Inscribed circle. A circle circumscribed about a triangle. A triangle inscribed in a circle

THEOREM ON A CIRCLE CIRCLE ABOUT A POLYGON: It is possible to circumscribe a circle around any regular polygon, and moreover, only one. THEOREM ABOUT THE WEAPON INSCRIBED IN A REGULAR POLYGON: Any regular polygon can be inscribed with a circle, and moreover, only one.

SPa4a4 rRN Calculate the area of a regular polygon, its side and inscribed circle radius and inscribed circle radius

Square of the Right Polygons Square of the Right Polygons Names and Polygon Square Number of Parties Name The Title of the Little Polygon Title 0.433A 2 4Chinth Cyagon1.000a 2 5598A 2 7 Cellifier 0.598A 2 7 Cellite 3,828A 2 9 Diggle6,182A 2 square

0 inscribed angles. Hippocrates of Chios The proof given in modern textbooks that an inscribed angle is measured by half the arc on which it rests is given in Euclid's Elements. However, Hippocrates of Chios (5th century BC) refers to this proposal in his work on “lunes”. The works of Hippocrates testify that already in the second half of the 5th century. BC e. a large number of theorems set forth in Euclid's Elements were known, and geometry reached a high level of development. The fact that an inscribed angle based on a diameter is a straight line was known to the Babylonians as early as 4,000 years ago. Its first proof is attributed to Pamphylia, a Roman writer from the time of Nero, to Thales of Miletus.

0 regular polygons Regular quadrangles, hexagons and octagons are found in Egyptian and Babylonian ancient monuments in the form of images on the walls and decorations carved from stone. Ancient Greek scientists began to show great interest in the correct figures since the time of Pythagoras. The division of a circle into a certain number of equal parts to construct regular polygons was important for the Pythagoreans, who claimed that numbers underlie all the phenomena of the world. The doctrine of regular polygons, begun in the school of Pythagoras, continued and developed in the 7th century. BC e., was systematized by Euclid and set forth in the IV book of the Beginnings. In addition to constructing a regular triangle, quadrilateral, pentagon and hexagon, Euclid also solves the problem of constructing a regular pentagon using only a compass and straightedge. This figure attracted the attention of the ancients, since it was noted that the arc of the angle of inclination of the ecliptic to the equator represents the entire circle, i.e., is contracted by the side of a regular pentagon.

ABC O1 O2 O1 is the center of the circumscribed circle, O2 is the center of the inscribed circle Necessity: Sufficiency: D AB + CD = BC + AD and, therefore, AB = CD = BAD = ADC, but BAD + ABC = 180 Hence ADC + ABC = 180 , and a circle can be inscribed around the trapezoid ABCD. Moreover, AB + CD = BC + AD and, therefore, a circle can be inscribed in ABCD. It is necessary and sufficient that the trapezoid be equilateral and the lateral side be equal to half the sum of the bases.

Definition 2

A polygon that satisfies the condition of Definition 1 is said to be inscribed about a circle.

Figure 1. Inscribed circle

Theorem 1 (on a circle inscribed in a triangle)

Theorem 1

In any triangle, you can inscribe a circle, and moreover, only one.

Proof.

Consider triangle $ABC$. Draw bisectors in it that intersect at the point $O$ and draw perpendiculars from it to the sides of the triangle (Fig. 2)

Figure 2. Illustration of Theorem 1

Existence: Draw a circle with center $O$ and radius $OK.\ $Since the point $O$ lies on three bisectors, it is equidistant from the sides of the triangle $ABC$. That is, $OM=OK=OL$. Consequently, the constructed circle also passes through the points $M\ and\ L$. Since $OM,OK\ and\ OL$ are perpendicular to the sides of the triangle, then by the tangent to circle theorem, the constructed circle touches all three sides of the triangle. Therefore, by virtue of the arbitrariness of a triangle, a circle can be inscribed in any triangle.

Uniqueness: Suppose that triangle $ABC$ can be inscribed with another circle centered at point $O"$. Its center is equidistant from the sides of the triangle, and therefore coincides with point $O$ and has a radius equal to the length of $OK$ But then this circle will coincide with the first one.

The theorem has been proven.

Corollary 1: The center of a circle inscribed in a triangle lies at the point of intersection of its bisectors.

