawn to the point of tangency.
2. If a line passes through the end of the radius of a circle and is perpendicular to this radius, then it is tangent to the circle.
Consequence
The tangent segments drawn from one point to a circle are equal.
Proof
Let us draw two tangents \(KA\) and \(KB\) to the circle from the point \(K\):
This means that \(OA\perp KA, OB\perp KB\) are like radii. Right Triangles\(\triangle KAO\) and \(\triangle KBO\) are equal in leg and hypotenuse, therefore, \(KA=KB\) .
Consequence
The center of the circle \(O\) lies on the bisector of the angle \(AKB\) formed by two tangents drawn from the same point \(K\) .
\[(\Large(\text(Theorems related to angles)))\]
Theorem on the angle between secants
The angle between two secants drawn from the same point is equal to the half-difference in degree measures of the larger and smaller arcs they cut.
Proof
Let \(M\) be the point from which two secants are drawn as shown in the figure:
Let's show that \(\angle DMB = \dfrac(1)(2)(\buildrel\smile\over(BD) - \buildrel\smile\over(CA))\).
\(\angle DAB\) – external corner triangle \(MAD\) , then \(\angle DAB = \angle DMB + \angle MDA\), where \(\angle DMB = \angle DAB - \angle MDA\), but the angles \(\angle DAB\) and \(\angle MDA\) are inscribed, then \(\angle DMB = \angle DAB - \angle MDA = \frac(1)(2)\buildrel\smile\over(BD) - \frac(1)(2)\buildrel\smile\over(CA) = \frac(1)(2)(\buildrel\smile\over(BD) - \buildrel\smile\over(CA))\), which was what needed to be proven.
Theorem on the angle between intersecting chords
The angle between two intersecting chords is equal to half the sum of the degree measures of the arcs they cut: \[\angle CMD=\dfrac12\left(\buildrel\smile\over(AB)+\buildrel\smile\over(CD)\right)\]
Proof
\(\angle BMA = \angle CMD\) as vertical.
From triangle \(AMD\) : \(\angle AMD = 180^\circ - \angle BDA - \angle CAD = 180^\circ - \frac12\buildrel\smile\over(AB) - \frac12\buildrel\smile\over(CD)\).
But \(\angle AMD = 180^\circ - \angle CMD\), from which we conclude that \[\angle CMD = \frac12\cdot\buildrel\smile\over(AB) + \frac12\cdot\buildrel\smile\over(CD) = \frac12(\buildrel\smile\over(AB) + \buildrel\ smile\over(CD)).\]
Theorem on the angle between a chord and a tangent
The angle between the tangent and the chord passing through the point of tangency is equal to half the degree measure of the arc subtended by the chord.
Proof
Let the straight line \(a\) touch the circle at the point \(A\), \(AB\) is the chord of this circle, \(O\) is its center. Let the line containing \(OB\) intersect \(a\) at the point \(M\) . Let's prove that \(\angle BAM = \frac12\cdot \buildrel\smile\over(AB)\).
Let's denote \(\angle OAB = \alpha\) . Since \(OA\) and \(OB\) are radii, then \(OA = OB\) and \(\angle OBA = \angle OAB = \alpha\). Thus, \(\buildrel\smile\over(AB) = \angle AOB = 180^\circ - 2\alpha = 2(90^\circ - \alpha)\).
Since \(OA\) is the radius drawn to the tangent point, then \(OA\perp a\), that is, \(\angle OAM = 90^\circ\), therefore, \(\angle BAM = 90^\circ - \angle OAB = 90^\circ - \alpha = \frac12\cdot\buildrel\smile\over(AB)\).
Theorem on arcs subtended by equal chords
Equal chords subtend equal arcs, smaller semicircles.
And vice versa: equal arcs are subtended by equal chords.
Proof
1) Let \(AB=CD\) . Let us prove that the smaller semicircles of the arc .
