Bisectors of a triangle. Bisectors Bisectors aa1 and bb1 of a triangle

It is known from the school that three bisectors of the interior angles of a triangle intersect at one point - the center of a circle inscribed in this triangle.

Theorem 1. Angle bisector BUT triangle ABC the intersection point of the bisectors is divisible by , counting from the side where a, b, c- side lengths BC, AC, AB respectively.

Proof. Let be AA 1 and BB 1 - angle bisectors BUT and AT respectively in a triangle ABC, L is their point of intersection, a, b, c- side lengths BC, AC, AB respectively (Fig.62). Then by the bisector theorem applied to the triangle ABC will have

Or b VA 1 = ac - with VA 1 , or VA 1 (b + c)= ac, means, VA 1 = with. By the same theorem, applied to the triangle AVA 1 we get BUT 1 L : LA = : with, or = .

Theorem 2. If a L ABC circle, then

Ð ALV= 90° + R C.

Proof. Given that the sum of the angles of a triangle is 180° and that the center L the inscribed circle is the intersection point of the bisectors of the triangle, we will have (Fig. 62):

Ð ALV= 180° - ( Ð ABL + Ð BAL) = 180° – ( Ð ABC + Ð YOU) =

180° - (180° – Р С) = 180° – 90° + R C= 90° + R C.

Theorem 3. If a L- point on the bisector of the angle With triangle ABC such that Ð ALV= 90° + R C, then L- the center of an inscribed triangle ABC circle.

Proof. Let us prove that none of the points L 1 between C and L cannot be the center of an inscribed circle (Fig. 62a).

We have R AL 1 With 1 < Ð ALC 1 , since the outer corner of the triangle AL 1 L greater than any interior angle not adjacent to it. Same way Р ВL 1 With < Ð BLC 1 .

So R AL 1 AT < Ð ALV= 90° + R C. Means, L 1 is not the center of the inscribed circle, since the condition for the sign of the center of the inscribed circle is not satisfied (see Theorem 2).

If the point L 2 on the bisector SS 1 does not belong to the segment CL, then R AL 2 AT > Ð ALV= 90° + R C and again the condition of the sign of the center of the inscribed circle is not met. So the center of the inscribed circle is the point L.

Theorem 4. The distance from the vertex of the triangle to the point of contact of the inscribed circle with the side passing through this vertex is equal to the half-perimeter of this triangle, reduced by the opposite side.

Proof. Let be BUT 1 , AT 1 , With 1 - points of contact of the inscribed circle with the sides of the triangle ABC(fig. 63), a, b, c- side lengths BC, AC, AB respectively.

Let be AC 1 = X, Then AB 1 = x, sun 1 = s - x = VA 1 , AT 1 With = b - x \u003d CA 1 ,

a = BC = BA 1 + SA 1 = (c - x) + (b - x) \u003d c + b – 2 X.

Then a + a = a + b + c – 2 X, or 2 a = 2 R – 2 X, or x = p - a.

Theorem 5. In any triangle ABC through a point L the intersection of the bisectors of its two outer angles passes the bisector of the third angle, while the point L is at equal distances from the lines containing the sides of the triangle.

Proof. Let be L- the point of intersection of two external corners AT and With trianglea ABC(Fig. 64). Since each point of the bisector is at the same distance from the sides of the angle, then the point L AB and Sun, since it belongs to the bisector BL. It is also at the same distance from the straight lines. Sun and AC, since it belongs to the bisector CL. Therefore, the point L is at the same distance from the straight lines AND YOU and Sun. Since the point L is at the same distance from the lines AB and AC, then JSC- angle bisector YOU.

A circle that touches a side of a triangle and the extensions of the other two sides is called a circle exscribed in this triangle.

Consequence 1. The centers of the circles exscribed in the triangle are located at the intersection points of pairs of bisectors of its external angles.

Theorem 6. The radius of a circle inscribed in a triangle is equal to the ratio of the side of this triangle and the cosine of half of the opposite angle, multiplied by the sines of the halves of the other two angles.

"Types of triangles" - Types of triangles. According to the comparative length of the sides, the following types of triangles are distinguished. The following types are distinguished by the size of the angles. Points are called vertices, and segments are called sides.

"Angles of a triangle" - An acute triangle. Can a triangle have two right angles? Equilateral triangle. Isosceles triangle. Right triangle. Obtuse triangle. Can a triangle have two obtuse angles? In an equilateral triangle, the angles are 600. In a right-angled isosceles triangle, the acute angles are 450 each.

