Today will be a very easy lesson. We will consider just one object - the angle bisector - and prove its most important property, which will be very useful to us in the future.

Just don’t relax: sometimes students who want to get a high score on the same Unified State Exam or Unified State Exam cannot even accurately formulate the definition of a bisector in the first lesson.

And instead of doing really interesting tasks, we waste time on such simple things. So read, watch, and adopt it. :)

To begin with, a slightly strange question: what is an angle? That's right: an angle is simply two rays emanating from the same point. For example:

Examples of angles: acute, obtuse and right

As you can see from the picture, angles can be acute, obtuse, straight - it doesn’t matter now. Often, for convenience, an additional point is marked on each ray and they say that in front of us is the angle $AOB$ (written as $\angle AOB$).

Captain Obviousness seems to be hinting that in addition to the rays $OA$ and $OB$, it is always possible to draw a bunch of more rays from the point $O$. But among them there will be one special one - he is called a bisector.

Definition. The bisector of an angle is the ray that comes out from the vertex of that angle and bisects the angle.

For the above angles, the bisectors will look like this:

Examples of bisectors for acute, obtuse and right angles

Since in real drawings it is not always obvious that a certain ray (in our case it is the $OM$ ray) splits the original angle into two equal ones, in geometry it is customary to mark equal angles with the same number of arcs (in our drawing this is 1 arc for an acute angle, two for obtuse, three for straight).

Okay, we've sorted out the definition. Now you need to understand what properties the bisector has.

The main property of an angle bisector

In fact, the bisector has a lot of properties. And we will definitely look at them in the next lesson. But there is one trick that you need to understand right now:

Theorem. The bisector of an angle is the locus of points equidistant from the sides of a given angle.

Translated from mathematical into Russian, this means two facts at once:

1. Any point lying on the bisector of a certain angle is at the same distance from the sides of this angle.
2. And vice versa: if a point lies at the same distance from the sides of a given angle, then it is guaranteed to lie on the bisector of this angle.

Before proving these statements, let's clarify one point: what, exactly, is called the distance from a point to the side of an angle? Here the good old determination of the distance from a point to a line will help us:

Definition. The distance from a point to a line is the length of the perpendicular drawn from a given point to this line.

For example, consider a line $l$ and a point $A$ that does not lie on this line. Let us draw a perpendicular to $AH$, where $H\in l$. Then the length of this perpendicular will be the distance from point $A$ to straight line $l$.

Graphic representation of the distance from a point to a line

Since an angle is simply two rays, and each ray is a piece of a straight line, it is easy to determine the distance from a point to the sides of an angle. These are just two perpendiculars:

Determine the distance from the point to the sides of the angle

That's all! Now we know what a distance is and what a bisector is. Therefore, we can prove the main property.

As promised, we will split the proof into two parts:

1. The distances from the point on the bisector to the sides of the angle are the same

Consider an arbitrary angle with vertex $O$ and bisector $OM$:

Let us prove that this very point $M$ is at the same distance from the sides of the angle.

Proof. Let us draw perpendiculars from point $M$ to the sides of the angle. Let's call them $M((H)_(1))$ and $M((H)_(2))$:

Draw perpendiculars to the sides of the angle

We obtained two right triangles: $\vartriangle OM((H)_(1))$ and $\vartriangle OM((H)_(2))$. They have a common hypotenuse $OM$ and equal angles:

1. $\angle MO((H)_(1))=\angle MO((H)_(2))$ by condition (since $OM$ is a bisector);
2. $\angle M((H)_(1))O=\angle M((H)_(2))O=90()^\circ $ by construction;
3. $\angle OM((H)_(1))=\angle OM((H)_(2))=90()^\circ -\angle MO((H)_(1))$, since the sum The acute angles of a right triangle are always 90 degrees.

Consequently, the triangles are equal in side and two adjacent angles (see signs of equality of triangles). Therefore, in particular, $M((H)_(2))=M((H)_(1))$, i.e. the distances from point $O$ to the sides of the angle are indeed equal. Q.E.D.:)

2. If the distances are equal, then the point lies on the bisector

Now the situation is reversed. Let an angle $O$ be given and a point $M$ equidistant from the sides of this angle:

Let us prove that the ray $OM$ is a bisector, i.e. $\angle MO((H)_(1))=\angle MO((H)_(2))$.

Proof. First, let’s draw this very ray $OM$, otherwise there will be nothing to prove:

Conducted $OM$ beam inside the corner

Again we get two right triangles: $\vartriangle OM((H)_(1))$ and $\vartriangle OM((H)_(2))$. Obviously they are equal because:

1. Hypotenuse $OM$ - general;
2. Legs $M((H)_(1))=M((H)_(2))$ by condition (after all, the point $M$ is equidistant from the sides of the angle);
3. The remaining legs are also equal, because by the Pythagorean theorem $OH_(1)^(2)=OH_(2)^(2)=O((M)^(2))-MH_(1)^(2)$.