On three sides, therefore, \(\angle AOB=\angle COD\) . But because \(\angle AOB, \angle COD\) - central angles supported by arcs \(\buildrel\smile\over(AB), \buildrel\smile\over(CD)\) accordingly, then \(\buildrel\smile\over(AB)=\buildrel\smile\over(CD)\).
2) If \(\buildrel\smile\over(AB)=\buildrel\smile\over(CD)\), That \(\triangle AOB=\triangle COD\) on two sides \(AO=BO=CO=DO\) and the angle between them \(\angle AOB=\angle COD\) . Therefore, and \(AB=CD\) .
Theorem
If the radius bisects the chord, then it is perpendicular to it.
The converse is also true: if the radius is perpendicular to the chord, then at the point of intersection it bisects it.
Proof
1) Let \(AN=NB\) . Let us prove that \(OQ\perp AB\) .
Consider \(\triangle AOB\) : it is isosceles, because \(OA=OB\) – radii of the circle. Because \(ON\) is the median drawn to the base, then it is also the height, therefore, \(ON\perp AB\) .
2) Let \(OQ\perp AB\) . Let us prove that \(AN=NB\) .
Similarly, \(\triangle AOB\) is isosceles, \(ON\) is the height, therefore, \(ON\) is the median. Therefore, \(AN=NB\) .
\[(\Large(\text(Theorems related to the lengths of segments)))\]
Theorem on the product of chord segments
If two chords of a circle intersect, then the product of the segments of one chord is equal to the product of the segments of the other chord.
Proof
Let the chords \(AB\) and \(CD\) intersect at the point \(E\) .
Consider the triangles \(ADE\) and \(CBE\) . In these triangles, angles \(1\) and \(2\) are equal, since they are inscribed and rest on the same arc \(BD\), and angles \(3\) and \(4\) are equal as vertical. Triangles \(ADE\) and \(CBE\) are similar (based on the first criterion of similarity of triangles).
Then \(\dfrac(AE)(EC) = \dfrac(DE)(BE)\), from which \(AE\cdot BE = CE\cdot DE\) .
Tangent and secant theorem
Square of tangent segment equal to the product secant to its outer part.
Proof
Let the tangent pass through the point \(M\) and touch the circle at the point \(A\) . Let the secant pass through the point \(M\) and intersect the circle at the points \(B\) and \(C\) so that \(MB< MC\) . Покажем, что \(MB\cdot MC = MA^2\) .
Consider the triangles \(MBA\) and \(MCA\) : \(\angle M\) is common, \(\angle BCA = 0.5\cdot\buildrel\smile\over(AB)\). According to the theorem about the angle between a tangent and a secant, \(\angle BAM = 0.5\cdot\buildrel\smile\over(AB) = \angle BCA\). Thus, triangles \(MBA\) and \(MCA\) are similar at two angles.
From the similarity of triangles \(MBA\) and \(MCA\) we have: \(\dfrac(MB)(MA) = \dfrac(MA)(MC)\), which is equivalent to \(MB\cdot MC = MA^2\) .
Consequence
The product of a secant drawn from the point \(O\) by its external part does not depend on the choice of the secant drawn from the point \(O\) .
“And I fell in love with the circle and settled on it.”
Information and educational project.
Topic: circle
Project goal: To study the properties, types different circles and theorems associated with them.
I began my work by studying the properties of a circle in a school geometry course using A.V. Pogorelov’s textbook “Geometry 7-9” and material beyond school course. When collecting information from various sources and while working on the project, I expanded my knowledge and will continue to study this topic and share knowledge with classmates and everyone who is interested.
Circle - locus points of the plane equidistant from a given point, called the center, at a given non-zero distance, called its radius. Vicious circle having no internal space. 
Other definitions
A circle of diameter AB is a figure consisting of points A, B and all points of the plane from which segment AB is visible at right angles.
