A circle circumscribed around a triangle. A triangle inscribed in a circle. Theorem of sines

Definition

A circle \(S\) is circumscribed about a polygon \(P\) if all the vertices of the polygon \(P\) lie on the circle \(S\) .

In this case, the polygon \(P\) is said to be inscribed in a circle.

Definition

The perpendicular bisector of a segment is a line passing through the middle of a given segment perpendicular to it.

Theorem

Each point of the perpendicular bisector of a segment is equidistant from the ends of that segment.

Proof

Consider the segment \(AB\) and the perpendicular bisector \(a\) to it. Let us prove that for any point \(X\in a\) the following holds: \(AX=BX\) .

Consider \(\triangle AXB\) : the segment \(XO\) is the median and altitude, therefore \(\triangle AXB\) is isosceles, therefore \(AX=BX\) .

Theorem

The perpendicular bisectors to the sides of a triangle intersect at one point.

Proof

Consider \(\triangle ABC\) . Let's draw perpendicular bisectors to the sides \(AB\) and \(AC\). They will intersect at the point \(O\) .

According to the previous theorem, for the perpendicular bisector \(C_1O\) the following holds: \(AO=BO\) , and for \(B_1O\) - \(AO=CO\) . Therefore, \(BO=CO\) . This means that \(\triangle BOC\) is isosceles, therefore, the height \(OA_1\) drawn to the base \(BC\) will also be the median. This means that \(OA_1\) is the perpendicular bisector to the segment \(BC\) .

Thus, all three perpendicular bisectors intersect at one point \(O\) .

Consequence

If a point is equidistant from the ends of a segment, then it lies on its perpendicular bisector.

Theorem

A single circle can be circumscribed around any triangle, and the center of the circumscribed circle is the point of intersection of the perpendicular bisectors to the sides of the triangle.

Proof

From the theorem proven above it follows that \(AO=BO=CO\) . This means that all the vertices of the triangle are equidistant from the point \(O\), therefore, they lie on the same circle.

There is only one such circle. Let us assume that another circle can be described around \(\triangle ABC\). Then its center should coincide with the point \(O\) (since this is the only point equidistant from the vertices of the triangle), and the radius should be equal to the distance from the center to some of the vertices, i.e. \(OA\) . Because If these circles have the same center and radius, then these circles also coincide.

Inscribed triangle area theorem

If \(a, b, c\) are the sides of the triangle, and \(R\) is the radius of the circle circumscribed around it, then the area of the triangle \

Proof*
It is recommended that you familiarize yourself with the proof of this theorem after studying the topic “Theorem of Sines”.

Let's denote the angle between sides \(a\) and \(c\) as \(\alpha\) . Then \(S_(\triangle)=\frac12 ac\cdot \sin \alpha\).

By the theorem of sines \(\dfrac b(\sin\alpha)=2R\) , whence \(\sin \alpha=\dfrac b(2R)\) . Hence, \(S_(\triangle)=\dfrac(abc)(4R)\).

Theorem

A circle can be described around a quadrilateral if and only if the sum of its opposite angles is equal to \(180^\circ\) .

Proof

Necessity.

If a circle can be described around a quadrilateral \(ABCD\), then \(\buildrel\smile\over(ABC) + \buildrel\smile\over(ADC) = 360^\circ\), where \(\angle ABC + \angle ADC = \frac(1)(2)\buildrel\smile\over(ABC) + \frac(1)(2)\buildrel\smile\over(ADC) = \frac(1 )(2)(\buildrel\smile\over(ABC) + \buildrel\smile\over(ADC)) = 180^\circ\). For angles \(BCD\) and \(BAD\) it is similar.

Adequacy.

Let us describe a circle around the triangle \(ABC\) . Let the center of this circle be the point \(O\) . On the line passing through the points \(O\) and \(D\), we mark the point \(D"\) of intersection of this line and the circle. Let us assume that the points \(D\) and \(D"\) do not coincide, then consider the quadrilateral \(CD"AD\) .

Angles \(CD"A\) and \(CDA\) complement angle \(ABC\) to \(180^\circ\) (\(\angle CDA\) complements by condition, and \(\angle CD"A \) as proven above), therefore, they are equal, but then the sum of the angles of the quadrilateral \(AD"CD\) is greater than \(360^\circ\), which cannot be (the sum of the angles of this quadrilateral is the sum of the angles of two triangles), therefore, the points \(D\) and \(D"\) coincide.

