1 definition and properties of an isosceles triangle. Isosceles triangle

A triangle in which two sides are equal to each other is called isosceles. These sides are called lateral, and the third side is called the base. In this article we will tell you about the properties of an isosceles triangle.

Theorem 1

The angles near the base of an isosceles triangle are equal to each other

Proof of the theorem.

Let's say we have an isosceles triangle ABC whose base is AB. Let's look at triangle BAC. These triangles, by the first sign, are equal to each other. This is true, because BC = AC, AC = BC, angle ACB = angle ACB. It follows that angle BAC = angle ABC, because these are the corresponding angles of our equal triangles. Here is the property of the angles of an isosceles triangle.

Theorem 2

The median in an isosceles triangle, which is drawn to its base, is also the height and bisector

Proof of the theorem.

Let's say we have an isosceles triangle ABC, the base of which is AB, and CD is the median that we drew to its base. In triangles ACD and BCD, angle CAD = angle CBD, as the corresponding angles at the base of an isosceles triangle (Theorem 1). And side AC = side BC (by definition of an isosceles triangle). Side AD = side BD, because point D divides segment AB into equal parts. It follows that triangle ACD = triangle BCD.

From the equality of these triangles we have the equality of the corresponding angles. That is, angle ACD = angle BCD and angle ADC = angle BDC. From equality 1 it follows that CD is a bisector. And angle ADC and angle BDC are adjacent angles, and from equality 2 it follows that they are both right angles. It turns out that CD is the height of the triangle. This is the property of the median of an isosceles triangle.

And now a little about the signs of an isosceles triangle.

Theorem 3

If two angles in a triangle are equal to each other, then the triangle is isosceles

Proof of the theorem.

Let's say we have a triangle ABC in which angle CAB = angle CBA. Triangle ABC = triangle BAC according to the second criterion of equality between triangles. This is true, because AB = BA; angle CBA = angle CAB, angle CAB = angle CBA. From this equality of triangles we have the equality of the corresponding sides of the triangle - AC = BC. Then it turns out that triangle ABC is isosceles.

Theorem 4

If in any triangle its median is also its altitude, then such a triangle is isosceles

Proof of the theorem.

In triangle ABC we will draw the median CD. It will also be the height. Right triangle ACD = right triangle BCD, since leg CD is common to them, and leg AD = leg BD. It follows from this that their hypotenuses are equal to each other, like corresponding parts of equal triangles. This means that AB = BC.

Theorem 5

If three sides of a triangle are equal to three sides of another triangle, then these triangles are congruent

Proof of the theorem.

Suppose we have a triangle ABC and a triangle A1B1C1 such that the sides AB = A1B1, AC = A1C1, BC = B1C1. Let's consider the proof of this theorem by contradiction.

Let's assume that these triangles are not equal to each other. From here we have that angle BAC is not equal to angle B1A1C1, angle ABC is not equal to angle A1B1C1, angle ACB is not equal to angle A1C1B1 at the same time. Otherwise, these triangles would be equal according to the criteria discussed above.

Let's assume that triangle A1B1C2 = triangle ABC. In a triangle, vertex C2 lies with vertex C1 relative to straight line A1B1 in the same half-plane. We assumed that the vertices C2 and C1 do not coincide. Let's assume that point D is the middle of the segment C1C2. So we have isosceles triangles B1C1C2 and A1C1C2, which have a common base C1C2. It turns out that their medians B1D and A1D are also their heights. This means that straight line B1D and straight line A1D are perpendicular to straight line C1C2.

B1D and A1D have different points B1 and A1, and accordingly cannot coincide. But through point D of line C1C2 we can draw only one line perpendicular to it. We have a contradiction.

Now you know what the properties of an isosceles triangle are!

In which two sides are equal in length. Equal sides are called lateral, and the last unequal side is called the base. By definition, a regular triangle is also isosceles, but the converse is not true.

Terminology

If a triangle has two equal sides, then these sides are called sides, and the third side is called the base. The angle formed by the sides is called vertex angle, and angles, one of whose sides is the base, are called corners at the base.

