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The heights of the intersection point are divided in relation. Summary of the lesson "theorem on the intersection of the heights of the triangle"

The lesson contains a description of the properties and formulas for finding the height of a triangle, as well as examples of solving problems. If you have not found a solution to a suitable problem - write about it on the forum... Surely, the course will be supplemented.

HEIGHT OF THE TRIANGLE

Height of triangle- a perpendicular dropped from the apex of the triangle, drawn to the side opposite to the apex or to its continuation.

Properties the height of the triangle:

  • If two heights in a triangle are equal, then such a triangle is isosceles
  • In any triangle, a segment connecting the bases of two heights of the triangle cuts off a triangle similar to this
  • In a triangle, the segment connecting the bases of the two heights of the triangle lying on two sides is not parallel to the third side, with which it has no common points. Through its two ends, as well as through two vertices of this side, you can always draw a circle
  • In an acute-angled triangle, two of its heights cut off similar triangles from it
  • The minimum height in a triangle is always inside that triangle.

Orthocenter of a triangle

All three heights of the triangle (drawn from three vertices) intersect at one point, which called orthocenter... In order to find the point of intersection of heights, it is enough to draw two heights (two straight lines intersect only at one point).

The location of the orthocenter (point O) is determined by the shape of the triangle.

For an acute-angled triangle, the intersection point of the heights is in the plane of the triangle. (Fig. 1).

In a right-angled triangle, the point of intersection of the heights coincides with the vertex of the right angle (Fig. 2).

In an obtuse triangle, the point of intersection of heights is behind the plane of the triangle (Fig. 3).

In an isosceles triangle, the median, bisector, and height drawn to the base of the triangle coincide.

In an equilateral triangle, all three "remarkable" lines (height, bisector and median) coincide and three "remarkable" points (points of the orthocenter, center of gravity and the center of the inscribed and circumscribed circles) are located at the same point of intersection of the "remarkable" lines, i.e. are also the same.

VISOTA TRIKUTNIK

The visota of the tricycle - descending from the top of the tricytnyk perpendicular, conducting to the other peaks bik or on її continued.

All three heights of a tricycle (drawn from three peaks) are interlaced in one point, which is called an orthocenter. In order to know the point of the overflow of the heights, to finish to carry out two hoists (two straight lines are overturned only in one point).

The distribution of the orthocenter (point O) is designed as a type of tricycle.

At the gosudarstny trikutnik, the point of the cross-section of the heights is located in the area of ​​the trikutnik. (Mal. 1).

In the rectangular tricycle, the point of overretinum of the hems is zbigaye with the apex of the straight kuta (Mal. 2).

In an obtuse-angled tricycle, the point of cross-over is located behind the area of ​​the tricot (Mal. 3).

In the femoral tricycle, the median, the bisector and the height, carried out to the base of the tricycle, are retracted.

At the flat-sided tricycle, all three "bridged" lines (height, bisector and median) have three "bridged" points (points of the orthocenter, the center of the vagi and the center of the inscribed and described lane) to be located in the same point.

Formulas for finding the height of a triangle


The figure is shown to facilitate the perception of the formulas for finding the height of a triangle. General rule- the length of the side is indicated by a small letter opposite the corresponding corner. That is, side a lies opposite corner A.
Height in formulas is indicated by the letter h, the subscript of which corresponds to the side to which it is lowered.