Therefore, the triangles $\vartriangle OM((H)_(1))$ and $\vartriangle OM((H)_(2))$ on three sides. In particular, their angles are equal: $\angle MO((H)_(1))=\angle MO((H)_(2))$. And this just means that $OM$ is a bisector.

To conclude the proof, we mark the resulting equal angles with red arcs:

The bisector splits the angle $\angle ((H)_(1))O((H)_(2))$ into two equal ones

As you can see, nothing complicated. We have proven that the bisector of an angle is the locus of points equidistant to the sides of this angle. :)

Now that we have more or less decided on the terminology, it’s time to move to the next level. In the next lesson we will look at more complex properties of the bisector and learn how to apply them to solve real problems.

Bisector of a triangle – a segment of the bisector of an angle of a triangle, enclosed between the vertex of the triangle and the side opposite to it.

Properties of the bisector

1. The bisector of a triangle bisects the angle.

2. The bisector of an angle of a triangle divides the opposite side in a ratio equal to the ratio of the two adjacent sides ()

3. The bisector points of an angle of a triangle are equidistant from the sides of that angle.

4. The bisectors of the interior angles of a triangle intersect at one point - the center of the circle inscribed in this triangle.

Some formulas related to the bisector of a triangle

(proof of formula – )
, Where
- the length of the bisector drawn to the side,
- sides of the triangle opposite the vertices, respectively,
- the lengths of the segments into which the bisector divides the side,

I invite you to watch video tutorial, which demonstrates the application of all the above properties of the bisector.

Tasks covered in the video:
1. In triangle ABC with sides AB = 2 cm, BC = 3 cm, AC = 3 cm, a bisector VM is drawn. Find the lengths of the segments AM and MC
2. The bisector of the internal angle at vertex A and the bisector of the external angle at vertex C of triangle ABC intersect at point M. Find angle BMC if angle B is 40 degrees, angle C is 80 degrees
3. Find the radius of a circle inscribed in a triangle, considering the sides of square cells equal to 1

You might also be interested in a short video tutorial where one of the properties of the bisector is applied

Do you know what the midpoint of a segment is? Of course you do. What about the center of the circle? Same.

What is the midpoint of an angle?

You can say that this doesn't happen. But why can a segment be divided in half, but an angle cannot? It’s quite possible - just not a dot, but…. line.

Do you remember the joke: a bisector is a rat that runs around the corners and divides the corner in half. So, the real definition of a bisector is very similar to this joke:

Bisector of a triangle- this is the bisector segment of an angle of a triangle connecting the vertex of this angle with a point on the opposite side.

Once upon a time, ancient astronomers and mathematicians discovered many interesting properties of the bisector. This knowledge has greatly simplified people's lives.

The first knowledge that will help with this is...

By the way, do you remember all these terms? Do you remember how they differ from each other? No? Not scary. Let's figure it out now.

• Base of an isosceles triangle- this is the side that is not equal to any other. Look at the picture, which side do you think it is? That's right - this is the side.
• The median is a line drawn from the vertex of a triangle and dividing the opposite side (that's it again) in half. Notice we don't say, "Median of an isosceles triangle." Do you know why? Because a median drawn from a vertex of a triangle bisects the opposite side in ANY triangle.
• Height is a line drawn from the top and perpendicular to the base. You noticed? We are again talking about any triangle, not just an isosceles one. The height in ANY triangle is always perpendicular to the base.

So, have you figured it out? Almost.

To understand even better and forever remember what a bisector, median and height are, you need them compare with each other and understand how they are similar and how they differ from each other.

At the same time, in order to remember better, it is better to describe everything in “human language”.

Then you will easily operate in the language of mathematics, but at first you don’t understand this language and you need to comprehend everything in your own language.

So, how are they similar?

The bisector, the median and the altitude - they all “come out” from the vertex of the triangle and rest on the opposite side and “do something” either with the angle from which they come out, or with the opposite side.

I think it's simple, no?

How are they different?

• The bisector divides the angle from which it emerges in half.
• The median divides the opposite side in half.
• The height is always perpendicular to the opposite side.

That's it. It's easy to understand. And once you understand, you can remember.

Now the next question.

Why, in the case of an isosceles triangle, is the bisector both the median and the altitude?

You can simply look at the figure and make sure that the median divides into two absolutely equal triangles.

That's all! But mathematicians do not like to believe their eyes. They need to prove everything.

Scary word?