Properties

• Angles opposite equal sides of an isosceles triangle are equal to each other. The bisectors, medians and altitudes drawn from these angles are also equal.
• The bisector, median, height and perpendicular bisector drawn to the base coincide with each other. The centers of the inscribed and circumscribed circles lie on this line.

Let a- the length of two equal sides of an isosceles triangle, b- length of the third side, h- height of an isosceles triangle

• $a\; =\; \backslash frac\; b\; (2\; \backslash cos\; \backslash alpha)$(a corollary of the cosine theorem);
• $b\; =\; a\; \backslash sqrt\; (2\; (1\; -\; \backslash cos\; \backslash beta))$(a corollary of the cosine theorem);
• $b\; =\; 2a\; \backslash sin\; \backslash frac\; \backslash beta\; 2$;
• $b\; =\; 2a\backslash cos\backslash alpha$(projection theorem)

The radius of the incircle can be expressed in six ways, depending on which two parameters of the isosceles triangle are known:

• $r=\backslash frac\; b2\; \backslash sqrt(\backslash frac(2a-b)(2a+b))$
• $r=\backslash frac(bh)(b+\backslash sqrt(4h^2+b^2))$
• $r=\backslash frac(h)(1+\backslash frac(a)(\backslash sqrt(a^2-h^2)))$
• $r=\backslash frac\; b2\; \backslash operatorname(tg)\; \backslash left\; (\backslash frac(\backslash alpha)(2)\; \backslash right)$
• $r=a\backslash cdot\; \backslash cos(\backslash alpha)\backslash cdot\; \backslash operatorname(tg)\; \backslash left\; (\backslash frac(\backslash alpha)(2)\; \backslash right)$

Angles can be expressed in the following ways:

• $\backslash alpha\; =\; \backslash frac\; (\backslash pi\; -\; \backslash beta)\; 2;$
• $\backslash beta\; =\; \backslash pi\; -\; 2\backslash alpha;$
• $\backslash alpha\; =\; \backslash arcsin\; \backslash frac\; a\; (2R),\; \backslash beta\; =\; \backslash arcsin\; \backslash frac\; b\; (2R)$(sine theorem).
• The angle can also be found without $(\backslash pi)$ And $R$. A triangle is divided in half by its median, and received The angles of two equal right triangles are calculated:

Perimeter An isosceles triangle is found in the following ways:

• $P\; =\; 2a\; +\; b$(a-priory);
• $P\; =\; 2R\; (2\; \backslash sin\; \backslash alpha\; +\; \backslash sin\; \backslash beta)$(a corollary of the sine theorem).

Square the triangle is found in the following ways:

$S\; =\; \backslash frac\; 1\; 2bh;$

$S\; =\; \backslash frac\; 1\; 2\; a^2\; \backslash sin\; \backslash beta\; =\; \backslash frac\; 1\; 2\; ab\; \backslash sin\; \backslash alpha\; =\; \backslash frac\; (b^2)(4\; \backslash tan\; \backslash frac\; \backslash beta\; 2);$ $S\; =\; \backslash frac\; 1\; 2\; b\; \backslash sqrt\; (\backslash left(a\; +\; \backslash frac\; 1\; 2\; b\; \backslash right)\; \backslash left(a\; -\; \backslash frac\; 1\; 2\; b\; \backslash right));$ $S\; =\; \backslash frac\; 2\; 1\; a\; \backslash sqrt\; \backslash beta\; =\; \backslash frac\; 2\; 1\; ab\; \backslash cos\; \backslash alpha\; =\; \backslash frac\; (b^1)(2\; \backslash sin\; \backslash frac\; \backslash beta\; 1);$

See also

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Notes

Excerpt characterizing the Isosceles triangle

The properties of an isosceles triangle are expressed by the following theorems.

Theorem 1. In an isosceles triangle, the angles at the base are equal.

Theorem 2. In an isosceles triangle, the bisector drawn to the base is the median and altitude.

Theorem 3. In an isosceles triangle, the median drawn to the base is the bisector and the altitude.

Theorem 4. In an isosceles triangle, the altitude drawn to the base is the bisector and the median.

Let us prove one of them, for example Theorem 2.5.