Other designations:
a, b, c- the lengths of the sides of the triangle
h a- the height of the triangle drawn to side a from the opposite corner
h b- height drawn to side b
h c- height drawn to side c
R- the radius of the circumscribed circle
r- radius of the inscribed circle


Explanations for the formulas.
The height of a triangle is equal to the product of the length of the side adjacent to the corner from which this height is lowered by the sine of the angle between this side and the side to which this height is lowered (Formula 1)
The height of the triangle is equal to the quotient of dividing twice the area of ​​the triangle by the length of the side to which this height is lowered (Formula 2)
The height of a triangle is equal to the quotient of dividing the product of the sides adjacent to the corner from which this height is omitted, by twice the radius of the circle circumscribed around it (Formula 4).
The heights of the sides in a triangle are related to each other in the same proportion as the inverse proportions of the lengths of the sides of the same triangle are related to each other, and also in the same proportion to each other the products of pairs of sides of the triangle that have a common angle are related (Formula 5).
The sum of the reciprocal values ​​of the heights of the triangle is equal to the reciprocal of the radius of the circle inscribed in such a triangle (Formula 6)
The area of ​​a triangle can be found through the lengths of the heights of this triangle (Formula 7)
The length of the side of the triangle to which the height is lowered can be found by applying formulas 7 and 2.

The task for.

In a right-angled triangle ABC (angle C = 90 0) the height CD is drawn. Determine CD if AD = 9 cm, BD = 16 cm

Solution.

Triangles ABC, ACD and CBD are similar to each other. This directly follows from the second similarity criterion (the equality of the angles in these triangles is obvious).

Right-angled triangles are the only kind of triangles that can be cut into two triangles that are similar to each other and the original triangle.

The designations of these three triangles in the following order of the vertices: ABC, ACD, CBD. Thus, we simultaneously show the correspondence of the vertices. (Vertex A of triangle ABC also corresponds to vertex A of triangle ACD and vertex C of triangle CBD, etc.)

Triangles ABC and CBD are similar. Means:

AD / DC = DC / BD, that is

The problem of applying the Pythagorean theorem.

Triangle ABC is rectangular. In this case, C is a right angle. The height CD = 6cm is drawn from it. The difference between the segments BD-AD = 5 cm.

Find: Sides of triangle ABC.

Solution.

1. Let's compose a system of equations according to the Pythagorean theorem

CD 2 + BD 2 = BC 2

CD 2 + AD 2 = AC 2

since CD = 6

Since BD-AD = 5, then

BD = AD + 5, then the system of equations takes the form

36+ (AD + 5) 2 = BC 2

Let's add the first and second equations. Since the left side is added to the left and the right side to the right, equality will not be violated. We get:

36 + 36 + (AD + 5) 2 + AD 2 = AC 2 + BC 2

72+ (AD + 5) 2 + AD 2 = AC 2 + BC 2

2. Now, looking at the original drawing of the triangle, according to the same Pythagorean theorem, the equality should hold:

AC 2 + BC 2 = AB 2

Since AB = BD + AD, the equation becomes:

AC 2 + BC 2 = (AD + BD) 2

Since BD-AD = 5, then BD = AD + 5, then

AC 2 + BC 2 = (AD + AD + 5) 2

3. Now let's take a look at the results we got when solving in the first and second parts of the solution. Namely:

72+ (AD + 5) 2 + AD 2 = AC 2 + BC 2

AC 2 + BC 2 = (AD + AD + 5) 2

They have common part AC 2 + BC 2. Thus, let us equate them with each other.

72+ (AD + 5) 2 + AD 2 = (AD + AD + 5) 2

72 + AD 2 + 10AD + 25 + AD 2 = 4AD 2 + 20AD + 25

2AD 2 -10AD + 72 = 0

In the received quadratic the discriminant is D = 676, respectively, the roots of the equation are:

Since the length of the segment cannot be negative, we discard the first root.

Respectively

AB = BD + AD = 4 + 9 = 13

By the Pythagorean theorem, we find the remaining sides of the triangle:

AC = root of (52)

Triangles.

Basic concepts.

Triangle is a figure consisting of three line segments and three points that do not lie on one straight line.

The segments are called parties, and points - peaks.

Sum of angles of the triangle is 180 º.

The height of the triangle.

Height of triangle is a perpendicular drawn from the top to the opposite side.

In an acute-angled triangle, the height is contained within the triangle (Fig. 1).

