Relationships in a triangle formula. Similar triangles

The simplest polygon that is studied in school is a triangle. It is more understandable for students and encounters fewer difficulties. Despite the fact that there are different kinds triangles that have special properties.

What shape is called a triangle?

Formed by three points and segments. The first ones are called vertices, the second ones are called sides. Moreover, all three segments must be connected so that angles are formed between them. Hence the name of the “triangle” figure.

Differences in names across corners

Since they can be acute, obtuse and straight, the types of triangles are determined by these names. Accordingly, there are three groups of such figures.

• First. If all the angles of a triangle are acute, then it will be called acute. Everything is logical.

• Second. One of the angles is obtuse, which means the triangle is obtuse. It couldn't be simpler.

• Third. There is an angle equal to 90 degrees, which is called a right angle. The triangle becomes rectangular.

Differences in names on the sides

Depending on the characteristics of the sides, the following types of triangles are distinguished:

the general case is scalene, in which all sides are of arbitrary length;

isosceles, two sides of which have the same numerical values;

equilateral, the lengths of all its sides are the same.

If the problem does not specify a specific type of triangle, then you need to draw an arbitrary one. In which all the corners are sharp, and the sides have different lengths.

Properties common to all triangles

1. If you add up all the angles of a triangle, you get a number equal to 180º. And it doesn't matter what type it is. This rule always applies.
2. The numerical value of any side of a triangle is less than the other two added together. Moreover, it is greater than their difference.
3. Each external angle has a value that is obtained by adding two internal angles that are not adjacent to it. Moreover, it is always larger than the internal one adjacent to it.
4. The smallest angle is always opposite the smaller side of the triangle. And vice versa, if the side is large, then the angle will be the largest.

These properties are always valid, no matter what types of triangles are considered in the problems. All the rest follow from specific features.

Properties of an isosceles triangle

• The angles that are adjacent to the base are equal.
• The height, which is drawn to the base, is also the median and bisector.
• The altitudes, medians and bisectors, which are built to the lateral sides of the triangle, are respectively equal to each other.

Properties of an equilateral triangle

If there is such a figure, then all the properties described a little above will be true. Because an equilateral will always be isosceles. But not vice versa; an isosceles triangle will not necessarily be equilateral.

• All its angles are equal to each other and have a value of 60º.
• Any median of an equilateral triangle is its altitude and bisector. Moreover, they are all equal to each other. To determine their values, there is a formula that consists of the product of the side and the square root of 3 divided by 2.

Properties of a right triangle

• Two acute angles add up to 90º.
• The length of the hypotenuse is always greater than that of any of the legs.
• The numerical value of the median drawn to the hypotenuse is equal to its half.
• The leg is equal to the same value if it lies opposite an angle of 30º.
• The height, which is drawn from the vertex with a value of 90º, has a certain mathematical dependence on the legs: 1/n 2 = 1/a 2 + 1/b 2. Here: a, b - legs, n - height.

Problems with different types of triangles

No. 1. Given an isosceles triangle. Its perimeter is known and equal to 90 cm. We need to find out its sides. As an additional condition: the side side is 1.2 times smaller than the base.

The value of the perimeter directly depends on the quantities that need to be found. The sum of all three sides will give 90 cm. Now you need to remember the sign of a triangle, according to which it is isosceles. That is, the two sides are equal. You can create an equation with two unknowns: 2a + b = 90. Here a is the side, b is the base.

Now it's time for an additional condition. Following it, the second equation is obtained: b = 1.2a. You can substitute this expression into the first one. It turns out: 2a + 1.2a = 90. After transformations: 3.2a = 90. Hence a = 28.125 (cm). Now it is easy to find out the basis. This is best done from the second condition: b = 1.2 * 28.125 = 33.75 (cm).

To check, you can add three values: 28.125 * 2 + 33.75 = 90 (cm). That's right.

