Circle - geometric figure, consisting of all points of the plane located at a given distance from a given point.
This point (O) is called circle center.
Circle radius is a line segment that connects the center to a point on the circle. All radii have the same length (by definition).
Chord A line segment that connects two points on a circle. The chord passing through the center of the circle is called diameter. The center of a circle is the midpoint of any diameter.
Any two points on the circle divide it into two parts. Each of these parts is called circular arc. The arc is called semicircle if the segment connecting its ends is a diameter.
The length of a unit semicircle is denoted by π
.
The sum of the degree measures of two circular arcs with common ends is 360º.
The part of the plane bounded by a circle is called around.
circular sector- a part of a circle bounded by an arc and two radii connecting the ends of the arc with the center of the circle. The arc that bounds the sector is called sector arc.
Two circles that have a common center are called concentric.
Two circles that intersect at right angles are called orthogonal.
Mutual arrangement of a straight line and a circle
- If the distance from the center of the circle to the straight line is less than the radius of the circle ( d), then the line and the circle have two common points. In this case, the line is called secant in relation to the circle.
- If the distance from the center of the circle to the line is equal to the radius of the circle, then the line and the circle have only one common point. Such a line is called tangent to circle, and their common point is called point of contact between a line and a circle.
- If the distance from the center of the circle to the line is greater than the radius of the circle, then the line and the circle do not have common points .
Central and inscribed angles
Central corner is the angle with the vertex at the center of the circle.
Inscribed angle An angle whose vertex lies on the circle and whose sides intersect the circle.
Inscribed angle theorem
An inscribed angle is measured by half the arc it intercepts.
- Consequence 1.
Inscribed angles subtending the same arc are equal. - Consequence 2.
An inscribed angle that intersects a semicircle is a right angle.
Theorem on the product of segments of intersecting chords.
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.
Basic formulas
- Circumference:
- Arc length:
- Diameter:
- Arc length:
Where α - degree measure of the length of an arc of a circle)
- Area of a circle:
- Circular sector area:
Circle equation
- In a rectangular coordinate system, the equation for a circle of radius r centered on a point C(x o; y o) has the form:
- The equation for a circle of radius r centered at the origin is:
Let's first understand the difference between a circle and a circle. To see this difference, it is enough to consider what both figures are. This is an infinite number of points in the plane, located at an equal distance from a single central 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 bounds it (o-circle (g)ness), and an uncountable 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 line segment that connects two points on a circle is 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 called that part of it, which is located between two of its points. These two points define two arcs of a circle. The chord CD subtends two arcs: CMD and CLD. The same chords subtend the same arcs.
Central corner is the angle between two radii.
arc length can be found using the formula:
- Using degrees: CD = \frac(\pi R \alpha ^(\circ))(180^(\circ))
- Using a radian measure: CD = \alpha R
The diameter, which is perpendicular to the chord, bisects the chord and the arcs it spans.
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 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 points in common, it is called secant.
If you draw a radius at the point of contact, 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 segments of the tangents 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 we draw a tangent and a secant to the circle from our point. We get that the square of the length of the tangent segment will be equal to the product of the entire secant segment by its outer part.
AC^(2) = CD \cdot BC
We can conclude: the product of an integer segment of the first secant by its outer part is equal to the product of an integer segment of the second secant by its outer 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, straight.
\angle CBD = \angle CED = \angle CAD = 90^ (\circ)
Inscribed angles that lean on the same arc are identical.
The inscribed angles based on the same chord are identical or their sum equals 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 the circle and located between two chords is identical to half the sum of the angular magnitudes of the arcs of the circle that are inside the given and vertical angles.
\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 magnitudes of the arcs of a circle that are 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 the polygon.
At the point where the bisectors of the angles of the 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 semiperimeter of the polygon,
r is the radius of the inscribed circle.
It follows that the radius of the inscribed circle is:
r = \frac(S)(p)
The sums of the lengths of opposite sides will be identical if the circle is inscribed in a convex quadrilateral. And vice versa: a circle is inscribed in a convex quadrilateral if the sums of the lengths of opposite sides in it are identical.