"Geometry lessons in grade 7" - Problem solving. Legs BC and SA. Work according to ready-made drawings. The sum of the angles of a triangle. New material. Right triangle. Task number 1. Solution of problems according to ready-made drawings. No. 232 (oral), No. 231. Prove that angle ABC is less than angle ADC. oral test. Geometry lesson in 7th grade. Hypotenuse AB.

"Right Triangle" - Information about Euclid is extremely scarce. A triangle is a polygon with three sides (or three corners). Euclid is the author of works on astronomy, optics, music, etc. The external angle of a triangle is equal to the sum of the internal angles that are not adjacent to it. Euclid is the first mathematician of the Alexandrian school. Definitions. Control test.

"Isosceles triangle and its properties" - Name the base and sides of these triangles. Find the value of angle 1 if the value of angle 2 is 40 degrees? A, C - angles at the base of an isosceles triangle. Are the triangles equal? Where in life are isosceles triangles found? AM is the median. A TRIANGLE all sides of which are equal is called an equilateral triangle.

"Geometry Right Triangle" - Egyptian Numbers: Calculate the area of ​​the plot of the triangular shape of an Egyptian peasant. Surveyors. What did the Egyptians call a right triangle? Egyptian builders: Leg and hypotenuse in Egypt Pythagoreans: Leg and hypotenuse in geometry. Surveyors' questions: - The leg is larger than the hypotenuse. The leg opposite the 60 degree angle is equal to half of the hypotenuse.

  • repeat and generalize the studied theorems;
  • consider their application in solving a number of problems;
  • preparing students for university entrance exams;
  • educate the aesthetic execution of drawings for tasks.

Equipment: multimedia projector. Appendix 1 .

During the classes:

1. Organizational moment.

2. Checking homework:

  • proof of theorems - 2 students + 2 students - consultants (verifiers);
  • solving home problems - 3 students;
  • work with the class - oral problem solving:

Point C 1 divides side AB of triangle ABC in relation to 2: 1. point B 1 lies on the continuation of side AC beyond point C, and AC \u003d CB 1. In what ratio does line B 1 C 1 divide side BC? (on slide 2).

Solution: By conditionUsing the theorem of Menelaus, we find: .

In the triangle ABC AD is the median, point O is the midpoint of the median. Line BO intersects side AC at point K.

In what ratio does point K divide AC, counting from point A? (on slide 3).

Decision: Let ВD = DC = a, AO = ОD = m. Line BK intersects two sides and the extension of the third side of triangle ADC. According to Menelaus' theorem .

In triangle ABC on side BC point N is taken so that NC = 3BN; on the extension of side AC, point M is taken as point A so that MA = AC. Line MN intersects side AB at point F. Find the ratio. (on slide 4).

Solution: According to the condition of the problem, MA = AC, NC = 3 BN. Let MA = AC = b, BN = k, NC = 3k. Line MN intersects two sides of the triangle ABC and the extension of the third. According to Menelaus' theorem

Point N is taken on side PQ of triangle PQR, and point L is taken on side PR, and NQ = LR. The intersection point of the segments QL and NR divides the QR in the ratio m: n, counting from the point Q. Find PN: PR. (on slide 5).

Solution: By condition NQ = LR, . Let NA = LR = a, QF =km, LF = kn. Line NR intersects two sides of triangle PQL and the extension of the third. According to Menelaus' theorem

3. Development of practical skills.

1. Problem solving:

Prove the theorem: The medians of a triangle intersect at one point; the point of intersection divides each of them in a ratio of 2:1, counting from the top. (Figure 1 slide 6).

Proof: Let AM 1 , VM 2 , CM 3 be the medians of triangle ABC. To prove that these segments intersect at one point, it suffices to show that Then, by the (inverse) Ceva theorem, the segments AM 1 , VM 2 and CM 3 intersect at one point. We have:

So, it is proved that the medians of a triangle intersect at one point.

Let O be the point of intersection of the medians. The line M 3 C intersects two sides of the triangle AVM 2 and the continuation of the third side of this triangle. According to Menelaus' theorem

or .

Considering the Menelaus theorem for triangles AM 1 C and AM 2 C, we get that

. The theorem has been proven.

Prove the theorem: The bisectors of a triangle intersect at one point.(Figure 2 slide 6).

Proof: It suffices to show that . Then, by the (inverse) Ceva theorem, AL 1, BL 2, CL 3 intersect at one point. According to the property of the bisectors of a triangle:

. Multiplying the equalities obtained term by term, we obtain: . So, for the bisectors of a triangle, Ceva's equality is satisfied, therefore, they intersect at one point. The theorem has been proven.

Task 7

Prove the theorem: The heights of an acute triangle intersect at one point.(Figure 3 slide 6).