Here are some more facts related to the concept of an inscribed circle:

Not every quadrilateral can be inscribed in a circle.

In any circumscribed quadrilateral, the sums of opposite sides are equal.

If the sums of opposite sides of a convex quadrilateral are equal, then a circle can be inscribed in it.

Definition 3

If all the vertices of the polygon lie on the circle, then the circle is called circumscribed near the polygon (Fig. 3).

Definition 4

A polygon that satisfies the condition of Definition 2 is called inscribed in a circle.

Figure 3. Circumscribed circle

Theorem 2 (on a circle circumscribed about a triangle)

Theorem 2

Near any triangle it is possible to circumscribe a circle, and moreover, only one.

Proof.

Consider triangle $ABC$. Let's draw the midperpendiculars in it, intersecting at the point $O$, and connect it with the vertices of the triangle (Fig. 4)

Figure 4. Illustration of Theorem 2

Existence: Let's construct a circle with center $O$ and radius $OC$. The point $O$ is equidistant from the vertices of the triangle, i.e. $OA=OB=OC$. Therefore, the constructed circle passes through all the vertices of the given triangle, which means that it is described around this triangle.

Uniqueness: Assume that around the triangle $ABC$ one more circle can be circumscribed with the center at the point $O"$. Its center is equidistant from the vertices of the triangle, and therefore coincides with the point $O$ and has a radius equal to the length of $OC. $ But then this circle will coincide with the first one.

The theorem has been proven.

Corollary 1: The center of the circle circumscribed about the triangle coincides with the point of intersection of its perpendicular bisectors.

Here are a few more facts related to the concept of the circumscribed circle:

It is not always possible to describe a circle around a quadrilateral.

In any inscribed quadrilateral, the sum of opposite angles is equal to $(180)^0$.

If the sum of the opposite angles of a quadrilateral is $(180)^0$, then a circle can be circumscribed around it.

An example of a problem on the concepts of an inscribed and circumscribed circle

Example 1

In an isosceles triangle, the base is 8 cm, the side is 5 cm. Find the radius of the inscribed circle.

Solution.

Consider triangle $ABC$. By Corollary 1, we know that the center of the inscribed circle lies at the intersection of the bisectors. Let us draw the bisectors $AK$ and $BM$, which intersect at the point $O$. Draw a perpendicular $OH$ from the point $O$ to the side $BC$. Let's draw a picture:

Figure 5

Since the triangle is isosceles, $BM$ is both the median and the altitude. By the Pythagorean theorem $(BM)^2=(BC)^2-(MC)^2,\ BM=\sqrt((BC)^2-\frac((AC)^2)(4))=\sqrt (25-16)=\sqrt(9)=3$. $OM=OH=r$ -- the desired radius of the inscribed circle. Since $MC$ and $CH$ are segments of intersecting tangents, by the intersecting tangent theorem, we have $CH=MC=4\ cm$. Therefore, $BH=5-4=1\ cm$. $BO=3-r$. From the triangle $OHB$, by the Pythagorean theorem, we get:

\[((3-r))^2=r^2+1\] \ \ \

Answer:$\frac(4)(3)$.

In this lesson, we will recall the foundations on which the theory of inscribed and circumscribed circles is based, recall the signs of inscribed and circumscribed quadrangles. In addition, we derive formulas for finding the radii of the circumscribed and inscribed circles in various cases.

Theme: Circle

Lesson: Inscribed and circumscribed circles

First of all, we are talking about inscribed and circumscribed circles relative to a triangle. We are prepared for this topic, as we have studied the properties of the bisectors and perpendicular bisectors of a triangle.

A circle can be inscribed in any triangle (see Fig. 1).

Rice. one

Proof:

We know that all the bisectors of a triangle intersect at one point - let's say at point O. Let's draw the bisectors AO, BO, CO. Their intersection point O is equidistant from the sides of the triangle. It is equidistant from the sides of the angle - AC and AB, since it belongs to the bisector of this angle. Similarly, it is equidistant from the sides of the corners and thus from the three sides of the triangle.

Let's drop the perpendiculars from the point O to the sides of the triangle - OM to the AC side, OL - to the BC, OK - to the AB. These perpendiculars will be the distances from the point O to the sides of the triangle, and they are equal:

Let's denote the distance from point O to the sides of the triangle as r and consider a circle with center at point O and radius r.