A circle is a figure consisting of all points of the plane, for each of which the ratio of the distances to two given points is equal given number, different from unity. (see Circle of Apollonius)
Also a figure consisting of all such points, for each of which the sum of the squares of the distances to two given points is equal to given value, more than half square of the distance between these points.
Radius- not only the distance, but also a segment connecting the center of the circle with one of its points.
A segment connecting two points on a circle is called its chord. A chord passing through the center of a circle is called diameter.
The circle is called single, if its radius equal to one. The unit circle is one of the main objects of trigonometry.
Any two divergent points on a circle divide it into two parts. Each of these parts is called arc of a circle. The arc is called semicircle, if the segment connecting its ends is a diameter.
Ptolemy's theorem.
Claudius Ptolemy(), who lived at the end of the first - beginning of the second century AD, was an ancient Greek astronomer, mathematician, astrologer, geographer, optician and music theorist. He is known as a commentator on Euclid. Ptolemy tried to prove the famous Fifth Postulate. The main work of Ptolemy is “Almagest”, in which he presented information on astronomy. Included “Almagest” and a catalog of the starry sky.
Ptolemy's theorem. A circle can be described around a quadrilateral if and only if the product of its diagonals is equal to the sum of the products of its opposite sides.
Proof of Necessity. Since a quadrilateral is inscribed in a circle, then
From the triangle, using the cosine theorem, we find
Similarly from a triangle:
The sum of these cosines is zero:
From here we express:
Let's look at the triangles and find:
Q.E.D.
Along the way, we proved one more statement. For a quadrilateral inscribed in a circle,
Proof of sufficiency. Let the equality hold
Let us prove that a circle can be circumscribed around a quadrilateral.
Let us denote by the radius of the circle described around . From a point we drop perpendiculars onto lines and and and denote the points of intersection of these lines and perpendiculars to them through and, respectively. Using the Sins theorem for a triangle, we obtain (the diameter of the circumscribed circle for this triangle is equal to):
By the law of sines for a triangle we have
Hence,
In the same way, considering triangles, we obtain the relations
Hence, substituting these expressions into the original equality, we have
whence it follows that the points and lie on the same straight line.
Let us now prove that it follows from this that a circle can be described around a quadrilateral ( sufficient condition Simson's theorem).
Let's construct circles on segments and as on diameters. The first of them passes through the points and (angles and straight lines), and the second - through the points and ( 
). Angles and are equal to vertical angles, which means that , and therefore . From here , and a circle can be drawn around the quadrilateral.
Euler's formula named after Leonhard Euler, who introduced it, and relates the complex exponent to trigonometric functions. 
Euler's formula states that for any real number x the following equality holds:
Where e- base natural logarithm,
i- imaginary unit.
The angle formed by an arc of a circle equal in length to the radius is taken as 1 radian
The length of a unit semicircle is denoted by π.
The geometric locus of points in the plane, the distance from which to a given point is not greater than a given non-zero distance, is called all around .
A straight line that has exactly one common point with a circle is called tangent to a circle, and their common point is called the tangency point of the line and the circle.
A straight line passing through two various points on a circle is called secant
.
Central angle - an angle with a vertex at the center of the circle. The central angle is equal to the degree measure of the arc on which it rests.
In this case, angle AOB is central. 
Inscribed angle
- an angle whose vertex lies on a circle and whose sides intersect this circle. The inscribed angle is equal to half the degree measure of the arc on which it rests. In this case angle ABC is inscribed. 
Two circles having general center, are called concentric
.
Two circles whose radii intersect at right angles are called
orthogonal.
Circumference: C = 2∙π∙R = π∙D
Circle radius: R = C/(2∙π) = D/2
Circle diameter: D = C/π = 2∙R
Two circles given by equations:
are concentric (that is, having a common center) if and only if A1 = A2 and B1 = B2.
Two circles are orthogonal (that is, intersecting at right angles) if and only if the condition
Inscribed circle
A circle is called inscribed in an angle if it lies inside the angle and touches its sides. The center of a circle inscribed in an angle lies on the bisector of that angle.