Comment. In the figure, the point \(D\) lies outside the circle bounded by the circle circumscribed by \(\triangle ABC\), however, in the case when \(D\) lies inside, the proof also remains valid.

Theorem

A circle can be described around a convex quadrilateral \(ABCD\) if and only if \(\angle ABD=\angle ACD\) .

Proof

Necessity. If a circle is circumscribed around \(ABCD\), then the angles \(\angle ABD\) and \(\angle ACD\) are inscribed and rest on one arc \(\buildrel\smile\over(AD)\), therefore, they are equal.

Adequacy. Let \(\angle ABD=\angle ACD=\alpha\). Let us prove that a circle can be described around \(ABCD\).

Let's describe a circle around \(\triangle ABD\) . Let the straight line \(CD\) intersect this circle at the point \(C"\). Then \(\angle ABD=\angle AC"D \Rightarrow \angle AC"D=\angle ACD\).

Hence, \(\angle CAD=\angle C"AD=180^\circ-\angle ADC-\angle AC"D\), that is \(\triangle AC"D=\triangle ACD\) along a common side \(AD\) and two adjacent angles (\(\angle C"AD=\angle CAD\) , \(\angle ADC"=\angle ADC\) – common). This means \(DC"=DC\), that is, the points \(C"\) and \(C\) coincide.

Theorems

1. If a circle is circumscribed around a parallelogram, then it is a rectangle (Fig. 1).

2. If a circle is described around a rhombus, then it is a square (Fig. 2).

3. If a circle is described around a trapezoid, then it is isosceles (Fig. 3).

The converse statements are also true: around a rectangle, rhombus and isosceles trapezoid, one can describe a circle, and only one.

Proof

1) Let a circle be circumscribed around the parallelogram \(ABCD\). Then the sums of its opposite angles are equal \(180^\circ: \quad \angle A+\angle C=180^\circ\). But in a parallelogram, opposite angles are equal, because \(\angle A=\angle C\) . Hence, \(\angle A=\angle C=90^\circ\). So, by definition, \(ABCD\) is a rectangle.

2) Let a circle be circumscribed around the rhombus \(MNKP\). Similar to the previous point (since a rhombus is a parallelogram), it is proved that \(MNKP\) is a rectangle. But all sides of this rectangle are equal (since it is a rhombus), which means \(MNKP\) is a square.

The opposite statement is obvious.

3) Let a circle be circumscribed around the trapezoid \(QWER\). Then \(\angle Q+\angle E=180^\circ\). But from the definition of a trapezoid it follows that \(\angle Q+\angle W=180^\circ\). Therefore, \(\angle W=\angle E\) . Because the angles at the base \(WE\) of the trapezoid are equal, then it is isosceles.

The opposite statement is obvious.

This article contains the minimum set of information about the circle required to successfully pass the Unified State Exam in mathematics.

Circumference is a set of points located at the same distance from a given point, which is called the center of the circle.

For any point lying on the circle, the equality is satisfied (The length of the segment is equal to the radius of the circle.

A line segment connecting two points on a circle is called chord.

A chord passing through the center of a circle is called diameter circle() .

Circumference:

Area of a circle:

Arc of a circle:

The part of a circle enclosed between two points is called arc circles. Two points on a circle define two arcs. The chord subtends two arcs: and . Equal chords subtend equal arcs.

The angle between two radii is called central angle :

To find the arc length, we make a proportion:

a) the angle is given in degrees:

b) the angle is given in radians:

Diameter perpendicular to chord , divides this chord and the arcs it subtends in half:

If chords And circles intersect at a point , then the products of the chord segments into which they are divided by a point are equal to each other:

Tangent to a circle.

A straight line that has one common point with a circle is called tangent to the circle. A straight line that has two points in common with a circle is called secant

A tangent to a circle is perpendicular to the radius drawn to the point of tangency.

If two tangents are drawn from a given point to a circle, then tangent segments are equal to each other and the center of the circle lies on the bisector of the angle with the vertex at this point:

If a tangent and a secant are drawn from a given point to a circle, then the square of the length of a tangent segment is equal to the product of the entire secant segment and its outer part :

Consequence: the product of the entire segment of one secant and its external part is equal to the product of the entire segment of another secant and its external part:

Angles in a circle.