The circle touches the straight line AB, because has a common point K with it, and the radius OK drawn to this point is perpendicular to the line AB. Similarly, the circle touches the lines AC and BC. Thus, the circle touches all those sides of the triangle, which means that it is inscribed in the triangle.

So, the three bisectors of a triangle intersect at a point that is the center of the inscribed circle.

Consider another theorem, it concerns the point of intersection of the perpendicular bisectors of a triangle. We know that they intersect at one point, and this point coincides with the center of the circle circumscribed about the triangle.

A circle can be circumscribed about any triangle.

So, a triangle is given. Let's draw the midperpendicular p 1 to the side of the triangle BC, p 2 - to the side AB, p 3 - to the side AC (see Fig. 2).

According to the theorem on the properties of perpendicular bisectors, a point belonging to the perpendicular bisector of a segment is equidistant from the ends of the segment. From here, because the point Q belongs to the perpendicular bisector of the segment AC. Likewise and . Thus, the point Q is equidistant from the vertices of the triangle. Hence QA, QB, QC - radii

Rice. 2

a circle circumscribed about a triangle. Let's denote the radius as R. The point O of the intersection of the medial perpendiculars is the center of the circumscribed circle.

Consider a circle inscribed in a certain quadrilateral and the properties of this quadrilateral (see Fig. 3).

Recall the properties of a point lying on the bisector of an angle.

An angle is given, its bisector is AL, point M lies on the bisector.

If the point M lies on the bisector of the angle, then it is equidistant from the sides of the angle, that is, the distances from the point M to AC and to the BC of the sides of the angle are equal.

Rice. 3

The distance from a point to a line is the length of the perpendicular. Draw from the point M perpendiculars MK to side AB and MP to side AC.

Consider triangles and . These are right-angled triangles, and they are equal, because. have a common hypotenuse AM, and the angles and are equal, since AL is the bisector of angle . Thus, right-angled triangles are equal in hypotenuse and acute angle, hence it follows that , which was required to be proved. Thus, a point on the bisector of an angle is equidistant from the sides of that angle.

In addition, legs. Thus, the segments of tangents drawn to the circle from one point are equal.

So, back to the quadrilateral. The first step is to draw a bisector in it.

All bisectors of a quadrilateral intersect at one point - point O, the center of the inscribed circle.

From the point O we lower the perpendiculars to the sides of the quadrangle to the points K, L, M, N and determine the points of contact (see Fig. 3).

The tangents drawn to the circle from one point are equal to each other, thus, a pair of equal tangents comes out of each vertex: , , , .

Rice. 3

If a circle can be inscribed in a quadrilateral, then the sums of its opposite sides are equal. This is easy to prove:

Let's expand the brackets:

Thus, we have proved a simple but important theorem.

If a circle can be inscribed in a quadrilateral, then the sums of its opposite sides are equal.

The converse theorem is true.

If the sums of opposite sides in a quadrilateral are equal, then a circle can be inscribed in it.

Consider a circle circumscribed about a quadrilateral.

Given a circle with center O and an arbitrary quadrilateral ABCD. Consider the properties of this quadrilateral. All four perpendicular bisectors of a given quadrilateral intersect at one point: this point is the center of the circumscribed circle.

It would be tedious to prove that all four perpendicular bisectors intersect at one point. There is another sign. Consider the angle ےА, this is the inscribed angle of the circle, it rests on the arc and is measured by half the degree measure of this arc (see Fig. 4). Denote the angle ےА for , then the arc . Similarly, we denote the opposite angle ےС for , it is inscribed in a circle and rests on an arc. Hence the arc.

Rice. 4

Arcs and make up a full circle. From here:

We divide the resulting expression by two, we get:

So, we have proved the direct theorem.

Theorem

If a circle is circumscribed about a quadrilateral, the sum of its opposite angles is .

This is a necessary and sufficient sign, that is, the converse theorem is true.

If the sum of the opposite angles of a quadrilateral is , then a circle can be circumscribed around this quadrilateral.

Based on these theorems, we note that a circle cannot be described around a parallelogram, since its opposite angles are equal, and their sum is not equal (see Fig. 5).

Rice. 5

A circle could be described near a parallelogram if its opposite angles were equal to 90 °, that is, if it were a rectangle, so a circle can be described near a rectangle (see Fig. 6).