A circle is said to be inscribed in convex polygon if she lies inside given polygon and touches all straight lines passing through it sides.
In a triangle
Properties of the inscribed circle:
the bisectors of T are the perpendicular bisectors of T 1
Let T 2 - orthotriangle T 1 . Then its sides are parallel to the sides of the original triangle T.
Let T 3 - middle triangle T 1 . Then the bisectors of T are the heights of T 3 .
Each triangle can fit a circle, and only one.
If a line passing through point O parallel to side AB intersects sides BC and CA at points A 1 and B 1 , That A 1 B 1 = A 1 B + AB 1 .
The tangent points of a circle inscribed in a triangle T are connected by segments - a triangle T is obtained 1
The center O of the incircle is called the incenter; it is equidistant from all sides and is the point of intersection of the bisectors of the triangle.
The radius of a circle inscribed in a triangle is equal to
In a polygon
If a circle can be inscribed in a given convex polygon, then the bisectors of all angles of the given polygon intersect at one point, which is the center of the inscribed circle.
Radius of a circle inscribed in a polygon equal to the ratio its area to semiperimeter
Circumscribed circle.
Circumcircle
- a circle containing all the vertices of a polygon. The center is a point (usually denoted
O
) the intersection of perpendicular bisectors to the sides of the polygon.
Properties
Circumcenter convex n-gon lies at the point of intersection of the perpendicular bisectors to its sides. As a consequence: if a circle is circumscribed next to an n-gon, then all the perpendicular bisectors to its sides intersect at one point (the center of the circle).
Around anyone regular polygon You can describe a circle, and only one.
For a triangle
:
Around any triangle you can describe a circle, and only one. Its center will be the intersection point perpendicular bisectors.
U acute triangle the center of the circumcircle lies inside the triangle, for an obtuse triangle it lies outside the triangle, and for a rectangular circle it lies in the middle of the hypotenuse.
3 of 4 circles circumscribed about the medial triangles (formed the middle lines of the triangle) intersect at one point inside the triangle. This point is the circumcenter of the main triangle.
The center of the circle circumscribed about the triangle serves as orthocenter triangle with vertices at the midpoints of the sides of the given triangle.
Distance from the vertex of the triangle to orthocenter twice the distance from the center circumcircle to the opposite side.
Radius
The radius of the circumcircle can be found using the formulas
Where:
a , b , c - sides of the triangle,
α - angle opposite to the side a ,
S - area of a triangle.
Position of the circumcircle center
Let the radius vectors of the vertices of the triangle be the radius vector of the center of the circumscribed circle. Then
Where
Circumcision equation
Let the coordinates of the vertices of the triangle in some Cartesian system coordinates on the plane, 
- coordinates of the center of the circumscribed circle. Then
and the equation of the circumcircle has the form
For points lying inside the circle, the determinant is negative, and for points outside it, it is positive.
Euler's formula: If d - the distance between the centers of the inscribed and circumscribed circles, and their radii are equal r And R accordingly, then d 2 = R 2 − 2 Rr .
For a quadrilateral.
An inscribed simple (without self-intersection) quadrilateral is necessarily convex.
A circle can be described around a convex quadrilateral if and only if the sum of its interior opposite corners equal to 180° (π radians).
You can describe a circle around:
any rectangle ( special case square)
any isosceles trapezoid
For a quadrilateral inscribed in a circle, the product of the lengths of the diagonals is equal to the sum of the products of the lengths of pairs of opposite sides:
|AC|·|BD| = |AB|·|CD| + |BC|·|AD|
Circle of Apollonius - the geometric locus of points on the plane, the ratio of the distances from which to two given points is a constant value, not equal to unity.
Bipolar coordinates - orthogonal system coordinates on a plane, based on Apollonius circles.