Answer: The sides of the triangle are 28.125 cm, 28.125 cm, 33.75 cm.

No. 2. The side of an equilateral triangle is 12 cm. You need to calculate its height.

Solution. To find the answer, it is enough to return to the moment where the properties of the triangle were described. This is the formula for finding the height, median and bisector of an equilateral triangle.

n = a * √3 / 2, where n is the height and a is the side.

Substitution and calculation give the following result: n = 6 √3 (cm).

There is no need to memorize this formula. It is enough to remember that the height divides the triangle into two rectangular ones. Moreover, it turns out to be a leg, and the hypotenuse in it is the side of the original one, the second leg is half of the known side. Now you need to write down the Pythagorean theorem and derive a formula for height.

Answer: height is 6 √3 cm.

No. 3. Given MKR is a triangle, in which angle K makes 90 degrees. The sides MR and KR are known, they are equal to 30 and 15 cm, respectively. We need to find out the value of angle P.

Solution. If you make a drawing, it becomes clear that MR is the hypotenuse. Moreover, it is twice as large as the side of the KR. Again you need to turn to the properties. One of them has to do with angles. From it it is clear that the KMR angle is 30º. This means that the desired angle P will be equal to 60º. This follows from another property, which states that the sum of two acute angles must equal 90º.

Answer: angle P is 60º.

No. 4. We need to find all the angles of an isosceles triangle. It is known about it that the external angle from the angle at the base is 110º.

Solution. Since only the external angle is given, this is what you need to use. It forms an unfolded angle with the internal one. This means that in total they will give 180º. That is, the angle at the base of the triangle will be equal to 70º. Since it is isosceles, the second angle has the same value. It remains to calculate the third angle. According to a property common to all triangles, the sum of the angles is 180º. This means that the third will be defined as 180º - 70º - 70º = 40º.

Answer: the angles are 70º, 70º, 40º.

No. 5. It is known that in an isosceles triangle the angle opposite the base is 90º. There is a point marked on the base. The segment connecting it to a right angle divides it in the ratio of 1 to 4. You need to find out all the angles of the smaller triangle.

Solution. One of the angles can be determined immediately. Since the triangle is right-angled and isosceles, those that lie at its base will be 45º each, that is, 90º/2.

The second of them will help you find the relation known in the condition. Since it is equal to 1 to 4, the parts into which it is divided are only 5. This means that to find out the smaller angle of a triangle you need 90º/5 = 18º. It remains to find out the third. To do this, you need to subtract 45º and 18º from 180º (the sum of all angles of the triangle). The calculations are simple, and you get: 117º.

Standard designations

Triangle with vertices A, B And C is designated as (see figure). A triangle has three sides:

The lengths of the sides of a triangle are indicated by lowercase Latin letters (a, b, c):

A triangle has the following angles:

The angle values ​​at the corresponding vertices are traditionally denoted by Greek letters (α, β, γ).

Signs of equality of triangles

A triangle on the Euclidean plane can be uniquely determined (up to congruence) by the following triplets of basic elements:

1. a, b, γ (equality on two sides and the angle lying between them);
2. a, β, γ (equality on the side and two adjacent angles);
3. a, b, c (equality on three sides).

Signs of equality of right triangles:

1. along the leg and hypotenuse;
2. on two legs;
3. along the leg and acute angle;
4. along the hypotenuse and acute angle.

Some points in the triangle are “paired”. For example, there are two points from which all sides are visible at either an angle of 60° or an angle of 120°. They're called Torricelli dots. There are also two points whose projections onto the sides lie at the vertices of a regular triangle. This - Apollonius points. Points and such are called Brocard points.

Direct

In any triangle, the center of gravity, the orthocenter and the center of the circumcircle lie on the same straight line, called Euler's line.

