Angle bisectors aa1 and bb1. Bisectors of a triangle

“Types of triangles” - Types of triangles. Based on the comparative length of the sides, the following types of triangles are distinguished. Based on the size of the angles, the following types are distinguished. The points are called vertices, and the segments are called sides.

“Angles of a triangle” - 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 equal to 600. In a right isosceles triangle, the acute angles are equal to 450.

“Geometry lessons in 7th grade” - Problem solving.” Legs BC and SA. Work according to ready-made drawings. “Sum of the angles of a triangle. New material. Right triangle. Task No. 1. Solving problems using ready-made drawings. No. 232 (oral), No. 231. Prove: 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 angles). 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 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 – median. A TRIANGLE, all sides of which are equal, is called EQUILATERAL.

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

From school we know that the three bisectors of the interior angles of a triangle intersect at one point - the center of the circle inscribed in this triangle.

Theorem 1. Angle bisector A triangle ABC the intersection point of the bisectors is divided in the ratio , counting from the side where a, b, c– side lengths BC, AC, AB respectively.

Proof. Let AA 1 and BB 1 – angle bisectors A And IN respectively in a triangle ABC, L– their point of intersection, a, b, c– side lengths BC, AC, AB accordingly (Fig. 62). Then, by the bisector theorem applied to a 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 a triangle AVA 1 we get A 1 L : LA = : With, or = .

Theorem 2. If L ABC circle, then

Ð ALB= 90° + R S.

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

Ð ALB= 180° – ( Ð ABL + Ð VAL) = 180° – ( Ð ABC + Ð VAS) =

180° – (180° – Ð C) = 180° – 90° + R S= 90° + R S.

Theorem 3. If L– point on the bisector of the angle WITH triangle ABC such that Ð ALB= 90° + R S, That L– center of 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 Ð AL 1 WITH 1 < Ð ALC 1, since the external angle of the triangle AL 1 L greater than any interior angle not adjacent to it. Also Ð VL 1 WITH < Р ВЛС 1 .

That's why Ð AL 1 IN < Ð ALB= 90° + R S. 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, That Ð AL 2 IN > Ð ALB= 90° + R S and again the condition for the sign of the center of the inscribed circle is not met. This means that the center of the inscribed circle is the point L.

Theorem 4. The distance from the vertex of the triangle to the point of tangency 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 A 1 , IN 1 , WITH 1 – points of tangency of the inscribed circle with the sides of the triangle ABC(Fig. 63), a, b, c– side lengths BC, AC, AB respectively.

Let AC 1 = X, Then AB 1 = x, sun 1 = c – x = VA 1 , IN 1 WITH = b – x = CA 1 ,

a = BC = VA 1 + SA 1 = (c – x) + (b – x) = 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 the point L the intersection of the bisectors of its two external angles passes the bisector of the third angle, and the point L is at equal distances from the lines containing the sides of the triangle.

Proof. Let L– the point of intersection of two external angles IN 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 ВL. It is located at the same distance from the straight lines Sun And AC, since it belongs to the bisector CL. Therefore point L is at the same distance between straight lines AND YOU And Sun. Since the point L is at the same distance from the lines AB And AC, That JSC– angle bisector YOU.

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

Corollary 1. The centers of circles excentric to a triangle are located at the points of intersection 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 the opposite angle, multiplied by the sines of the halves of the other two angles.

• repeat and generalize the theorems studied;
• consider their use in solving a number of problems;
• preparing students for entrance exams to universities;
• cultivate aesthetic execution of drawings for tasks.

Equipment: multimedia projector. Annex 1.

During the classes:

1. Organizational moment.

2. Checking homework:

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

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

Solution: By condition Using Menelaus' theorem, we find: .

In triangle ABC, AD is the median, point O is the middle of the median. Straight line BO intersects side AC at point K.

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

Solution: Let ВD = DC = a, АО = ОD = m. Straight line BK intersects two sides and the continuation of the third side of triangle ADC. According to Menelaus' theorem .

In triangle ABC, on side BC, point N is taken so that NC = 3ВN; on the continuation 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 conditions of the problem, MA = AC, NC = 3 ВN. Let MA = AC = b, BN = k, NC = 3k. Straight line MN intersects two sides of triangle ABC and the continuation 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 QR in the ratio m: n, counting from 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 continuation of the third. According to Menelaus' theorem

3. Practicing 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 vertex. (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 is enough to show that Then, according to Cheva’s (converse) theorem, the segments AM 1, VM 2 and CM 3 intersect at one point. We have:

So, it has been proven that the medians of a triangle intersect at one point.

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

or .