Proof: Let AH 1 , AH 2, AH 3 be the heights of triangle ABC with sides a, b, c. From right-angled triangles ABH 2 and BCH 2, according to the Pythagorean theorem, we express, respectively, the square of the common leg BH 2, denoting AH 2 = x, CH 2 = b - x.

(BH 2) 2 \u003d c 2 - x 2 and (BH 2) 2 \u003d a 2 - (b - x) 2. equating the right parts of the obtained equalities, we obtain with 2 - x 2 \u003d a 2 - (b - x) 2, whence x \u003d.

Then b –x = b - = .

So, AH 2 = , CH 2 = .

Arguing similarly for right triangles ACH 2 and BCH 3 , VAN 1 and SAN 1 , we get AN 3 = , VN 3 = and VN 1 = ,

To prove the theorem, it suffices to show that . Then, by the Ceva (inverse) theorem, the segments AN 1 , VN 2 and CH 3 intersect at one point. Substituting in the left side of the equation the expressions for the lengths of the segments AN 3, VN 3, VN 1, CH 1, CH 2 and AN 2 through a, b, c, we make sure that Ceva's equality for the heights of the triangle is fulfilled. The theorem has been proven.

Tasks 5 - 7 independent decision of 3 students. (screen drawings).

2. others:

Prove the theorem: If a circle is inscribed in a triangle, then the segments connecting the vertices of the triangle with the points of contact of opposite sides intersect at one point. (in figure 4 slide 6).

Proof: Let A 1 , B 1 and C 1 be the tangent points of the inscribed circle of triangle ABC. In order to prove that the segments AA 1, BB 1 and CC 1 intersect at one point, it is enough to show that the Ceva equality is satisfied:

. Using the property of tangents drawn from one point, we introduce the notation: BC 1 = BA 1 = x, CA 1 = CB 1 = y, AB 1 = AC 1 = z.

. Ceva's equality is satisfied, which means that the indicated segments (triangle bisectors) intersect at one point. This point is called the Gergon point. The theorem has been proven.

3. Analysis of tasks 5, 6, 7.

Task 9

Let AD be the median of triangle ABC. A point K is taken on side AD so that AK: KD = 3: 1. Straight line BK splits triangle ABC into two. Find the ratio of the areas of these triangles. (on slide 7 figure 1)

Solution: Let AD = DC = a, KD = m, then AK = 3m. Let P be the point of intersection of the line VC with the side AC. You need to find a relationship. Since triangles ABP and PBC have equal heights drawn from vertex B, then = . According to the Menelaus theorem, for the triangle ADC and the secant PB we have: . So = .

Task 10

In a triangle ABC circumscribed about a circle, AB \u003d 8, BC \u003d 5, AC \u003d 4. A 1 and C 1 are the touch points belonging to the sides BC and BA, respectively. P - the point of intersection of the segments AA 1 and SS 1. The point P lies on the bisector BB 1 . Find AR: RA 1 .

(on slide 7 figure 2)

Solution: The point of contact of the circle with the side AC does not coincide with B 1, since the triangle ABC is scalene. Let C 1 B \u003d x, then, using the property of tangents drawn to the circle from one point, we introduce the notation (see figure) 8 - x + 5 - x \u003d 4, x \u003d.

Hence, C 1 B \u003d VA 1 \u003d, A 1 C \u003d 5 - \u003d, AC 1 \u003d 8 - \u003d.

In the triangle ABA 1, the line C 1 C intersects two of its sides and the extension of the third side. According to Menelaus' theorem .

Answer: 70:9.

The sides of the triangle are 5, 6 and 7. Find the ratio of the segments into which the bisector of the larger angle of this triangle is divided by the center of the circle inscribed in the triangle. (on slide 7).

Solution: Let AB = 5, BC = 7, AC = 6 in the triangle ABC. The angle BAC lies opposite the larger side in the triangle ABC, which means that the angle BAC is the larger angle of the triangle. The center of the triangle's inscribed circle lies at the intersection of the bisectors. Let O be the intersection point of the bisectors. It is necessary to find AO: OD. Since AD ​​is the bisector of the triangle ABC, then , that is, BD = 5k, DC = 6k. since BF is the bisector of the triangle ABC, then , that is, AF = 5m, FC = 7m. Line BF intersects two sides and the extension of the third side of triangle ADC. According to Menelaus' theorem .

4. Independent solution of problems 9, 10, 11.– 3 students.

Task 12 (for all remaining students in the class):

The bisectors BE and AD of the triangle ABC intersect at point Q. Find the area of ​​the triangle ABC if the area of ​​the triangle is BQD = 1, 2AC = 3 AB, 3BC = 4 AB. (Figure 4 on slide 7).