Rice. 6

It is also impossible to circumscribe a circle around a rhombus, but it can be inscribed, since all sides of the rhombus are equal, and thus, the sums of opposite sides of the rhombus are equal.

In addition, in a rhombus, each diagonal is a bisector, the intersection point of the bisectors is equidistant from all sides of the rhombus (see Fig. 7).

Rice. 7

So, we have proved that a circle can be inscribed in any triangle, and the center of this circle coincides with the intersection point of the bisectors of the triangle. We also proved that a circle can be circumscribed about any triangle, and its center will coincide with the point of intersection of the perpendicular bisectors. In addition, we have seen that it is possible to inscribe a circle in some quadrilaterals, and for this it is necessary that the sums of the opposite sides of the quadrilateral be equal. We have also shown that a circle can be circumscribed around some quadrilaterals, and a necessary and sufficient condition for this is the equality of the sum of opposite angles .

Bibliography

Aleksandrov A.D. etc. Geometry, grade 8. - M.: Education, 2006.
Butuzov V.F., Kadomtsev S.B., Prasolov V.V. Geometry, 8th grade. - M.: Education, 2011.
Merzlyak A.G., Polonsky V.B., Yakir S.M. Geometry, 8th grade. - M.: VENTANA-GRAF, 2009.

Uztest.ru ().
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Homework

Definition 2

A polygon that satisfies the condition of Definition 1 is said to be inscribed about a circle.

Figure 1. Inscribed circle

Theorem 1 (on a circle inscribed in a triangle)

Theorem 1

In any triangle, you can inscribe a circle, and moreover, only one.

Proof.

Consider triangle $ABC$. Draw bisectors in it that intersect at the point $O$ and draw perpendiculars from it to the sides of the triangle (Fig. 2)

Figure 2. Illustration of Theorem 1

The theorem has been proven.

Corollary 1: The center of a circle inscribed in a triangle lies at the point of intersection of its bisectors.

Here are some more facts related to the concept of an inscribed circle:

Not every quadrilateral can be inscribed in a circle.

In any circumscribed quadrilateral, the sums of opposite sides are equal.

If the sums of opposite sides of a convex quadrilateral are equal, then a circle can be inscribed in it.

Definition 3

If all the vertices of the polygon lie on the circle, then the circle is called circumscribed near the polygon (Fig. 3).

Definition 4

A polygon that satisfies the condition of Definition 2 is called inscribed in a circle.

Figure 3. Circumscribed circle

Theorem 2 (on a circle circumscribed about a triangle)

Theorem 2

Near any triangle it is possible to circumscribe a circle, and moreover, only one.

Proof.

Consider triangle $ABC$. Let's draw the midperpendiculars in it, intersecting at the point $O$, and connect it with the vertices of the triangle (Fig. 4)

Figure 4. Illustration of Theorem 2

The theorem has been proven.

Corollary 1: The center of the circle circumscribed about the triangle coincides with the point of intersection of its perpendicular bisectors.

Here are a few more facts related to the concept of the circumscribed circle:

It is not always possible to describe a circle around a quadrilateral.

In any inscribed quadrilateral, the sum of opposite angles is equal to $(180)^0$.

If the sum of the opposite angles of a quadrilateral is $(180)^0$, then a circle can be circumscribed around it.

An example of a problem on the concepts of an inscribed and circumscribed circle

Example 1

In an isosceles triangle, the base is 8 cm, the side is 5 cm. Find the radius of the inscribed circle.

Solution.

Figure 5

\[((3-r))^2=r^2+1\] \ \ \

Answer:$\frac(4)(3)$.

And it applies to all aspects of it.