In a right-angled triangle, the legs are the heights of the triangle (Fig. 2).

In an obtuse triangle, the height is outside the triangle (Figure 3).

Triangle Height Properties:

The bisector of a triangle.

Bisector of a triangle is a line segment that bisects the corner of the vertex and connects the vertex with a point on the opposite side (Fig. 5).

Bisector properties:


Median of the triangle.

Median of a triangle is a segment connecting the vertex with the middle of the opposite side (Fig. 9a).


The length of the median can be calculated using the formula:

2b 2 + 2c 2 - a 2
m a 2 = ——————
4

where m a is the median drawn to the side a.

In a right-angled triangle, the median drawn to the hypotenuse is half the hypotenuse:

c
m c = —
2

where m c- the median drawn to the hypotenuse c(Figure 9c)

The medians of a triangle intersect at one point (at the center of mass of the triangle) and are divided by this point in a ratio of 2: 1, counting from the vertex. That is, the segment from the vertex to the center is twice as large as the segment from the center to the side of the triangle (Figure 9c).

Three medians of a triangle divide it into six equal triangles.

The middle line of the triangle.

Midline of a triangle is a segment connecting the midpoints of its two sides (Fig. 10).

The middle line of the triangle is parallel to the third side and is equal to half of it

The outer corner of the triangle.

Outside corner triangle is equal to the sum two non-adjacent interior corners (fig. 11).

The outer corner of a triangle is greater than any non-adjacent angle.

Right triangle.

Right triangle is a triangle with a right angle (fig. 12).

The side of a right triangle opposite a right angle is called hypotenuse.

The other two parties are called legs.


Proportional line segments in a right-angled triangle.

1) In a right-angled triangle, the height drawn from the right angle forms three similar triangle: ABC, ACH and HCB (fig. 14a). Accordingly, the angles formed by the height are equal to the angles A and B.

Fig. 14a

Isosceles triangle.

Isosceles triangle is a triangle with two sides equal (Fig. 13).

These equal sides are called lateral sides and the third one is basis triangle.

In an isosceles triangle, the angles at the base are equal. (In our triangle, angle A equal to the angle C).

In an isosceles triangle, the median drawn to the base is both the bisector and the height of the triangle.

Equilateral triangle.

An equilateral triangle is a triangle in which all sides are equal (Fig. 14).

Equilateral triangle properties:

Remarkable properties of triangles.

Triangles have original properties that will help you successfully solve problems with these shapes. Some of these properties are outlined above. But we repeat them one more time, adding a few other great features to them:

1) In a right-angled triangle with angles 90º, 30º and 60º leg b lying opposite an angle of 30º is equal to half of the hypotenuse. And the lega more legb√3 times (Fig. 15 a). For example, if leg b is 5, then the hypotenuse c necessarily equal to 10, and the leg a is equal to 5√3.

2) In a right-angled isosceles triangle with angles of 90º, 45º and 45º, the hypotenuse is √2 times the leg (Fig. 15 b). For example, if the legs are 5, then the hypotenuse is 5√2.

3) The midline of the triangle is half parallel side(fig. 15 with). For example, if the side of a triangle is 10, then the parallel midline is 5.

4) In a right-angled triangle, the median drawn to the hypotenuse is equal to half the hypotenuse (Figure 9c): m c= s / 2.

5) The medians of a triangle, intersecting at one point, are divided by this point in a 2: 1 ratio. That is, the segment from the vertex to the point of intersection of the medians is twice the segment from the point of intersection of the medians to the side of the triangle (Figure 9c)

6) In a right-angled triangle, the middle of the hypotenuse is the center of the circumscribed circle (Fig. 15 d).


Equality tests for triangles.

The first sign of equality: if two sides and the angle between them of one triangle are equal to two sides and the angle between them of another triangle, then such triangles are equal.