AB+DC=AD+BC
It is possible to inscribe a circle in any of the triangles. Only one single. At the point where the bisectors of the inner angles of the figure intersect, 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)
Circumscribed circle
If a circle passes through every vertex of a polygon, then such a circle is called circumscribed about a polygon.
The center of the circumscribed circle will be at the point of intersection of the perpendicular bisectors of the sides of this figure.
The radius can be found by calculating it as the radius of a circle that is circumscribed about a triangle defined by any 3 vertices of the polygon.
There is the following condition: a circle can be circumscribed 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)
Near any triangle it is possible to describe a circle, and one 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 by the formulas:
R = \frac(a)(2 \sin A) = \frac(b)(2 \sin B) = \frac(c)(2 \sin C)
R = \frac(abc)(4S)
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 the diagonals is identical to the sum of the products of the opposite sides of an inscribed quadrilateral.
AC \cdot BD = AB \cdot CD + BC \cdot AD
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Preview:
Related lesson:
"Theorem on the product of segments of intersecting chords»
Subject : geometry
Grade : 8
teacher b: Herat Ludmila Vasilievna
School : MOBU "Druzhbinskaya secondary school" Sol-Iletsk district, Orenburg region
Lesson type: Lesson "discovery" of new knowledge.
Forms of work: individual, frontal, group.
Teaching methods:verbal, visual, practical, problematic.
Equipment: computer class, multimedia projector,
Handout (cards), presentation.
Lesson Objectives:
- educational- study the theorem on the product of intersecting chords, and show its application in solving problems.
Improve the skills of solving problems on the application of the inscribed angle theorem and its consequences.
- developing - develop creativity and mental activity students in the classroom; to develop the intellectual qualities of the personality of schoolchildren, such as independence, flexibility, the ability for evaluative actions, generalization; to promote the formation of skills of collective and independent work; develop the ability to clearly and clearly express their thoughts.
- educational - to instill in students an interest in the subject through the use of information technologies(using a computer); to form the ability to accurately and competently perform mathematical records, draw up a picture for the problem.
Educational activities are aimed at improving the effectiveness, productivity of pedagogical work by transferring students from the position object teacher's activities in position the subject of the doctrine , contributes to the development of the potential of each child, the disclosure of the possibilities inherent in it.
The upbringing (development) of subjectness is possible only in activities,in which the subject is involved, in which he himself: a) sets goals; b) concentrates volitional effort to achieve the goal; c) reflects on the progress and results of their work. Reflection is the most powerful tool for personal self-development(self-construction of personality).
The problem of the development of the student's subjectivitycannot be fully resolved by one-time measures. This quality developsconsistently by including the student in the educational and cognitive activity (ideally in every lesson) that he/she performs himself, applying his own efforts, their on their own, with minimal outside help, all actions in their logical sequence. The lesson provides students with reflection on all 4 stages of work plus outcomes, fully meeting the requirementsactivity approach in education.
Through the proposed design of the lesson and the use of computer technology, the development goals are pursued:
- intellectual culture;
- Information culture;
- Cultures of self-organization;
- Research culture;
The activities of students should be organized in such a way as to provide students with internal goals-motives; the need for search is the most important task of training and education, for this it is necessary to create situations of success, search situations that cause positive emotions.
Lesson plan
1. Proof of the inscribed angle theorem (3 cases); card work,
Solution of problems according to ready-made drawings.
2. Work in pairs.
3. Study of the theorem on the product of segments of intersecting chords.
4. Solving problems for fixing the theorem.
During the classes.
- Updating students' knowledge on the topic under study.
Three students are called to the board to prove theorems, two students receive task cards, the rest of the students solve problems on ready-made drawings. The proof of theorems is heard by the whole class after students solve problems on finished drawings.
Card number 1..
1. Insert the missing words "An angle is called inscribed if its vertex lies on …………….., and the sides of the angle are……………………………..".
2. Find and write down the inscribed angles shown in the figure:
3. Find the degree measure of the angle ABC shown in the figure if the degree measure of the arc ABC = 270.
Card number 2.