Encyclopedic YouTube

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Inscribed circle properties:
r = (− a + b + c) (a − b + c) (a + b − c) 4 (a + b + c) ; (\displaystyle r=(\sqrt (\frac ((-a+b+c)(a-b+c)(a+b-c))(4(a+b+c))));) 1 r = 1 ha + 1 hb + 1 hc (\displaystyle (\frac (1)(r))=(\frac (1)(h_(a)))+(\frac (1)(h_(b) ))+(\frac (1)(h_(c))))
where a , b , c (\displaystyle a,b,c)- sides of a triangle h a , h b , h c (\displaystyle h_(a),h_(b),h_(c))- heights drawn to the respective sides;
r = S p = (p − a) (p − b) (p − c) p (\displaystyle r=(\frac (S)(p))=(\sqrt (\frac ((pa)(pb) (pc))(p))))
Where S (\displaystyle S) is the area of the triangle, and p (\displaystyle p) is its semiperimeter.
- If A B (\displaystyle AB)- the base of an isosceles triangle, then the circle tangent to the sides of the angle ∠ A C B (\displaystyle \angle ACB) at points A (\displaystyle A) and B (\displaystyle B), passes through the center of the inscribed circle of the triangle △ A B C (\displaystyle \triangle ABC).
- Euler's theorem: R 2 − 2 R r = | O I | 2 (\displaystyle R^(2)-2Rr=|OI|^(2)), where R (\displaystyle R) is the radius of the circumscribed circle around the triangle, r (\displaystyle r) is the radius of the circle inscribed in it, O (\displaystyle O)- the center of the circumscribed circle, I (\displaystyle I)- the center of the inscribed circle.
- If a line passing through point I parallel to side AB intersects sides BC and CA at points A 1 and B 1 , then A 1 B 1 = A 1 B + A B 1 (\displaystyle A_(1)B_(1)=A_(1)B+AB_(1)).
- If the tangent points of an inscribed triangle T (\displaystyle T) connect the circles with segments, then you get a triangle T 1 with the properties:
  - The bisectors of T are mid perpendiculars of T 1
  - Let T 2 be an orthotriangle T 1 . Then its sides are parallel to the sides of the original triangle T.
  - Let T 3 be the middle triangle of T 1 . Then the bisectors of T are the heights of T 3 .
  - Let T 4 be an orthotriangle of T 3 , then the bisectors of T are the bisectors of T 4 .
- The radius of a circle inscribed in a right triangle with legs a, b and hypotenuse c is a + b − c 2 (\displaystyle (\frac (a+b-c)(2))).
- The distance from the vertex C of the triangle to the point where the inscribed circle touches the side is d = a + b − c 2 = p − c (\displaystyle d=(\frac (a+b-c)(2))=p-c).
- The distance from the vertex C to the center of the inscribed circle is l c = r sin ⁡ (γ 2) (\displaystyle l_(c)=(\frac (r)(\sin((\frac (\gamma )(2)))))), where r is the radius of the inscribed circle and γ is the angle of the vertex C.
- The distance from the vertex C to the center of the inscribed circle can also be found using the formulas l c = (p − c) 2 + r 2 (\displaystyle l_(c)=(\sqrt ((p-c)^(2)+r^(2)))) and l c = a b − 4 R r (\displaystyle l_(c)=(\sqrt (ab-4Rr)))
- The theorem about the trident or shamrock theorem: If D- point of intersection of the angle bisector A with the circumscribed circle of a triangle ABC, I and J- respectively, the centers of the inscribed and excircle tangent to the side BC, then | D I | = | D B | = | D C | = | D J | (\displaystyle |DI|=|DB|=|DC|=|DJ|).
- Verriere's lemma: let the circle V (\displaystyle V) concerns the parties A B (\displaystyle AB), A C (\displaystyle AC) and arcs B C (\displaystyle BC) the circumscribed circle of the triangle. Then the tangent points of the circle V (\displaystyle V) with sides and center inscribed circle triangle A B C (\displaystyle ABC) lie on the same line.
- Feuerbach's theorem. The circle nine points touches all three excircles, as well as inscribed circle. touch point circle Euler and inscribed circle known as the Feuerbach point.
Relation of the inscribed circle to the circumscribed circle

R R = 4 S 2 p a b c = cos ⁡ α + cos ⁡ β + cos ⁡ γ − 1 ; (\displaystyle (\frac (r)(R))=(\frac (4S^(2))(pabc))=\cos \alpha +\cos \beta +\cos \gamma -1;)

Inscribed circle. A circle circumscribed about a triangle. A triangle inscribed in a circle

Theorem 1 (on a circle inscribed in a triangle)

Theorem 2 (on a circle circumscribed about a triangle)

An example of a problem on the concepts of an inscribed and circumscribed circle

Theorem 1 (on a circle inscribed in a triangle)

Theorem 2 (on a circle circumscribed about a triangle)

An example of a problem on the concepts of an inscribed and circumscribed circle

Encyclopedic YouTube

Relation of the inscribed circle to the circumscribed circle