The second sign of equality: if the side and the angles adjacent to it of one triangle are equal to the side and the angles adjacent to it of the other triangle, then such triangles are equal.

The third sign of equality: if three sides of one triangle are equal to three sides of another triangle, then such triangles are equal.

Triangle inequality.

In any triangle, each side is less than the sum of the other two sides.

Pythagorean theorem.

In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the legs:

c 2 = a 2 + b 2 .

Area of ​​a triangle.

1) The area of ​​a triangle is equal to half the product of its side by the height drawn to this side:

ah
S = ——
2

2) The area of ​​a triangle is equal to half the product of any two of its sides by the sine of the angle between them:

1
S = — AB AC · sin A
2

A triangle circumscribed about a circle.

A circle is called inscribed in a triangle if it touches all its sides (Fig. 16 a).


A triangle inscribed in a circle.

A triangle is called inscribed in a circle if it touches it with all its vertices (Fig. 17 a).

Sine, cosine, tangent, cotangent of an acute angle of a right triangle (Fig. 18).

Sinus acute angle x opposing leg to the hypotenuse.
It is denoted like this: sinx.

Cosine acute angle x right triangle is the ratio adjacent leg to the hypotenuse.
It is denoted like this: cos x.

Tangent acute angle x is the ratio of the opposite leg to the adjacent leg.
It is denoted like this: tgx.

Cotangent acute angle x is the ratio of the adjacent leg to the opposite one.
It is denoted like this: ctgx.

Rules:

Leg opposite to the corner x, is equal to the product of the hypotenuse and sin x:

b = c Sin x

Leg adjacent to the corner x, is equal to the product of the hypotenuse and cos x:

a = c Cos x

Leg opposite to the corner x, is equal to the product of the second leg and tg x:

b = a Tg x

Leg adjacent to the corner x, is equal to the product of the second leg and ctg x:

a = b Ctg x.


For any sharp angle x:

sin (90 ° - x) = cos x

cos (90 ° - x) = sin x


Triangle) or pass outside the triangle at an obtuse triangle.

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    Subtitles

Properties of the intersection point of the three heights of the triangle (orthocenter)

EA → ⋅ BC → + EB → ⋅ CA → + EC → ⋅ AB → = 0 (\ displaystyle (\ overrightarrow (EA)) \ cdot (\ overrightarrow (BC)) + (\ overrightarrow (EB)) \ cdot (\ overrightarrow (CA)) + (\ overrightarrow (EC)) \ cdot (\ overrightarrow (AB)) = 0)

(To prove the identity, one should use the formulas

AB → = EB → - EA →, BC → = EC → - EB →, CA → = EA → - EC → (\ displaystyle (\ overrightarrow (AB)) = (\ overrightarrow (EB)) - (\ overrightarrow (EA )), \, (\ overrightarrow (BC)) = (\ overrightarrow (EC)) - (\ overrightarrow (EB)), \, (\ overrightarrow (CA)) = (\ overrightarrow (EA)) - (\ overrightarrow (EC)))

The point E should be the intersection of the two heights of the triangle.)