1. Insert the missing words: "The inscribed angle is measured by ………….".
- Given: OA=AB. Find the degree measure of the arc AB.
Solution of problems according to ready-made drawings.
Fig.1. Find Fig.2. Fig.3. Fig.4. Fig.5.
AOD, ACD Find ABC Find BCD Find BAC Find BCD
II. Work in pairs.
The proof of the theorem on segments of intersecting chords is carried out in the form of a problem:
Prove that if two chords AB and CD of a circle intersect at a point E, then
AE * BE = CE * DE
The task is proposed to be solved independently in pairs, and then discuss its solution. In notebooks and on the board, write down the outline of the proof of the theorem.
Outline plan
a) ACE TWO (A = D as inscribed angles based on arc BC;
AEC = DEB as vertical).
Issues for discussion:
What can you say about the angles CAB and CDB? About the angles AEC and DEB?
What are triangles ACE and DBE? What is the ratio of their sides, which are segments of the tangent chords?
What equality can be written from the equality of two ratios using the basic property of proportions?
IV. Consolidation of the studied material.
Solve the problem: The chords of the circle RT and KM intersect at point E. Find ME if
KE = 4cm, TE = 6cm, PE = 2cm.
Solution: AE * BE = CE * DE
AE * 4 = 2 * 6
AE = 3cm.
No. 666 b. x*x =16*9
X * x \u003d 144
X = 12
V. Reflection. (using stickers in three colors)
VI. Homework.
p. 71, No. 666 a, c; 667.
Part 3. Circles
I. Reference materials.
I. Properties of tangents, chords and secants. Inscribed and central angles.
Circle and circle
1. If from one point lying outside the circle, draw two tangents to it, then
a) the lengths of the segments from a given point to the points of contact are equal;
b) the angles between each tangent and secant passing through the center of the circle are equal.
2. If from one point lying outside the circle, draw a tangent and a secant to it, then the square of the tangent is equal to the product of the secant by its outer part
3. If two chords intersect at one point, then the product of the segments of one chord is equal to the product of the segments of the other.
4. Circumference С=2πR;
5. Arc length L =πRn/180˚
6. Circle area S=πR 2
7. Sector area S c=πR 2 n/360
The measure of an inscribed angle is half the measure of the arc it intercepts.
Theorem 1. The measure of the angle between a tangent and a chord having a common point on a circle is equal to half the degree measure of the arc enclosed between its sides
Theorem 2(about tangent and secant). If a tangent and a secant are drawn from the point M to the circle, then the square of the segment of the tangent from the point M to the point of contact is equal to the product of the lengths of the segments of the secant from the point M to the points of its intersection with the circle.
Theorem 3. If two chords of a circle intersect, then the product of the lengths of the segments of one chord is equal to the product of the lengths of the segments of the other chord, that is, if the chords AB and SD intersect at the point M, then AB MV \u003d CM MD.
Properties of circle chords:
A diameter perpendicular to a chord bisects it. Conversely: the diameter passing through the middle of the chord is perpendicular to it.
Equal chords of a circle are equidistant from the center of the circle. Conversely, there are equal chords at an equal distance from the center of the circle.
Circular arcs enclosed between parallel chords are equal.
circles that have a common point and a common tangent at this point are called) tangent. If the circles are located on the same side of the common tangent, then they are called internally tangent, and if on opposite sides of the tangent, then they are called externally tangent.
II. Additional materials
Properties of some angles.
Theorem.
1) The angle (ABC), the vertex of which lies inside the circle, is the half-sum of two arcs (AC and DE), of which one is enclosed between its sides, and the other between the extensions of the sides.
2) the angle (ABC), the vertex of which lies outside the circle and the sides intersect with the circle, is the half-difference of two arcs (AC and ED) enclosed between its sides
Proof .