Nothing like that - it's simple! Look: both have equal sides and, they generally have a common side and. (- bisector!) And so it turns out that two triangles have two equal sides and an angle between them.

We recall the first sign of equality of triangles (if you don’t remember, look in the topic) and conclude that, and therefore = and.

This is already good - it means it turned out to be the median.

But what is it?

Let's look at the picture - . And we got it. So, too! Finally, hurray! And.

Did you find this proof a bit heavy? Look at the picture - two identical triangles speak for themselves.

In any case, remember firmly:

Now it’s more difficult: we’ll count angle between bisectors in any triangle! Don't be afraid, it's not that tricky. Look at the picture:

Let's count it. Do you remember that the sum of the angles of a triangle is?

Let's apply this amazing fact.

On the one hand, from:

That is.

Now let's look at:

But bisectors, bisectors!

Let's remember about:

Now through the letters

Isn't it surprising?

It turned out that the angle between the bisectors of two angles depends only on the third angle!

Well, we looked at two bisectors. What if there are three of them??!! Will they all intersect at one point?

Or will it be like this?

How do you think? So mathematicians thought and thought and proved:

Isn't that great?

Do you want to know why this happens?

Move to the next level - you are ready to conquer new heights of knowledge about the bisector!

BISECTOR. AVERAGE LEVEL

Do you remember what a bisector is?

A bisector is a line that bisects an angle.

Did you encounter a bisector in the problem? Try to apply one (or sometimes several) of the following amazing properties.

1. Bisector in an isosceles triangle.

Aren't you afraid of the word "theorem"? If you are afraid, then it is in vain. Mathematicians are accustomed to calling a theorem any statement that can somehow be deduced from other, simpler statements.

So, attention, theorem!

Let's prove this theorem, that is, let us understand why this happens? Look at the isosceles.

Let's look at them carefully. And then we will see that

1. - general.

And this means (quickly remember the first sign of equality of triangles!) that.

So what? Do you want to say that? And the fact is that we have not yet looked at the third sides and the remaining angles of these triangles.

Now let's see. Once, then absolutely, and even in addition, .

So it turned out that

1. divided the side in half, that is, it turned out to be the median
2. , which means they are both like (look again at the picture).

So it turned out to be a bisector and a height too!

Hooray! We proved the theorem. But guess what, that's not all. Also faithful converse theorem:

Proof? Are you really interested? Read the next level of theory!

And if you're not interested, then remember firmly:

Why remember this firmly? How can this help? But imagine that you have a task:

Given: .

Find: .

You immediately realize, bisector and, lo and behold, she divided the side in half! (by condition…). If you firmly remember that this happens only in an isosceles triangle, then you draw a conclusion, which means, you write the answer: . Great, right? Of course, not all tasks will be so easy, but knowledge will definitely help!

And now the next property. Ready?

2. The bisector of an angle is the locus of points equidistant from the sides of the angle.

Scared? It's really no big deal. Lazy mathematicians hid four in two lines. So, what does it mean, “Bisector - locus of points"? This means that they are executed immediately twostatements:

1. If a point lies on a bisector, then the distances from it to the sides of the angle are equal.
2. If at some point the distances to the sides of the angle are equal, then this point Necessarily lies on the bisector.

Do you see the difference between statements 1 and 2? If not very much, then remember the Hatter from “Alice in Wonderland”: “So what else will you say, as if “I see what I eat” and “I eat what I see” are the same thing!”

So we need to prove statements 1 and 2, and then the statement: “a bisector is the locus of points equidistant from the sides of an angle” will be proven!

Why is 1 true?

Let's take any point on the bisector and call it .

Let us drop perpendiculars from this point to the sides of the angle.

And now...get ready to remember the signs of equality of right triangles! If you have forgotten them, then take a look at the section.

So...two right triangles: and. They have:

• General hypotenuse.
• (because it is a bisector!)

This means - by angle and hypotenuse. Therefore, the corresponding legs of these triangles are equal! That is.

We proved that the point is equally (or equally) distant from the sides of the angle. Point 1 is dealt with. Now let's move on to point 2.

Why is 2 true?

And let's connect the dots and.

This means that it lies on the bisector!

That's all!

How can all this be applied when solving problems? For example, in problems there is often the following phrase: “A circle touches the sides of an angle...”. Well, you need to find something.

Then you quickly realize that

And you can use equality.

3. Three bisectors in a triangle intersect at one point

From the property of a bisector to be the locus of points equidistant from the sides of an angle, the following statement follows:

How exactly does it come out? But look: two bisectors will definitely intersect, right?

And the third bisector could go like this:

But in reality, everything is much better!

Let's look at the intersection point of two bisectors. Let's call it .