Considering Menelaus' theorem for triangles AM 1 C and AM 2 C, we obtain that

. The theorem has been proven.

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

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

. Multiplying the resulting equalities term by term, we get: . So, for the bisectors of a triangle, Cheva’s equality is satisfied, therefore, they intersect at one point. The theorem has been proven.

Problem 7

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

Proof: Let AH 1, AH 2, AH 3 be the altitudes of triangle ABC with sides a, b, c. Using the Pythagorean theorem, from right-angled triangles ABN 2 and BSN 2, we express, respectively, the square of the common leg BN 2, denoting AH 2 = x, CH 2 = b – x.

(VN 2) 2 = c 2 – x 2 and (VN 2) 2 = a 2 – (b – x) 2. equating the right-hand sides of the resulting equalities, we obtain c 2 – x 2 = a 2 – (b – x) 2, whence x =.

Then b –x = b - = .

So, AN 2 = , CH 2 = .

Reasoning similarly for right triangles ASN 2 and VSN 3, VAN 1 and SAN 1, we obtain AN 3 =, VN 3 = and VN 1 =,

To prove the theorem it is enough to show that . Then, according to Cheva’s (converse) theorem, the segments AN 1, VN 2 and CH 3 intersect at one point. Substituting into the left side of the equality 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 are convinced that Cheva’s equality for the altitudes of the triangle is satisfied. The theorem has been proven.

Problems 5 – 7 independent solution by 3 students. (drawings on the screen).

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. (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 Cheva’s equality holds:

. 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.

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

3. Analysis of problems 5, 6, 7.

Problem 9

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

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

Problem 10

In triangle ABC, circumscribed about a circle, AB = 8, BC = 5, AC = 4. A 1 and C 1 are points of tangency, belonging to sides BC and BA, respectively. P – point of intersection of segments AA 1 and CC 1. 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 side AC does not coincide with B1, since triangle ABC is scalene. Let C 1 B = x, then, using the property of tangents drawn to a circle from one point, we introduce the notation (see figure) 8 – x + 5 – x = 4, x = .

This means C 1 B = VA 1 = , A 1 C = 5 - = , AC 1 = 8 - = .

In triangle ABA 1, line C 1 C intersects its two sides and the continuation 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 triangle ABC. Angle BAC lies opposite the larger side in triangle ABC, which means angle BAC is the larger angle of the triangle. The center of the incircle of a triangle lies at the intersection of the bisectors. Let O be the point of intersection of the bisectors. You need to find AO: OD. Since AD ​​is the bisector of triangle ABC, that is, BD = 5k, DC = 6k. since BF is the bisector of triangle ABC, 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.

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

Bisectors BE and AD of triangle ABC intersect at point Q. Find the area of ​​triangle ABC if the area of ​​triangle 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 , that is, BD = 2p, DC = 3p. BE is the bisector of triangle ABC, then , AE = 3 k, EC = 4k. In triangle BEC, line AD intersects two of its sides and the extension of the third side. According to Menelaus' theorem . . . that is, EQ = 9m, QB = 14m. Triangles QBD and EBC have a common angle, which means , S EBC = .

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

5. Analysis of problems 9, 10, 11.

Problem solving – workshop:

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

Prove that

Proof:

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

According to the law of sines: , from where CA 1 = CA.,

, from where A 1 B = AB. , ,

whence AB 1 = AB. , , from where B 1 C = BC. , since CA = BC by condition. Substituting the resulting equalities into equality (1) we obtain:

Q.E.D.

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

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

. By condition hence ,

since 0.5. (-2) . x = 1, - 2x = - 2, x = 1.

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

means, - , from where, .

8. Independent problem solving: Option 1:

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

2. On the median CC 1 of triangle ABC, point M is taken. Straight lines AM and BM 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 points C 1, A 1 and B 1 be taken respectively on the continuation of sides AB, BC and AC of triangle ABC. Prove that points A 1, B 1, C 1 lie on the same straight line if and only if the equality holds . (4 points).

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

7 . Let 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 points A 1, B 1, C 1, D 1 lie in the same plane if and only if , when the equality is satisfied (5 points).

Option 2:

1. Points A 1 and B 1 divide 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 equal to 1. Find the area of ​​triangle OBC. (3 points).

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

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

4. Using Ceva's theorem, prove that the altitudes 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 excircles intersect at one point (Nagel point). (A circle is called excircle in a triangle if it touches one side of this triangle and the extensions of its two other sides). (5 points).

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

7. Let 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 points A 1, B 1, C 1, D 1 lie in the same plane then and only when equality is satisfied (5 points).

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