Drawing the chord AD (on both drawings), we get ∆ABD,
relative to which the considered angle ABC serves as outer when its vertex lies inside the circle, and as inner when its vertex lies outside the circle. So in the first case: ; in the second case:
But the angles ADC and DAE, as inscribed, are measured by half arcs AC and DE; therefore, the angle ABC is measured: in the first case, by the sum: ½ ﬞ AC + 1/2 ﬞ DE, which is equal to 1
/
2
(ﮟ
AC+ﮞ
DE), and in the second case, the difference is 1/2 ﬞ AC- 1/2 ﬞ DE, which is equal to 1/2 (ﬞ AC- ﬞ DE). Theorem. An angle (ACD) composed by a tangent and a chord is measured by half the arc enclosed within it. Now take the general case when the chord CD does not pass through the center. Having then drawn the diameter CE, we will have: At Proportional lines in a circle Theorem. If any chord (AB) and a diameter (CD) are drawn through a point (M) taken inside the circle, then the product of the segments of the chord (AM MB) is equal to the product of the segments of the diameter (MV MC). Proof. P AM: MD=MS: MB, whence AM MB=MD MC. Consequence. If any number of chords (AB, EF, KL, ...) are drawn through a point (M) taken inside the circle, then the product of segments of each chord is a constant number for all chords, since for each cord this product is equal to the product of segments of diameter CD passing through the taken point M. Theorem. If from a point (M) taken outside the circle, some secant (MA) and tangent (MC) are drawn to it, then the product of the secant by its outer part is equal to the square of the tangent (it is assumed that the secant is limited by the second intersection point, and the tangent is touch point). Let's draw auxiliary chords AC and BC; then we get two triangles MAC and MBC (covered in the figure with strokes), which are similar, because they have a common angle M and the angles MSV and CAB are equal, since each of them is measured by half the arc BC. Let us take the sides MA and MC in ∆MAS; similar sides in ∆MVS will be MC and MB; therefore MA: MS=MS: MB, whence MA MB=MS 2 . Consequence. If from a point (M) taken outside the circle, any number of secants (MA, MD, ME, ...) are drawn to it, then the product of each secant by its outer part is a constant number for all secants, since for each secant it is the product is equal to the square of the tangent (MS 2) drawn from the point M. III. introductory tasks. Task 1. IN Solution 1) The radius of a circle circumscribed about a trapezoid is the same as the radius of a circle circumscribed about a triangle whose vertices are any three vertices of the trapezoid. Find the radius R of a circle circumscribed about a triangle ABD. 2)
ABCD isosceles trapezoid, so AK = MD, KM =. In ∆ ABK AK
= AB cos A = · cos 60° = . Means, BK = AB sin A
=
· = . 3) By the law of cosines in ∆ ABD
BD 2
= AB 2
+ AD 2
– 2AB ·
AD cos A. BD 2 = () 2 + (3) 2 – 2 3 = 21 + 9 21 – 3 21 = 7 21; 4) S(∆ ABD)
= AD ·
BK; S(∆ ABD) = · · 3 = . Task 2. In an equilateral triangle ABC a circle is inscribed and a line segment is drawn NM, M
AC, N
BC, which is tangent to it and parallel to the side AB. Determine the perimeter of the trapezoid AMNB if the length of the segment MN equals 6. Solution. 1 2) MN- tangent to the circle, P is the point of contact, OD = 3) ∆CMN ∾
∆ CAB, so ∆ CMN- equilateral CM = CN = MN = = 6; P. And 3) BN = CB – CN = 18 – 6 = 12. 4) P ( AMNB)
= AM + MN
+ BN + AB
= 18 + 6 + 12 + 12 = 48. An isosceles trapezoid is described near the circle, the median line of which is 5, and the sine of the acute angle at the base is 0.8. Find the area of the trapezoid. Solution. FP is the midline of the trapezoid BC + AD = 2FP. Then AB = CD = FP = 5. ∆ABK- rectangular, BK = AB sin A; BK= 5 0.8 = 4. S( ABCD)
= FP · BK= 5 4 = 20. Answer:
20. The inscribed circle of triangle ABC touches the side BC at point K, and the excircle touches the side BC at point L. Prove that CK=BL=(a+b+c)/2 Proof: let M and N be the points of contact of the inscribed circle with the sides AB and BC. Then BK+AN=BM+AM=AB, so CK+CN= a+b-c. Let P and Q be the tangent points of the excircle with the extensions of the sides AB and BC. Then AP=AB+BP=AB+BL and AQ=AC+CQ=AC+CL. Therefore AP+AQ=a+b+c. Therefore, BL=BP=AP-AB=(a+b-c)/2. a) The continuation of the bisector of angle B of triangle ABC intersects the circumscribed circle at point M. O is the center of the inscribed circle. О В is the center of the excircle tangent to the side AC. Prove that points A, C, O and O B lie on a circle with center M. D b) The point O, which lies inside the triangle ABC, has the property that the lines AO, BO, CO pass through the centers of the circumscribed circles of the triangles BCO, ACO, ABO. Prove that O is the center of the inscribed circle of triangle ABC. IV. Additional tasks No. 1. The circle tangent to the hypotenuse of a right triangle and the extensions of its legs has radius R. Find the perimeter of the triangle R 1) ∆OAH = ∆OAF along the leg and hypotenuse =>HA=FA 2) ∆OCF=∆OCG =>CF=CG 3) P ABC=AB+AF+FC+BC=AB+AM+GC+BC+BH+BG=2R No. 2. Points C and D lie on a circle with diameter AB. AC ∩ BD = P and AD ∩ BC = Q. Prove that lines AB and PQ are perpendicular to Proof: A No. 3. In parallelogram ABCD, diagonal AC is greater than diagonal BD; М is a point of diagonal AC, BDCM is an inscribed quadrilateral. Prove that line BD is a common tangent to the circumscribed circles of triangles ABM and ADM P No. 4. H According to the introductory task 4, CM=(AC+CE-AE)/2 and CN=(BC+CE-BE)/2. Considering that AC=BC, we get МN=|CM-CN|=|AE-BE|/2 No. 5. The lengths of the sides of triangle ABC form an arithmetic progression, with a Let M be the midpoint of side AC and N be the tangent point of the inscribed circle with side BC. Then BN=p–b (introductory problem 4), so BN=AM, because p=3b/2 by condition. Besides, V
.Tasks for independent solution
No. 1. Quadrilateral ABCD has the property that there is a circle inscribed in angle BAD and tangent to the extensions of sides BC and CD. Prove that AB+BC=AD+DC. No. 2. The common internal tangent to circles with radii R and r intersects their common external tangents at points A and B and touches one of the circles at point C. Prove that AC∙CB=Rr No. 3. In triangle ABC, angle C is a right angle. Prove that r =(a+b-c)/2 and r c =(a+b+c)/2 No. 4. The two circles intersect at points A and B; MN is the common tangent to them. Prove that line AB bisects segment MN. No. 5. The extensions of the bisectors of the angles of the triangle ABC intersect the circumscribed circle at points A 1 , B 1 , C 1 . M is the point of intersection of the bisectors. Prove that: a) MA·MC/MB 1 =2r; b) MA 1 MC 1 /MB=R No. 6. The angle formed by two tangents drawn from the same point on the circle is 23° 15`. Calculate arcs enclosed between tangent points No. 7. Calculate the angle formed by the tangent and the chord if the chord divides the circle into two parts related as 3:7. VI. Control tasks. Option 1. Point M is outside the circle with center O. Three secants are drawn from point M: the first intersects the circle at points B and A (M-B-A), the second at points D and C (M-D-C), and the third intersects the circle at points F and E (M-F-E) and passes through the center of the circle, AB = 4, BM = 5, FM = 3. Prove that if AB = CD, then angles AME and CME are equal. Find the radius of the circle. Find the length of the tangent drawn from point M to the circle. Find the angle AEB. Option 2. AB is the diameter of the circle with center O. The chord EF intersects the diameter at the point K (A-K-O), EK = 4, KF = 6, OK = 5. Find the radius of the circle. Find the distance from the center of the circle to the chord BF. Find the acute angle between diameter AB and chord EF. What is the chord FM if EM is parallel to AB. Option 3. Into a right triangle ABC ( Option 4. AB is the diameter of a circle with center O. The radius of this circle is 4, O 1 is the middle of OA. With the center at point O 1, a circle is drawn tangent to the larger circle at point A. The chord CD of the larger circle is perpendicular to AB and intersects AB at point K. E and F are the points of intersection of CD with the smaller circle (C-E-K-F-D), AK=3. Find chords AE and AC. Find the degree measure of the arc AF and its length. Find the area of the part of the smaller circle cut off by the chord EF. Find the radius of the circle circumscribed about the triangle ACE.