The degree measure of the central angle is equal to the degree measure of the arc on which it rests:

An angle whose vertex lies on a circle and whose sides contain chords is called inscribed angle . An inscribed angle is measured by half the arc on which it subtends:

∠∠

The inscribed angle subtended by the diameter is right:

∠∠∠

Inscribed angles subtended by one arc are equal :

Inscribed angles subtending one chord are equal or their sum is equal

∠∠

The vertices of triangles with a given base and equal angles at the vertex lie on the same circle:

Angle between two chords (an angle with a vertex inside a circle) is equal to half the sum of the angular values of the arcs of a circle contained inside a given angle and inside a vertical angle.

∠ ∠∠(⌣ ⌣ )

Angle between two secants (an angle with a vertex outside the circle) is equal to the half-difference of the angular values of the arcs of the circle contained inside the angle.

∠ ∠∠(⌣ ⌣ )

Inscribed circle.

The circle is called inscribed in a polygon , if it touches its sides. Center of inscribed circle lies at the intersection point of the bisectors of the angles of the polygon.

Not every polygon can fit a circle.

Area of a polygon in which a circle is inscribed can be found using the formula

here is the semi-perimeter of the polygon, and is the radius of the inscribed circle.

From here inscribed circle radius equals

If a circle is inscribed in a convex quadrilateral, then the sums of the lengths of opposite sides are equal . Conversely: if in a convex quadrilateral the sums of the lengths of opposite sides are equal, then a circle can be inscribed in the quadrilateral:

You can inscribe a circle into any triangle, and only one. The center of the incircle lies at the point of intersection of the bisectors of the interior angles of the triangle.

Inscribed circle radius equal to . Here

Circumscribed circle.

The circle is called described about a polygon , if it passes through all the vertices of the polygon. The center of the circumcircle lies at the point of intersection of the perpendicular bisectors of the sides of the polygon. The radius is calculated as the radius of the circle circumscribed by the triangle defined by any three vertices of the given polygon:

A circle can be described around a quadrilateral if and only if the sum of its opposite angles is equal to .

Around any triangle you can describe a circle, and only one. Its center lies at the point of intersection of the perpendicular bisectors of the sides of the triangle:

Circumradius calculated using the formulas:

Where are the lengths of the sides of the triangle and is its area.

Ptolemy's theorem

In a cyclic quadrilateral, the product of the diagonals is equal to the sum of the products of its opposite sides:

The diameter of a circle is the straight line segment that connects the two points of the circle that are most distant from each other, passing through the center of the circle. The name diameter comes from the Greek language and literally means transverse. The diameter is indicated by the letter D of the Latin alphabet or the symbol O.

Circle diameter

In order to know how to find the diameter of a circle, you need to refer to the formulas. There are two basic formulas by which you can calculate the diameter of a circle. The first is D = 2R. Here the diameter is equal to twice the radius, where the radius is the distance from the center to any point on the circle (R). Let's consider an example: if the radius is known in the task and it is equal to 10 cm, then you can easily find the diameter. For this radius value, we substitute D = 2 * 10 = 20 cm into the formula

The second formula makes it possible to find the diameter along the circumference and it looks like this: D = L/P, where L is the value of the circumference, and P is the number Pi, which is approximately equal to 3.14. This formula is very convenient to use in practice. If you need to know the diameter of a hatch, tank cover, or some kind of pit, you just need to measure their circumference and divide it by 3.14. For example, the circumference is 600 cm, hence D = 600/3.14 = 191.08 cm.

Circumcircle diameter

The diameter of a circumscribed circle can also be found if it is circumscribed or inscribed in a triangle. To do this, you first need to find the radius for the inscribed circle using the formula: R = S/p, where S denotes the area of the triangle, and p is its semi-perimeter, p is equal to (a + b + c)/2. Once the radius is known, you need to use the first formula. Or immediately substitute all values into the formula D = 2S/p.

If you don't know how to find the diameter of a circumscribed circle, use the formula to find the radius of a circle circumscribed by a triangle. R = (a * b * c)/4 * S, S in the formula denotes the area of the triangle. Then, in the same way, substitute the value of the radius into the formula D = 2R.

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 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 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 the circle and located between two chords is identical to half the sum of the angular values of the arcs of the circle that are contained within 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 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 a 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 of the internal corners 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)

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 circumscribed circle.

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.

There is the following 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.