The straight line passing through the center of the circumcircle and the Lemoine point is called Brocard axis. The Apollonius points lie on it. The Torricelli point and the Lemoine point also lie on the same line. The bases of the external bisectors of the angles of a triangle lie on the same straight line, called axis of external bisectors. The intersection points of the lines containing the sides of an orthotriangle with the lines containing the sides of the triangle also lie on the same line. This line is called orthocentric axis, it is perpendicular to the Euler straight line.

If we take a point on the circumcircle of a triangle, then its projections onto the sides of the triangle will lie on the same straight line, called Simson's straight this point. Simson's lines of diametrically opposite points are perpendicular.

Triangles

• A triangle with vertices at the bases drawn through a given point is called cevian triangle this point.
• A triangle with vertices in the projections of a given point onto the sides is called sod or pedal triangle this point.
• A triangle with vertices at the second points of intersection of lines drawn through the vertices and a given point with the circumcircle is called circumferential triangle. The circumferential triangle is similar to the sod triangle.

Circles

• Inscribed circle- a circle touching all three sides of the triangle. She's the only one. The center of the inscribed circle is called incenter.
• Circumcircle- a circle passing through all three vertices of the triangle. The circumscribed circle is also unique.
• Excircle- a circle touching one side of the triangle and the continuation of the other two sides. There are three such circles in a triangle. Their radical center is the center of the inscribed circle of the medial triangle, called Spiker's point.

The midpoints of the three sides of a triangle, the bases of its three altitudes and the midpoints of the three segments connecting its vertices with the orthocenter lie on one circle called circle of nine points or Euler circle. The center of the nine-point circle lies on the Euler line. A circle of nine points touches an inscribed circle and three excircles. The point of tangency between the inscribed circle and the circle of nine points is called Feuerbach point. If from each vertex we lay outwards of the triangle on straight lines containing the sides, orthoses equal in length to the opposite sides, then the resulting six points lie on the same circle - Conway circle. Three circles can be inscribed in any triangle in such a way that each of them touches two sides of the triangle and two other circles. Such circles are called Malfatti circles. The centers of the circumscribed circles of the six triangles into which the triangle is divided by medians lie on one circle, which is called circumference of Lamun.

A triangle has three circles that touch two sides of the triangle and the circumcircle. Such circles are called semi-inscribed or Verrier circles. The segments connecting the points of tangency of the Verrier circles with the circumcircle intersect at one point called Verrier's point. It serves as the center of a homothety, which transforms a circumcircle into an inscribed circle. The points of contact of the Verrier circles with the sides lie on a straight line that passes through the center of the inscribed circle.

The segments connecting the points of tangency of the inscribed circle with the vertices intersect at one point called Gergonne point, and the segments connecting the vertices with the points of tangency of the excircles are in Nagel point.

Ellipses, parabolas and hyperbolas

Inscribed conic (ellipse) and its perspector

An infinite number of conics (ellipses, parabolas or hyperbolas) can be inscribed into a triangle. If we inscribe an arbitrary conic into a triangle and connect the tangent points with opposite vertices, then the resulting straight lines will intersect at one point called prospect bunks. For any point of the plane that does not lie on a side or on its extension, there is an inscribed conic with a perspector at this point.

The described Steiner ellipse and the cevians passing through its foci

You can inscribe an ellipse into a triangle, which touches the sides in the middle. Such an ellipse is called inscribed Steiner ellipse(its perspective will be the centroid of the triangle). The circumscribed ellipse, which touches the lines passing through the vertices parallel to the sides, is called described by the Steiner ellipse. If we transform a triangle into a regular triangle using an affine transformation (“skew”), then its inscribed and circumscribed Steiner ellipse will transform into an inscribed and circumscribed circle. The Chevian lines drawn through the foci of the described Steiner ellipse (Scutin points) are equal (Scutin's theorem). Of all the described ellipses, the described Steiner ellipse has the smallest area, and of all the inscribed ellipses, the inscribed Steiner ellipse has the largest area.

Brocard ellipse and its perspector - Lemoine point

An ellipse with foci at Brocard points is called Brocard ellipse. Its perspective is the Lemoine point.

