The sum of the angles of the triangle. The theorem on the sum of the angles of a triangle

A triangle is a polygon with three sides (three corners). Most often, the sides are denoted by small letters corresponding to capital letters denoting opposite vertices. In this article, we will get acquainted with the types of these geometric figures, a theorem that determines what the sum of the angles of a triangle is equal to.

Kinds of angles

There are the following types of polygon with three vertices:

• Angular, with all corners sharp;
• Rectangular, having one right angle, while the sides forming it are called legs, and the side that is placed opposite to the right angle is called the hypotenuse;
• Obtuse, when one corner is blunt ;
• Isosceles, in which the two sides are equal, and they are called lateral, and the third is the base of the triangle;
• Equilateral, having all three equal sides.

Properties

Allocate the basic properties that are characteristic for each type of triangle:

• Opposite the larger side there is always a larger angle, and vice versa;
• Opposite the equal sides are equal angles, and vice versa;
• Every triangle has two sharp angles;
• The outer angle is larger than any internal angle not adjacent to it;
• The sum of any two angles is always less than 180 degrees;
• The outer angle is equal to the sum of the remaining two angles, which do not interfere with it.

The theorem on the sum of the angles of a triangle

The theorem states that if you add all the angles of a given geometric figure that is located on the Euclidean plane, then their sum will be 180 degrees. Let us try to prove this theorem.

Let's have an arbitrary triangle with the vertices of the CMN. Through the vertex M we draw a straight line parallel to the straight line KN (this straight line is also called the Euclidean line). On it we mark the point A in such a way that the points K and A are located on opposite sides of the straight line MN. We obtain equal angles AMN and CNM, which, like internal ones, lie in a cross and form a secant MN together with straight lines KN and MA, which are parallel. It follows that the sum of the angles of the triangle located at the vertices of M and H is equal to the size of the angle of the MRA. All three angles constitute a sum that is equal to the sum of the angles of the MRA and MKN. Since these angles are internal one-sided with respect to parallel lines KN and MA with secant CM, their sum is 180 degrees. The theorem is proved.

Consequence

From the above theorem, the following corollary follows: every triangle has two acute angles. To prove this, let's assume that the given geometric figure has only one acute angle. It can also be assumed that none of the corners are sharp. In this case, there must be at least two angles, the value of which is equal to or greater than 90 degrees. But then the sum of the angles will be more than 180 degrees. And this can not be, because according to the theorem the sum of the angles of the triangle is 180 ° - no more and no less. That's what it was necessary to prove.

The property of external angles

What is the sum of the angles of the triangle that are external? The answer to this question can be obtained by applying one of two methods. The first is that it is necessary to find the sum of the angles, which are taken one at each vertex, that is, three corners. The second implies that we need to find the sum of all six corners at the vertices. First, we'll figure it out with the first option. So, the triangle contains six outer corners - two at each vertex. Each pair has equal angles because they are vertical:

∟1 = ∟4, ∟2 = ∟5, ∟3 = ∟6.

In addition, it is known that the outer angle of the triangle is equal to the sum of two internal ones that do not interfere with it. Consequently,

∟1 = ∟А + ∟С, ∟2 = ∟А + ∟В, ∟3 = ∟В + ∟С.

From this it turns out that the sum of the external angles, which are taken one at each vertex, will be equal to:

∟1 + ∟2 + ∟3 = ∟A + ∟С + ∟А + ∟В + ∟В + ∟С = 2 х (∟А + ∟В + ∟С).

Given that the sum of the angles is 180 degrees, we can state that ∟A + ∟B + ∟C = 180 °. And this means that ∟1 + ∟2 + ∟3 = 2 × 180 ° = 360 °. If the second option is applied, then the sum of the six corners will be, respectively, twice as large. That is, the sum of the outer corners of the triangle will be:

∟1 + ∟2 + ∟3 + ∟4 + ∟5 + ∟6 = 2 x (∟1 + ∟2 + ∟2) = 720 °.