Suppose first that the chord CD passes through the center O, i.e. that a chord is a diameter. Then the angle ACD-
straight and therefore equal to 90°. But half of the arc CmD is also equal to 90°, since the whole arc CmD, constituting a semicircle, contains 180°. So the theorem is justified in this particular case.
the head of the ACE, as composed by the tangent and the diameter, is measured, as proved, by half the arc of the CDE; Angle DCE, as an inscribed one, is measured by half of the arc CnED: the only difference in the proof is that this angle must be considered not as a difference, but as the sum of a right angle ALL and an acute angle ECD.
drawing two auxiliary chords AC and BD, we get two triangles AMC and MBD (covered in the figure with strokes), which are similar, since their angles A and D are equal, as inscribed, based on the same arc BC, angles C and B are equal, as inscribed, based on the same arc AD. From the similarity of triangles we deduce:
Proof.
an isosceles trapezoid with an acute angle of 60 °, the side is equal, and the smaller base is. Find the radius of the circle circumscribed about this trapezoid.
AD
=
.
BD
=
.
)
∆ABC- equilateral, point O- the point of intersection of the medians (bisectors, heights), which means that CO :
OD = 2 :
1.
=OP, Then CD= 3 CP.
Since the circle is inscribed in a quadrilateral, then BC + AD = AB + CD. This quadrilateral is an isosceles trapezoid, so BC + AD = 2AB.
proof: because
Proof: Let P be the center of the circumscribed circle of triangle ACO. Then
Solution: HOGB - square with side R
D – diameter => inscribed angle ADB=90 o (as based on diameter) => QD/QP=QN/QA; ∆QDP is similar to ∆QNA in 2 sides and the angle between them => QN is perpendicular to AB .
ust O - the point of intersection of the diagonals AC and BD. Then MO ·
OC=BO ·
OD. Whereas OS = OA and BO = BD, then MO ·
OA \u003d IN 2 and MO ·
OA=DO 2 . These equalities mean that OB is tangent to the circumscribed circle of triangle ADM
the point E is taken at the base AB of the isosceles triangle ABC, and the triangles ACE and ABE are inscribed with circles touching the segment CE at the points M and N. Find the length of the segment MN if the lengths AE and BE are known.
Chord in Greek means "string". This concept is widely used in various fields of science - in mathematics, biology and others.
In geometry, the definition for the term will be as follows: it is a straight line segment that connects two arbitrary points on the same circle. If such a segment intersects the center curve, it is called the diameter of the circumscribed circle.
In contact with
How to build a geometric chord
To build this segment, you first need to draw a circle. Designate two arbitrary points through which a secant line is drawn. The line segment that is located between the points of intersection with the circle is called a chord.
If we divide such an axis in half and draw a perpendicular line from this point, it will pass through the center of the circle. You can carry out the opposite action - from the center of the circle to draw a radius perpendicular to the chord. In this case, the radius will divide it into two identical halves.
If we consider the parts of the curve that are limited to two parallel equal segments, then these curves will also be equal to each other.
Properties
There are a number of regularities connecting the chords and the center of the circle:
Relationship with radius and diameter
The above mathematical concepts are interconnected by the following laws:
Chord and Radius
There are the following connections between these concepts:
Relations with inscribed angles
Angles inscribed in a circle obey the following rules:
Arc Interactions
If two segments contract sections of the curve that are the same in size, then such axes are equal to each other. The following patterns follow from this rule:
A chord that subtends exactly half of a circle is its diameter. If two lines on the same circle are parallel to each other, then the arcs that are enclosed between these segments will also be equal. However, one should not confuse enclosed arcs with those contracted by the same lines.