Properties of an inscribed parabola

Kiepert parabola

The prospects of the inscribed parabolas lie on the described Steiner ellipse. The focus of an inscribed parabola lies on the circumcircle, and the directrix passes through the orthocenter. A parabola inscribed in a triangle and having Euler's directrix as its directrix is ​​called Kiepert parabola. Its perspector is the fourth point of intersection of the circumscribed circle and the circumscribed Steiner ellipse, called Steiner point.

Kiepert's hyperbole

If the described hyperbola passes through the point of intersection of the heights, then it is equilateral (that is, its asymptotes are perpendicular). The intersection point of the asymptotes of an equilateral hyperbola lies on the circle of nine points.

Transformations

If the lines passing through the vertices and some point not lying on the sides and their extensions are reflected relative to the corresponding bisectors, then their images will also intersect at one point, which is called isogonally conjugate the original one (if the point lay on the circumscribed circle, then the resulting lines will be parallel). Many pairs of remarkable points are isogonally conjugate: the circumcenter and the orthocenter, the centroid and the Lemoine point, the Brocard points. The Apollonius points are isogonally conjugate to the Torricelli points, and the center of the inscribed circle is isogonally conjugate to itself. Under the action of isogonal conjugation, straight lines transform into circumscribed conics, and circumscribed conics into straight lines. Thus, the Kiepert hyperbola and the Brocard axis, the Jenzabek hyperbola and the Euler straight line, the Feuerbach hyperbola and the line of centers of the inscribed and circumscribed circles are isogonally conjugate. The circumcircles of the triangles of isogonally conjugate points coincide. The foci of inscribed ellipses are isogonally conjugate.

If, instead of a symmetrical cevian, we take a cevian whose base is as distant from the middle of the side as the base of the original one, then such cevians will also intersect at one point. The resulting transformation is called isotomic conjugation. It also converts straight lines into described conics. The Gergonne and Nagel points are isotomically conjugate. Under affine transformations, isotomically conjugate points are transformed into isotomically conjugate points. With isotomic conjugation, the described Steiner ellipse will go into the infinitely distant straight line.

If in the segments cut off by the sides of the triangle from the circumcircle, we inscribe circles touching the sides at the bases of the cevians drawn through a certain point, and then connect the tangent points of these circles with the circumcircle with opposite vertices, then such straight lines will intersect at one point. A plane transformation that matches the original point to the resulting one is called isocircular transformation. The composition of isogonal and isotomic conjugates is the composition of an isocircular transformation with itself. This composition is a projective transformation, which leaves the sides of the triangle in place, and transforms the axis of the external bisectors into a straight line at infinity.

If we continue the sides of a Chevian triangle of a certain point and take their points of intersection with the corresponding sides, then the resulting points of intersection will lie on one straight line, called trilinear polar starting point. The orthocentric axis is the trilinear polar of the orthocenter; the trilinear polar of the center of the inscribed circle is the axis of the external bisectors. Trilinear polars of points lying on a circumscribed conic intersect at one point (for a circumscribed circle this is the Lemoine point, for a circumscribed Steiner ellipse it is the centroid). The composition of an isogonal (or isotomic) conjugate and a trilinear polar is a duality transformation (if a point isogonally (isotomically) conjugate to a point lies on the trilinear polar of a point, then the trilinear polar of a point isogonally (isotomically) conjugate to a point lies on the trilinear polar of a point).

Cubes

Ratios in a triangle

Note: in this section, , are the lengths of the three sides of the triangle, and , are the angles lying respectively opposite these three sides (opposite angles).

Triangle inequality

In a non-degenerate triangle, the sum of the lengths of its two sides is greater than the length of the third side, in a degenerate triangle it is equal. In other words, the lengths of the sides of a triangle are related by the following inequalities:

The triangle inequality is one of the axioms of metrics.