Let two points be given on the plane A And B . Let's consider all points P this plane, for each of which
,
Where
k
- fixed positive number. At
k
= 1 these points fill the midperpendicular to the segment
AB
; in other cases, the specified geometric location is a circle called
Apollonius circle
.
Circles of Apollonius. Each blue circle intersects each red circle at right angles. Each red circle passes through two points (C and D) and each blue circle surrounds only one of these points
The radius of Apollonius' circles is
:
Unit circle is a circle with radius 1 and center at the origin. The concept of a unit circle can be easily generalized to n-dimensional space ( n 2). In this case, the term “unit sphere” is used.
For all points on the circle it is valid to agree with the Pythagorean theorem: x 2 + y 2 = 1.
Do not confuse the terms “circle” and “circle”!
Circle on given distance from a given point, on one plane - a curve.
Circle - geometric locus of points located no further than a circle , on one plane - a figure.
Also, a section of algebra, such as trigonometry, can be attributed to the unit circle.
Trigonometry.
Sine and cosine can be described in the following way: connecting any point (
x
,
y
) on the unit circle with the origin (0,0), we obtain a segment located at an angle α relative to the positive semi-axis of abscissa. Then indeed:
cos α = x
sin α = y
Substituting these values into the above equation x 2 + y 2 = 1, we get:
cos 2 α + sin 2 α = 1
Note the common spelling cos 2 x = (cos x ) 2 .
The frequency is also clearly described here. trigonometric functions, since the angle of the segment does not depend on the number of " full revolutions»:
sin( x + 2 π k ) = sin( x )
cos( x + 2 π k ) = cos( x )
for all integers k , in other words, k belongs Z .
Complex plane.
In the complex plane unit circle describes the set:
A bunch of G satisfies the conditions of a multiplicative group (with a neutral element e i 0 = 1).
The secant theorem - planimetry theorem. Formulated as follows:
If two secants are drawn from a point lying outside the circle, then the product of one secant and its external part is equal to the product of the other secant and its external part.
If we translate this statement into the language of letters (according to the figure on the right), we get the following:
A special case of the secant theorem is Tangent and secant theorem:
If a tangent and a secant are drawn to a circle from one point, then the product of the entire secant and its outer part is equal to the square of the tangent.
Internet resources used:
www.wikipedia.org
And also literature: Geometry grades 7-11 Definitions, properties, methods for solving problems in tables E.P. Nelin
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First, let's understand the difference between a circle and a circle. To see this difference, it is enough to consider what both figures are. These are an infinite number of points on the plane, located at equal distances from a single center point. But, if the circle also consists of internal space, then it does not belong to the circle. It turns out that a circle is both a circle that limits it (circle(r)), and an innumerable number of points that are inside the circle.
For any point L lying on the circle, the equality OL=R applies. (The length of the segment OL is equal to the radius of the circle).
A segment that connects two points on a circle is its chord.
A chord passing directly through the center of a circle is diameter this circle (D). The diameter can be calculated using the formula: D=2R
Circumference calculated by the formula: C=2\pi R
Area of a circle: S=\pi R^(2)
Arc of a circle is called that part of it that is located between its two points. These two points define two arcs of a circle. The chord CD subtends two arcs: CMD and CLD. Identical chords subtend equal arcs.
Central angle An angle that lies between two radii is called.
Arc length can be found using the formula:
- Using degree measure: CD = \frac(\pi R \alpha ^(\circ))(180^(\circ))
- Using radian measure: CD = \alpha R
The diameter, which is perpendicular to the chord, divides the chord and the arcs contracted by it in half.
If the chords AB and CD of the circle intersect at the point N, then the products of the segments of the chords separated by the point N are equal to each other.
AN\cdot NB = CN\cdot ND
Tangent to a circle
Tangent to a circle It is customary to call a straight line that has one common point with a circle.
If a line has two common points, it is called secant.
If you draw the radius to the tangent point, it will be perpendicular to the tangent to the circle.