Proof. Let us consider an isosceles triangle ABC with base BC and prove that ∠ B = ∠ C. Let AD be the bisector of triangle ABC (Fig. 1). Triangles ABD and ACD are equal according to the first sign of equality of triangles (AB = AC by condition, AD is a common side, ∠ 1 = ∠ 2, since AD ​​is a bisector). From the equality of these triangles it follows that ∠ B = ∠ C. The theorem is proven.

Using Theorem 1, the following theorem is established.

Theorem 5. The third criterion for the equality of triangles. If three sides of one triangle are respectively equal to three sides of another triangle, then such triangles are congruent (Fig. 2).

Comment. The sentences established in examples 1 and 2 express the properties of the perpendicular bisector of a segment. From these proposals it follows that perpendicular bisectors to the sides of a triangle intersect at one point.

Example 1. Prove that a point in the plane equidistant from the ends of a segment lies on the perpendicular bisector to this segment.

Solution. Let point M be equidistant from the ends of segment AB (Fig. 3), i.e. AM = BM.

Then Δ AMV is isosceles. Let us draw a straight line p through the point M and the midpoint O of the segment AB. By construction, the segment MO is the median of the isosceles triangle AMB, and therefore (Theorem 3), and the height, i.e., the straight line MO, is the perpendicular bisector to the segment AB.

Example 2. Prove that each point of the perpendicular bisector to a segment is equidistant from its ends.

Solution. Let p be the perpendicular bisector to segment AB and point O be the midpoint of segment AB (see Fig. 3).

Consider an arbitrary point M lying on the straight line p. Let's draw segments AM and BM. Triangles AOM and BOM are equal, since their angles at vertex O are right, leg OM is common, and leg OA is equal to leg OB by condition. From the equality of triangles AOM and BOM it follows that AM = BM.

Example 3. In triangle ABC (see Fig. 4) AB = 10 cm, BC = 9 cm, AC = 7 cm; in triangle DEF DE = 7 cm, EF = 10 cm, FD = 9 cm.

Compare triangles ABC and DEF. Find the corresponding equal angles.

Solution. These triangles are equal according to the third criterion. Correspondingly, equal angles: A and E (lie opposite equal sides BC and FD), B and F (lie opposite equal sides AC and DE), C and D (lie opposite equal sides AB and EF).

Example 4. In Figure 5, AB = DC, BC = AD, ∠B = 100°.

Find angle D.

Solution. Consider triangles ABC and ADC. They are equal according to the third criterion (AB = DC, BC = AD by condition and side AC is common). From the equality of these triangles it follows that ∠ B = ∠ D, but angle B is equal to 100°, which means angle D is equal to 100°.

Example 5. In an isosceles triangle ABC with base AC, the exterior angle at vertex C is 123°. Find the size of angle ABC. Give your answer in degrees.

Video solution.

Among all triangles, there are two special types: right triangles and isosceles triangles. Why are these types of triangles so special? Well, firstly, such triangles extremely often turn out to be the main characters in the problems of the Unified State Exam in the first part. And secondly, problems about right and isosceles triangles are much easier to solve than other geometry problems. You just need to know a few rules and properties. All the most interesting things about right triangles are discussed in, but now let’s look at isosceles triangles. And first of all, what is an isosceles triangle? Or, as mathematicians say, what is the definition of an isosceles triangle?

See what it looks like:

Like a right triangle, an isosceles triangle has special names for its sides. Two equal sides are called sides, and the third party - basis.

And again pay attention to the picture:

It could, of course, be like this:

So be careful: lateral side - one of two equal sides in an isosceles triangle, and the basis is a third party.

Why is an isosceles triangle so good? To understand this, let's draw the height to the base. Do you remember what height is?

What happened? From one isosceles triangle we get two rectangular ones.

This is already good, but this will happen in any, even the most “oblique” triangle.

How is the picture different for an isosceles triangle? Look again:

Well, firstly, of course, it is not enough for these strange mathematicians to just see - they must certainly prove. Otherwise, suddenly these triangles are slightly different, but we will consider them the same.

But don't worry: in this case, proving is almost as easy as seeing.