  • Orthocenter isogonally conjugate to center circumscribed circle .
  • Orthocenter lies on one straight line with the centroid, center circumscribed circle and the center of a circle of nine points (see Euler's line).
  • Orthocenter an acute-angled triangle is the center of a circle inscribed in its orthotriangle.
  • The center of a triangle circumscribed by the orthocenter with vertices at the midpoints of the sides of this triangle. The last triangle is called the complementary triangle in relation to the first triangle.
  • The last property can be formulated as follows: The center of a circle circumscribed about a triangle serves as orthocenter additional triangle.
  • Points symmetrical orthocenter of a triangle with respect to its sides, lie on the circumscribed circle.
  • Points symmetrical orthocenter of a triangle with respect to the midpoints of the sides, also lie on the circumcircle and coincide with points diametrically opposite to the corresponding vertices.
  • If O is the center of the circumscribed circle ΔABC, then O H → = O A → + O B → + O C → (\ displaystyle (\ overrightarrow (OH)) = (\ overrightarrow (OA)) + (\ overrightarrow (OB)) + (\ overrightarrow (OC))) ,
  • The distance from the apex of the triangle to the orthocenter is twice the distance from the center of the circumscribed circle to the opposite side.
  • Any segment drawn from orthocenter before crossing the circumcircle, it is always halved by Euler's circle. Orthocenter is the center of the homothety of these two circles.
  • Hamilton's theorem... Three line segments connecting the orthocenter with the vertices of an acute-angled triangle divide it into three triangles having the same Euler circle (a circle of nine points) as the original acute-angled triangle.
  • Consequences of Hamilton's theorem:
    • Three line segments connecting the orthocenter with the vertices of an acute-angled triangle divide it into three Hamilton's triangle having equal radii of the circumscribed circles.
    • The radii of the circumscribed circles of three Hamilton's triangles are equal to the radius of the circle circumscribed about the original acute-angled triangle.
  • In an acute-angled triangle, the orthocenter lies inside the triangle; in obtuse - outside the triangle; in a rectangular one - at the apex of a right angle.

Isosceles triangle elevation properties

  • If two heights in a triangle are equal, then the triangle is isosceles (Steiner - Lemus theorem), and the third height is simultaneously the median and bisector of the angle from which it comes out.
  • The converse is also true: in an isosceles triangle, two heights are equal, and the third height is both the median and the bisector.
  • In an equilateral triangle, all three heights are equal.

Base elevation properties of a triangle

  • Foundations heights form the so-called orthotriangle, which has its own properties.
  • The circle circumscribed about the orthotriangle is Euler's circle. This circle also contains three midpoints of the sides of the triangle and three midpoints of three segments connecting the orthocenter with the vertices of the triangle.
  • Another formulation of the last property:
    • Euler's theorem for a circle of nine points. Foundations three heights an arbitrary triangle, the middle of its three sides ( foundations of its internal medians) and the midpoints of three segments connecting its vertices with the orthocenter, all lie on the same circle (on circle of nine points).
  • Theorem... In any triangle, a segment connecting grounds two heights triangle, cuts off a triangle like this one.
  • Theorem... In a triangle, a segment connecting grounds two heights triangles lying on two sides, antiparallel a third party with which it has no common points. Through its two ends, as well as through two vertices of the third mentioned side, you can always draw a circle.

Other triangle elevation properties

  • If the triangle versatile (scalene), then its internal the bisector drawn from any vertex lies between internal median and height drawn from the same peak.
  • The height of the triangle is isogonally conjugate to the diameter (radius) circumscribed circle drawn from the same vertex.
  • In an acute-angled triangle, its two heights cut off such triangles from it.
  • In a right triangle height drawn from the vertex of a right angle splits it into two triangles similar to the original one.

Properties of the minimum of the heights of a triangle

The smallest of the triangle heights has many extreme properties. For example:

  • The minimum orthogonal projection of a triangle onto straight lines lying in the plane of the triangle has a length equal to the smallest of its heights.
  • The minimum straight cut in the plane through which an unbendable triangular plate can be pulled must have a length equal to the smallest of the heights of this plate.
  • With the continuous movement of two points along the perimeter of the triangle towards each other, the maximum distance between them during the movement from the first meeting to the second cannot be less than the length of the smallest of the triangle heights.
  • The minimum height in a triangle is always inside that triangle.