Let's draw two tangents from this point to our circle. It turns out that the tangent segments will be equal to one another, and the center of the circle will be located on the bisector of the angle with the vertex at this point.
AC = CB
Now let’s draw a tangent and a secant to the circle from our point. We obtain that the square of the length of the tangent segment will be equal to the product of the entire secant segment and its outer part.
AC^(2) = CD \cdot BC
We can conclude: the product of an entire segment of the first secant and its external part is equal to the product of an entire segment of the second secant and its external part.
AC\cdot BC = EC\cdot DC
Angles in a circle
The degree measures of the central angle and the arc on which it rests are equal.
\angle COD = \cup CD = \alpha ^(\circ)
Inscribed angle is an angle whose vertex is on a circle and whose sides contain chords.
You can calculate it by knowing the size of the arc, since it is equal to half of this arc.
\angle AOB = 2 \angle ADB
Based on a diameter, inscribed angle, right angle.
\angle CBD = \angle CED = \angle CAD = 90^ (\circ)
Inscribed angles that subtend the same arc are identical.
Inscribed angles resting on one chord are identical or their sum is equal to 180^ (\circ) .
\angle ADB + \angle AKB = 180^ (\circ)
\angle ADB = \angle AEB = \angle AFB
On the same circle are the vertices of triangles with identical angles and a given base.
An angle with a vertex inside a circle and located between two chords is identical to half the sum angular values arcs of a circle that are contained within a given and vertical angle.
\angle DMC = \angle ADM + \angle DAM = \frac(1)(2) \left (\cup DmC + \cup AlB \right)
An angle with a vertex outside the circle and located between two secants is identical to half the difference in the angular values of the arcs of the circle that are contained inside the angle.
\angle M = \angle CBD - \angle ACB = \frac(1)(2) \left (\cup DmC - \cup AlB \right)
Inscribed circle
Inscribed circle is a circle tangent to the sides of a polygon.
At the point where the bisectors of the corners of a polygon intersect, its center is located.
A circle may not be inscribed in every polygon.
The area of a polygon with an inscribed circle is found by the formula:
S = pr,
p is the semi-perimeter of the polygon,
r is the radius of the inscribed circle.
It follows that the radius of the inscribed circle is equal to:
r = \frac(S)(p)
The sums of the lengths of opposite sides will be identical if the circle is inscribed in convex quadrilateral. And vice versa: a circle fits into a convex quadrilateral if the sums of the lengths of opposite sides are identical.
AB + DC = AD + BC
It is possible to inscribe a circle in any of the triangles. Only one single one. At the point where the bisectors intersect internal corners figure, the center of this inscribed circle will lie.
The radius of the inscribed circle is calculated by the formula:
r = \frac(S)(p) ,
where p = \frac(a + b + c)(2)
Circumcircle
If a circle passes through each vertex of a polygon, then such a circle is usually called described about a polygon.
At the point of intersection of the perpendicular bisectors of the sides of this figure will be the center of the circumcircle.
The radius can be found by calculating it as the radius of the circle that is circumscribed about the triangle defined by any 3 vertices of the polygon.
Eat next condition: a circle can be described around a quadrilateral only if the sum of its opposite angles is equal to 180^( \circ) .
\angle A + \angle C = \angle B + \angle D = 180^ (\circ)
Around any triangle you can describe a circle, and only one. The center of such a circle will be located at the point where the perpendicular bisectors of the sides of the triangle intersect.
The radius of the circumscribed circle can be calculated using the formulas:
R = \frac(a)(2 \sin A) = \frac(b)(2 \sin B) = \frac(c)(2 \sin C)
R = \frac(abc)(4 S)
a, b, c are the lengths of the sides of the triangle,
S is the area of the triangle.
Ptolemy's theorem
Finally, consider Ptolemy's theorem.
Ptolemy's theorem states that the product of diagonals is identical to the sum of the products of opposite sides of a cyclic quadrilateral.
AC \cdot BD = AB \cdot CD + BC \cdot AD
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