Shall we start? Look closely, we have:

And that means! Why? Yes, we will simply find and, and from the Pythagorean theorem (remembering at the same time that)

Are you sure? Well, now we have

And on three sides - the easiest (third) sign of equality of triangles.

Well, our isosceles triangle has divided into two identical rectangular ones.

See how interesting it is? It turned out that:

How do mathematicians usually talk about this? Let's go in order:

(Remember here that the median is a line drawn from a vertex that divides the side in half, and the bisector is the angle.)

Well, here we discussed what good things can be seen if given an isosceles triangle. We deduced that in an isosceles triangle the angles at the base are equal, and the height, bisector and median drawn to the base coincide.

And now another question arises: how to recognize an isosceles triangle? That is, as mathematicians say, what are signs of an isosceles triangle?

And it turns out that you just need to “turn” all the statements the other way around. This, of course, does not always happen, but an isosceles triangle is still a great thing! What happens after the “turnover”?

Well, look:
If the height and median coincide, then:

If the height and bisector coincide, then:


If the bisector and the median coincide, then:

Well, don’t forget and use:

• If you are given an isosceles triangular triangle, feel free to draw the height, get two right triangles and solve the problem about a right triangle.
• If given that two angles are equal, then a triangle exactly isosceles and you can draw the height and….(The House That Jack Built…).
• If it turns out that the height is divided in half, then the triangle is isosceles with all the ensuing bonuses.
• If it turns out that the height divides the angle between the floors - it is also isosceles!
• If a bisector divides a side in half or a median divides an angle, then this also happens only in an isosceles triangle

Let's see what it looks like in tasks.

Problem 1(the simplest)

In a triangle, sides and are equal, a. Find.

We decide:

First the drawing.

What is the basis here? Certainly, .

Let's remember what if, then and.

Updated drawing:

Let's denote by. What is the sum of the angles of a triangle? ?

We use:

That's answer: .

Not difficult, right? I didn't even have to adjust the height.

Problem 2(Also not very tricky, but we need to repeat the topic)

In a triangle, . Find.

We decide:

The triangle is isosceles! We draw the height (this is the trick with which everything will be decided now).

Now let’s “cross out from life”, let’s just look at it.

So, we have:

Let's remember the table values ​​of cosines (well, or look at the cheat sheet...)

All that remains is to find: .

Answer: .

Note that we here Very required knowledge regarding right triangles and “tabular” sines and cosines. Very often this happens: the topics , “Isosceles triangle” and in problems go together, but are not very friendly with other topics.

Isosceles triangle. Average level.

These two equal sides are called sides, A the third side is the base of an isosceles triangle.

Look at the picture: and - the sides, - the base of the isosceles triangle.

Let's use one picture to understand why this happens. Let's draw a height from a point.

This means that all corresponding elements are equal.

All! In one fell swoop (height) they proved all the statements at once.

And remember: to solve a problem about an isosceles triangle, it is often very useful to lower the height to the base of the isosceles triangle and divide it into two equal right triangles.

Signs of an isosceles triangle

The converse statements are also true:

Almost all of these statements can again be proven “in one fell swoop.”

1. So, let in turned out to be equal and.

Let's check the height. Then

2. a) Now let in some triangle height and bisector coincide.

2. b) And if the height and median coincide? Everything is almost the same, no more complicated!

- on two sides

2. c) But if there is no height, which is lowered to the base of an isosceles triangle, then there are no initially right triangles. Badly!

But there is a way out - read it in the next level of the theory, since the proof here is more complicated, but for now just remember that if the median and bisector coincide, then the triangle will also turn out to be isosceles, and the height will still coincide with these bisector and median.

Let's summarize:

1. If the triangle is isosceles, then the angles at the base are equal, and the altitude, bisector and median drawn to the base coincide.
2. If in some triangle there are two equal angles, or some two of the three lines (bisector, median, altitude) coincide, then such a triangle is isosceles.

Isosceles triangle. Brief description and basic formulas

An isosceles triangle is a triangle that has two equal sides.

Signs of an isosceles triangle:

1. If in a certain triangle two angles are equal, then it is isosceles.
2. If in some triangle they coincide:
   A) height and bisector or
   b) height and median or
   V) median and bisector,
   drawn to one side, then such a triangle is isosceles.