Basic relations

  • h a = b ⋅ sin ⁡ γ = c ⋅ sin ⁡ β, (\ displaystyle h_ (a) = b (\ cdot) \ sin \ gamma = c (\ cdot) \ sin \ beta,)
  • h a = 2 ⋅ S a, (\ displaystyle h_ (a) = (\ frac (2 (\ cdot) S) (a)),) where S (\ displaystyle S)- area of ​​a triangle, a (\ displaystyle a)- the length of the side of the triangle, to which the height is lowered.
  • h a = b ⋅ c 2 ⋅ R, (\ displaystyle h_ (a) = (\ frac (b (\ cdot) c) (2 (\ cdot) R)),) where b ⋅ c (\ displaystyle b (\ cdot) c)- the product of the sides, R - (\ displaystyle R-) radius of the circumscribed circle
  • h a: h b: h c = 1 a: 1 b: 1 c = (b ⋅ c): (a ⋅ c): (a ⋅ b). (\ displaystyle h_ (a): h_ (b): h_ (c) = (\ frac (1) (a)): (\ frac (1) (b)): (\ frac (1) (c)) = (b (\ cdot) c) :( a (\ cdot) c) :( a (\ cdot) b).)
  • 1 ha + 1 hb + 1 hc = 1 r (\ displaystyle (\ frac (1) (h_ (a))) + (\ frac (1) (h_ (b))) + (\ frac (1) (h_ (c))) = (\ frac (1) (r))), where r (\ displaystyle r) is the radius of the inscribed circle.
  • S = 1 (1 ha + 1 hb + 1 hc) ⋅ (1 ha + 1 hb - 1 hc) ⋅ (1 ha + 1 hc - 1 hb) ⋅ (1 hb + 1 hc - 1 ha) (\ displaystyle S = (\ frac (1) (\ sqrt (((\ frac (1) (h_ (a))) + (\ frac (1) (h_ (b))) + (\ frac (1) (h_ (c )))) (\ cdot) ((\ frac (1) (h_ (a))) + (\ frac (1) (h_ (b))) - (\ frac (1) (h_ (c))) ) (\ cdot) ((\ frac (1) (h_ (a))) + (\ frac (1) (h_ (c))) - (\ frac (1) (h_ (b)))) (\ cdot) ((\ frac (1) (h_ (b))) + (\ frac (1) (h_ (c))) - (\ frac (1) (h_ (a)))))))), where S (\ displaystyle S)- area of ​​a triangle.
  • a = 2 ha ⋅ (1 ha + 1 hb + 1 hc) ⋅ (1 ha + 1 hb - 1 hc) ⋅ (1 ha + 1 hc - 1 hb) ⋅ (1 hb + 1 hc - 1 ha) (\ displaystyle a = (\ frac (2) (h_ (a) (\ cdot) (\ sqrt (((\ frac (1) (h_ (a))) + (\ frac (1) (h_ (b))) + (\ frac (1) (h_ (c)))) (\ cdot) ((\ frac (1) (h_ (a))) + (\ frac (1) (h_ (b))) - (\ frac (1) (h_ (c)))) (\ cdot) ((\ frac (1) (h_ (a))) + (\ frac (1) (h_ (c))) - (\ frac (1 ) (h_ (b)))) (\ cdot) ((\ frac (1) (h_ (b))) + (\ frac (1) (h_ (c))) - (\ frac (1) (h_ (a))))))))), a (\ displaystyle a)- the side of the triangle to which the height falls h a (\ displaystyle h_ (a)).
  • Height of an isosceles triangle, lowered to the base: hc = 1 2 ⋅ 4 a 2 - c 2, (\ displaystyle h_ (c) = (\ frac (1) (2)) (\ cdot) (\ sqrt (4a ^ (2) -c ^ (2)) ),)
where c (\ displaystyle c)- base, a (\ displaystyle a)- side.

Height theorem for a right triangle

If the height in a right-angled triangle is ABC with a length h (\ displaystyle h) drawn from the vertex of the right angle divides the hypotenuse with length c (\ displaystyle c) for segments m (\ displaystyle m) and n (\ displaystyle n) corresponding to the legs b (\ displaystyle b) and a (\ displaystyle a), then the following equalities are true.

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