Solution: Let AB = a, then AC = , BC = . AD is the bisector of triangle ABC, then , i.e. BD = 2p, DC = 3p. BE is the bisector of the triangle ABC, then , AE = 3k, EC = 4k. In triangle BEC, line AD intersects two of its sides and the extension of the third side. According to Menelaus' theorem , , , i.e. EQ = 9m, QB = 14m. Triangles QBD and EBC have a common angle, so , S EBC = .

Triangles ABC and BEC have equal heights drawn from vertex B, so , then S ABC = .

5. Analysis of problems 9, 10, 11.

Problem solving - workshop:

A. On the sides BC, CA, AB of an isosceles triangle ABC with base AB, points A 1, B 1, C 1 are taken, so that the lines AA 1, BB 1, CC 1 are competitive.

Prove that

Proof:

By Ceva's theorem, we have: (1).

By the law of sines: , whence CA 1 = CA.,

, whence A 1 B = AB. , ,

whence AB 1 = AB. , , whence B 1 C = BC. , since CA = BC by condition. Substituting the obtained equalities into equality (1) we get:

Q.E.D.

B. On the AC side of the triangle ABC, a point M is taken such that AM = ?AC, and on the extension of the side BC, a point N is taken such that BN = CB. In what relation does the point P - the point of intersection of the segments AB and MN divide each of these segments?

According to the Menelaus theorem, for the triangle ABC and the secant MN we have:

. By condition hence ,

since 0.5 . (-2) . x \u003d 1, - 2x \u003d - 2, x \u003d 1.

For the triangle MNC and the secant AB, according to the Menelaus theorem, we have: by condition

means - , from where, .

8. Independent problem solving: option 1:

1. On the extensions of the sides AB, BC, AC of the triangle ABC, points C 1, A 1, B 1 are taken, respectively, so that AB \u003d BC 1, BC \u003d CA 1, CA \u003d AB 1. Find the ratio in which the line AB 1 divides the side A 1 C 1 of the triangle A 1 B 1 C 1. (3 points).

2. Point M is taken on the median CC 1 of triangle ABC. Straight lines AM and VM intersect the sides of the triangle, respectively, at points A 1 and B 1. Prove that lines AB and A 1 B 1 are parallel. (3 points).

3. Let the points C 1 , A 1 and B 1 be taken respectively on the continuation of the sides AB, BC and AC of the triangle ABC. Prove that the points A 1 , B 1, C 1 lie on one straight line if and only if the equality . (4 points).

6. Let the points C 1 , A 1 and B 1 be taken on the sides AB, BC and AC of the triangle ABC, respectively, so that the lines AA 1 , BB 1 , CC 1 intersect at the point O. Prove that the equality . (5 points).

7 . Let the points A 1 , B 1 , C 1 , D 1 be taken on the edges AB, BC, CD and AD of the tetrahedron ABCD, respectively. Prove that the points A 1 , B 1 , C 1 , D 1 lie in the same plane if and only if when the equality (5 points).

Option 2:

1. Points A 1 and B 1 divide the sides BC and AC of triangle ABC in ratios 2: 1 and 1: 2. Lines AA 1 and BB 1 intersect at point O. The area of ​​triangle ABC is 1. Find the area of ​​triangle OBC. (3 points).

2. The segment MN connecting the midpoints of the sides AD and BC of the quadrilateral ABCD is divided diagonally into three equal parts. Prove that ABCD is a trapezoid, one of the bases of AB or CD that is twice the other. (3 points).

3. Let the points C 1 , A 1 and B 1 be taken respectively on the side AB and the continuation of the sides BC and AC of the triangle ABC. Prove that lines AA 1 , BB 1 , СС 1 intersect at one point or are parallel if and only if the equality . (4 points).

4. Using Ceva's theorem, prove that the heights of a triangle or their extensions intersect at one point. (4 points).

5. Prove that the lines passing through the vertices of the triangle and the tangent points of the excircles intersect at one point (the Nagel point). (A circle is said to be exscribed in a triangle if it touches one side of this triangle and the extensions of the other two of its sides.) (5 points).

6. Let the points C 1 , A 1 , B 1 be taken on the sides AB, BC and AC of the triangle ABC, respectively, so that the lines AA 1 , BB 1 and CC 1 intersect at the point O. Prove that the equality . (5 points).

7. Let the points A 1 , B 1 , C 1 , D 1 be taken on the edges AB, BC, CD and AD of the tetrahedron ABCD, respectively. Prove that the points A 1 , B 1 , C 1 , D 1 lie in the same plane then and only when equality is met (5 points).

9. Homework: textbook § 3, no. 855, no. 861, no. 859.

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