Triangle Angle Sum Theorem

Theorem of sines

,

where R is the radius of the circle circumscribed around the triangle. It follows from the theorem that if a< b < c, то α < β < γ.

Cosine theorem

Tangent theorem

Other ratios

Metric ratios in a triangle are given for:

Solving triangles

Calculating the unknown sides and angles of a triangle based on the known ones has historically been called “solving triangles.” The above general trigonometric theorems are used.

Area of ​​a triangle

Special cases Notation

For the area the following inequalities are valid:

Calculating the area of ​​a triangle in space using vectors

Let the vertices of the triangle be at points , , .

Let's introduce the area vector . The length of this vector is equal to the area of ​​the triangle, and it is directed normal to the plane of the triangle:

Let us set , where , , are the projections of the triangle onto the coordinate planes. Wherein

and similarly

The area of ​​the triangle is .

An alternative is to calculate the lengths of the sides (using the Pythagorean theorem) and then using Heron's formula.

Triangle theorems

Desargues's theorem: if two triangles are perspective (the lines passing through the corresponding vertices of the triangles intersect at one point), then their corresponding sides intersect on the same line.

Sonda's theorem: if two triangles are perspective and orthologous (perpendiculars drawn from the vertices of one triangle to the sides opposite the corresponding vertices of the triangle, and vice versa), then both centers of orthology (the points of intersection of these perpendiculars) and the center of perspective lie on the same straight line, perpendicular to the perspective axis (straight line from Desargues' theorem).

The science of geometry tells us what a triangle, square, and cube are. IN modern world it is studied in schools by everyone without exception. Also, the science that studies directly what a triangle is and what properties it has is trigonometry. She explores in detail all phenomena related to data. We will talk about what a triangle is today in our article. Their types will be described below, as well as some theorems associated with them.

What is a triangle? Definition

This is a flat polygon. It has three corners, as is clear from its name. It also has three sides and three vertices, the first of them are segments, the second are points. Knowing what two angles are equal to, you can find the third by subtracting the sum of the first two from the number 180.

What types of triangles are there?

They can be classified according to various criteria.

First of all, they are divided into acute-angled, obtuse-angled and rectangular. The former have acute angles, that is, those that are equal to less than 90 degrees. In obtuse angles, one of the angles is obtuse, that is, one that is equal to more than 90 degrees, the other two are acute. Acute triangles also include equilateral triangles. Such triangles have all sides and angles equal. They are all equal to 60 degrees, this can be easily calculated by dividing the sum of all angles (180) by three.

Right triangle

It is impossible not to talk about what a right triangle is.

Such a figure has one angle equal to 90 degrees (straight), that is, two of its sides are perpendicular. The remaining two angles are acute. They can be equal, then it will be isosceles. The Pythagorean theorem is related to the right triangle. Using it, you can find the third side, knowing the first two. According to this theorem, if you add the square of one leg to the square of the other, you can get the square of the hypotenuse. The square of the leg can be calculated by subtracting the square of the known leg from the square of the hypotenuse. Speaking about what a triangle is, we can also recall an isosceles triangle. This is one in which two of the sides are equal, and two angles are also equal.

What are leg and hypotenuse?

A leg is one of the sides of a triangle that forms an angle of 90 degrees. The hypotenuse is the remaining side that is opposite the right angle. You can lower a perpendicular from it onto the leg. The ratio of the adjacent side to the hypotenuse is called cosine, and the opposite side is called sine.

- what are its features?

It's rectangular. Its legs are three and four, and its hypotenuse is five. If you see that the legs of a given triangle are equal to three and four, you can rest assured that the hypotenuse will be equal to five. Also, using this principle, you can easily determine that the leg will be equal to three if the second is equal to four, and the hypotenuse is equal to five. To prove this statement, you can apply the Pythagorean theorem. If two legs are equal to 3 and 4, then 9 + 16 = 25, the root of 25 is 5, that is, the hypotenuse is equal to 5. An Egyptian triangle is also a right triangle whose sides are equal to 6, 8 and 10; 9, 12 and 15 and other numbers with the ratio 3:4:5.