Right triangle

What is the sum of the angles of a right triangle that are sharp? The answer to this question, again, follows from the theorem, which asserts that the angles in the triangle in the sum are 180 degrees. And our statement (property) sounds like this: in a rectangular triangle, sharp angles in the sum give 90 degrees. Let us prove its truthfulness. Let us give a triangle of CMN, for which ∟H = 90 °. It is necessary to prove that ∟K + ∟M = 90 °.

Thus, according to the theorem on the sum of the angles ∟K + ∟M + ∟H = 180 °. In our condition, it is said that ∟H = 90 °. So it turns out, ∟K + ∟M + 90 ° = 180 °. That is, ∟K + ∟M = 180 ° - 90 ° = 90 °. This is what we should prove.

In addition to the above-described properties of a right-angled triangle, you can add the following:

• The angles that lie against the legs are sharp;
• The hypotenuse is triangular more than any of the legs;
• The sum of the legs is greater than the hypotenuse;
• The catheter of the triangle, which lies opposite the angle of 30 degrees, is half the size of the hypotenuse, that is, equal to its half.

As another property of this geometric figure, one can distinguish the Pythagorean theorem. She claims that in the triangle with an angle of 90 degrees (rectangular), the sum of the squares of the legs is equal to the square of the hypotenuse.

The sum of the angles of an isosceles triangle

Earlier we said that an isosceles is a polygon with three vertices containing two equal sides. It is known such property of the given geometrical figure: angles at its basis are equal. Let us prove this.

Take the triangle CMN, which is isosceles, the CN is its base. We are required to prove that ∟K = ∟H. So, let's say that MA is the bisector of our triangle CMN. The triangle MKA with the first sign of equality is equal to the triangle MNA. Namely, by condition, it is given that CM = NM, MA is a common side, ∟1 = ∟2, since MA is a bisector. Using the fact of equality of these two triangles, we can assert that ∟K = ∟H. Hence, the theorem is proved.

But we are interested in the sum of the angles of a triangle (isosceles). Since in this respect it does not have its own singularities, we start from the theorem considered earlier. That is, we can state that ∟K + ∟M + ∟H = 180 °, or 2 × ∟K + ∟M = 180 ° (since ∟K = ∟H). We shall not prove this property, since the theorem on the sum of the angles of the triangle itself was proved earlier.

In addition to the properties considered about the angles of a triangle, such important statements also hold:

• In an isosceles triangle, the height that was lowered to the base is simultaneously a median, a bisector of the angle that lies between equal sides, and also the axis of symmetry of its base;
• Medians (bisectrixes, heights), which are drawn to the sides of such a geometric figure, are equal.

Equilateral triangle

It is also called right, it is the triangle, in which all sides are equal. And therefore angles are also equal. Each of them is 60 degrees. Let us prove this property.

Suppose that we have a triangle of CMN. We know that KM = HM = KH. And this means that, according to the property of the angles located at the base in an isosceles triangle, ∟K = ∟M = ∟H. Since according to the theorem the sum of the angles of the triangle ∟K + ∟M + ∟H = 180 °, then 3 × ∟K = 180 ° or ∟K = 60 °, ∟M = 60 °, ∟H = 60 °. Thus, the assertion is proved. As can be seen from the above proof on the basis of the theorem, the sum of the angles of an equilateral triangle, like the sum of the angles of any other triangle, is 180 degrees. It is not necessary to prove this theorem again.

There are also such properties characteristic of an equilateral triangle:

• Median, bisector, height in such a geometric figure coincide, and their length is calculated as (а х √3): 2;
• If we describe a circle around a given polygon, its radius will be (a x √ 3): 3;
• If we include a circle in an equilateral triangle, its radius will be (а х √3): 6;
• The area of this geometric figure is calculated by the formula: (a2 x √3): 4.

The obtuse triangle

According to the definition of an obtuse triangle, one of its angles is in the range from 90 to 180 degrees. But given that the other two corners of this geometric figure are sharp, we can conclude that they do not exceed 90 degrees. Consequently, the theorem on the sum of the angles of a triangle works when calculating the sum of the angles in an obtuse triangle. It turns out, we can safely assert, relying on the above theorem, that the sum of the angles of an obtuse triangle is 180 degrees. Again, this theorem does not need re-proof.