What else could a triangle be?

Triangles can also be inscribed or circumscribed. The figure around which the circle is described is called inscribed; all its vertices are points lying on the circle. A circumscribed triangle is one into which a circle is inscribed. All its sides come into contact with it at certain points.

How is it located?

The area of ​​any figure is measured in square units (sq. meters, sq. millimeters, sq. centimeters, sq. decimeters, etc.) This value can be calculated in a variety of ways, depending on the type of triangle. The area of ​​any figure with angles can be found by multiplying its side by the perpendicular dropped onto it from the opposite corner, and dividing this figure by two. You can also find this value by multiplying the two sides. Then multiply this number by the sine of the angle located between these sides, and divide this result by two. Knowing all the sides of a triangle, but not knowing its angles, you can find the area in another way. To do this you need to find half the perimeter. Then alternately subtract different sides from this number and multiply the resulting four values. Next, find from the number that came out. The area of ​​an inscribed triangle can be found by multiplying all the sides and dividing the resulting number by that circumscribed around it, multiplied by four.

The area of ​​a circumscribed triangle is found in this way: we multiply half the perimeter by the radius of the circle that is inscribed in it. If then its area can be found as follows: square the side, multiply the resulting figure by the root of three, then divide this number by four. In a similar way, you can calculate the height of a triangle in which all sides are equal; to do this, you need to multiply one of them by the root of three, and then divide this number by two.

Theorems related to triangle

The main theorems that are associated with this figure are the Pythagorean theorem described above and cosines. The second (of sines) is that if you divide any side by the sine of the angle opposite it, you can get the radius of the circle that is described around it, multiplied by two. The third (cosines) is that if from the sum of the squares of the two sides we subtract their product, multiplied by two and the cosine of the angle located between them, then we get the square of the third side.

Dali Triangle - what is it?

Many, when faced with this concept, at first think that this is some kind of definition in geometry, but this is not at all the case. Dali's triangle is common name three places that are closely related to life famous artist. Its “peaks” are the house in which Salvador Dali lived, the castle that he gave to his wife, as well as the museum of surrealist paintings. You can learn a lot during a tour of these places. interesting facts about this unique creative artist known throughout the world.

Generally, two triangles are considered similar if they have the same shape, even if they are different sizes, rotated, or even upside down.

The mathematical representation of two similar triangles A 1 B 1 C 1 and A 2 B 2 C 2 shown in the figure is written as follows:

ΔA 1 B 1 C 1 ~ ΔA 2 B 2 C 2

Two triangles are similar if:

1. Each angle of one triangle is equal to the corresponding angle of another triangle:
∠A 1 = ∠A 2 , ∠B 1 = ∠B 2 And ∠C 1 = ∠C 2

2. The ratios of the sides of one triangle to the corresponding sides of another triangle are equal to each other:
$\frac(A_1B_1)(A_2B_2)=\frac(A_1C_1)(A_2C_2)=\frac(B_1C_1)(B_2C_2)$

3. Relationships two sides one triangle to the corresponding sides of another triangle are equal to each other and at the same time
the angles between these sides are equal:
$\frac(B_1A_1)(B_2A_2)=\frac(A_1C_1)(A_2C_2)$ and $\angle A_1 = \angle A_2$
or
$\frac(A_1B_1)(A_2B_2)=\frac(B_1C_1)(B_2C_2)$ and $\angle B_1 = \angle B_2$
or
$\frac(B_1C_1)(B_2C_2)=\frac(C_1A_1)(C_2A_2)$ and $\angle C_1 = \angle C_2$

Do not confuse similar triangles with equal triangles. Equal triangles have equal corresponding side lengths. Therefore, for congruent triangles:

$\frac(A_1B_1)(A_2B_2)=\frac(A_1C_1)(A_2C_2)=\frac(B_1C_1)(B_2C_2)=1$

It follows from this that all equal triangles are similar. However, not all similar triangles are equal.

Although the above notation shows that to find out whether two triangles are similar or not, we must know the values ​​of the three angles or the lengths of the three sides of each triangle, to solve problems with similar triangles it is enough to know any three of the values ​​​​mentioned above for each triangle. These quantities can be in various combinations:

1) three angles of each triangle (you don’t need to know the lengths of the sides of the triangles).

Or at least 2 angles of one triangle must be equal to 2 angles of another triangle.
Since if 2 angles are equal, then the third angle will also be equal. (The value of the third angle is 180 - angle1 - angle2)

2) the lengths of the sides of each triangle (you don’t need to know the angles);

3) the lengths of the two sides and the angle between them.

Next we will look at solving some problems with similar triangles. We will first look at problems that can be solved by directly using the above rules, and then discuss some practical problems, which are solved using the method of similar triangles.

Practice problems with similar triangles

Example #1: Show that the two triangles in the figure below are similar.

Solution:
Since the lengths of the sides of both triangles are known, the second rule can be applied here:

$\frac(PQ)(AB)=\frac(6)(2)=3$ $\frac(QR)(CB)=\frac(12)(4)=3$ $\frac(PR)(AC )=\frac(15)(5)=3$

Example #2: Show that two given triangles are similar and determine the lengths of the sides PQ And PR.

Solution:
∠A = ∠P And ∠B = ∠Q, ∠C = ∠R(since ∠C = 180 - ∠A - ∠B and ∠R = 180 - ∠P - ∠Q)

It follows from this that the triangles ΔABC and ΔPQR are similar. Hence:
$\frac(AB)(PQ)=\frac(BC)(QR)=\frac(AC)(PR)$

$\frac(BC)(QR)=\frac(6)(12)=\frac(AB)(PQ)=\frac(4)(PQ) \Rightarrow PQ=\frac(4\times12)(6) = 8$ and
$\frac(BC)(QR)=\frac(6)(12)=\frac(AC)(PR)=\frac(7)(PR) \Rightarrow PR=\frac(7\times12)(6) = 14$

Example #3: Determine the length AB in this triangle.

Solution:

∠ABC = ∠ADE, ∠ACB = ∠AED And ∠A general => triangles ΔABC And ΔADE are similar.

$\frac(BC)(DE) = \frac(3)(6) = \frac(AB)(AD) = \frac(AB)(AB + BD) = \frac(AB)(AB + 4) = \frac(1)(2) \Rightarrow 2\times AB = AB + 4 \Rightarrow AB = 4$

Example #4: Determine length AD (x) geometric figure in the picture.

Triangles ΔABC and ΔCDE are similar because AB || DE and they have a common upper corner C.
We see that one triangle is a scaled version of the other. However, we need to prove this mathematically.

AB || DE, CD || AC and BC || E.C.
∠BAC = ∠EDC and ∠ABC = ∠DEC

Based on the above and taking into account the presence of a common angle C, we can claim that triangles ΔABC and ΔCDE are similar.

Hence:
$\frac(DE)(AB) = \frac(7)(11) = \frac(CD)(CA) = \frac(15)(CA) \Rightarrow CA = \frac(15 \times 11)(7 ) = 23.57$
x = AC - DC = 23.57 - 15 = 8.57

Practical examples

Example #5: The factory uses an inclined conveyor belt to transport products from level 1 to level 2, which is 3 meters higher than level 1, as shown in the figure. The inclined conveyor is serviced from one end to level 1 and from the other end to a workplace located at a distance of 8 meters from the level 1 operating point.

The factory wants to upgrade the conveyor to access the new level, which is 9 meters above level 1, while maintaining the conveyor's inclination angle.

Determine the distance at which the new work station must be installed to ensure that the conveyor will operate at its new end at level 2. Also calculate the additional distance the product will travel when moving to the new level.

Solution:

First, let's label each intersection point with a specific letter, as shown in the figure.

Based on the reasoning given above in the previous examples, we can conclude that the triangles ΔABC and ΔADE are similar. Hence,

$\frac(DE)(BC) = \frac(3)(9) = \frac(AD)(AB) = \frac(8)(AB) \Rightarrow AB = \frac(8 \times 9)(3 ) = 24 m$
x = AB - 8 = 24 - 8 = 16 m

Thus, the new point must be installed at a distance of 16 meters from the existing point.

And since the structure consists of right triangles, we can calculate the distance of movement of the product as follows:

$AE = \sqrt(AD^2 + DE^2) = \sqrt(8^2 + 3^2) = 8.54 m$

Similarly, $AC = \sqrt(AB^2 + BC^2) = \sqrt(24^2 + 9^2) = 25.63 m$
which is the distance that the product currently travels when it reaches the existing level.

y = AC - AE = 25.63 - 8.54 = 17.09 m
this is the additional distance that the product must travel to reach a new level.

Example #6: Steve wants to visit his friend who recently moved to new house. The road map to Steve and his friend's house, along with the distances known to Steve, is shown in the figure. Help Steve get to his friend's house in the shortest possible way.

Solution:

The road map can be represented geometrically in the following form, as shown in the figure.

We see that triangles ΔABC and ΔCDE are similar, therefore:
$\frac(AB)(DE) = \frac(BC)(CD) = \frac(AC)(CE)$

The problem statement states that:

AB = 15 km, AC = 13.13 km, CD = 4.41 km and DE = 5 km

Using this information we can calculate the following distances:

$BC = \frac(AB \times CD)(DE) = \frac(15 \times 4.41)(5) = 13.23 km$
$CE = \frac(AC \times CD)(BC) = \frac(13.13 \times 4.41)(13.23) = 4.38 km$

Steve can get to his friend's house using the following routes:

A -> B -> C -> E -> G, total distance is 7.5+13.23+4.38+2.5=27.61 km

F -> B -> C -> D -> G, total distance is 7.5+13.23+4.41+2.5=27.64 km

F -> A -> C -> E -> G, total distance is 7.5+13.13+4.38+2.5=27.51 km

F -> A -> C -> D -> G, total distance is 7.5+13.13+4.41+2.5=27.54 km

Therefore, route No. 3 is the shortest and can be offered to Steve.

Example 7:
Trisha wants to measure the height of the house, but she doesn't have the right tools. She noticed that there was a tree growing in front of the house and decided to use her resourcefulness and knowledge of geometry acquired at school to determine the height of the building. She measured the distance from the tree to the house, the result was 30 m. She then stood in front of the tree and began to move back until the top edge of the building became visible above the top of the tree. Trisha marked this place and measured the distance from it to the tree. This distance was 5 m.

The height of the tree is 2.8 m, and the height of Trisha's eye level is 1.6 m. Help Trisha determine the height of the building.

Solution:

The geometric representation of the problem is shown in the figure.

First we use the similarity of triangles ΔABC and ΔADE.

$\frac(BC)(DE) = \frac(1.6)(2.8) = \frac(AC)(AE) = \frac(AC)(5 + AC) \Rightarrow 2.8 \times AC = 1.6 \times (5 + AC) = 8 + 1.6 \times AC$

$(2.8 - 1.6) \times AC = 8 \Rightarrow AC = \frac(8)(1.2) = 6.67$

We can then use the similarity of triangles ΔACB and ΔAFG or ΔADE and ΔAFG. Let's choose the first option.

$\frac(BC)(FG) = \frac(1.6)(H) = \frac(AC)(AG) = \frac(6.67)(6.67 + 5 + 30) = 0.16 \Rightarrow H = \frac(1.6 )(0.16) = 10 m$