What did we use here both times? Yes paragraph 1, of course! If a point lies on a bisector, then it is equally distant from the sides of the angle.

And so it happened.

But look carefully at these two equalities! After all, it follows from them that and, therefore, .

And now it will come into play point 2: if the distances to the sides of an angle are equal, then the point lies on the bisector...what angle? Look at the picture again:

and are the distances to the sides of the angle, and they are equal, which means the point lies on the bisector of the angle. The third bisector passed through the same point! All three bisectors intersect at one point! And as an additional gift -

Radii inscribed circles.

(To be sure, look at another topic).

Well, now you'll never forget:

The point of intersection of the bisectors of a triangle is the center of the circle inscribed in it.

Let's move on to the next property... Wow, the bisector has many properties, right? And this is great, because the more properties, the more tools for solving bisector problems.

4. Bisector and parallelism, bisectors of adjacent angles

The fact that the bisector divides the angle in half in some cases leads to completely unexpected results. For example,

Case 1

Great, right? Let's understand why this is so.

On the one hand, we draw a bisector!

But, on the other hand, there are angles that lie crosswise (remember the theme).

And now it turns out that; throw out the middle: ! - isosceles!

Case 2

Imagine a triangle (or look at the picture)

Let's continue the side beyond the point. Now we have two angles:

• - internal corner
• - the outer corner is outside, right?

So, and now someone wanted to draw not one, but two bisectors at once: both for and for. What will happen?

Will it work out? rectangular!

Surprisingly, this is exactly the case.

Let's figure it out.

What do you think the amount is?

Of course, - after all, they all together make such an angle that it turns out to be a straight line.

Now remember that and are bisectors and see that inside the angle there is exactly half from the sum of all four angles: and - - that is, exactly. You can also write it as an equation:

So, incredible but true:

The angle between the bisectors of the internal and external angles of a triangle is equal.

Case 3

Do you see that everything is the same here as for the internal and external corners?

Or let's think again why this happens?

Again, as for adjacent corners,

(as corresponding with parallel bases).

And again, they make up exactly half from the sum

Conclusion: If the problem contains bisectors adjacent angles or bisectors relevant angles of a parallelogram or trapezoid, then in this problem certainly a right triangle is involved, or maybe even a whole rectangle.

5. Bisector and opposite side

It turns out that the bisector of an angle of a triangle divides the opposite side not just in some way, but in a special and very interesting way:

That is:

An amazing fact, isn't it?

Now we will prove this fact, but get ready: it will be a little more difficult than before.

Again - exit to “space” - additional formation!

Let's go straight.

For what? We'll see now.

Let's continue the bisector until it intersects with the line.

Is this a familiar picture? Yes, yes, yes, exactly the same as in point 4, case 1 - it turns out that (- bisector)

Lying crosswise

So, that too.

Now let's look at the triangles and.

What can you say about them?

They are similar. Well, yes, their angles are equal as vertical ones. So, in two corners.

Now we have the right to write the relations of the relevant parties.

And now in short notation:

Oh! Reminds me of something, right? Isn't this what we wanted to prove? Yes, yes, exactly that!

You see how great the “spacewalk” proved to be - the construction of an additional straight line - without it nothing would have happened! And so, we have proven that

Now you can safely use it! Let's look at one more property of the bisectors of the angles of a triangle - don't be alarmed, now the hardest part is over - it will be easier.

We get that

Theorem 1:

Theorem 2:

Theorem 3:

Theorem 4:

Theorem 5:

Theorem 6:

Well, the topic is over. If you are reading these lines, it means you are very cool.

Because only 5% of people are able to master something on their own. And if you read to the end, then you are in this 5%!

Now the most important thing.

You have understood the theory on this topic. And, I repeat, this... this is just super! You are already better than the vast majority of your peers.

The problem is that this may not be enough...

For what?

For successfully passing the Unified State Exam, for entering college on a budget and, MOST IMPORTANTLY, for life.

I won’t convince you of anything, I’ll just say one thing...

People who have received a good education earn much more than those who have not received it. This is statistics.

But this is not the main thing.

The main thing is that they are MORE HAPPY (there are such studies). Perhaps because many more opportunities open up before them and life becomes brighter? Don't know...

But think for yourself...

What does it take to be sure to be better than others on the Unified State Exam and ultimately be... happier?

GAIN YOUR HAND BY SOLVING PROBLEMS ON THIS TOPIC.

You won't be asked for theory during the exam.

You will need solve problems against time.

And, if you haven’t solved them (A LOT!), you’ll definitely make a stupid mistake somewhere or simply won’t have time.

It's like in sports - you need to repeat it many times to win for sure.

Find the collection wherever you want, necessarily with solutions, detailed analysis and decide, decide, decide!

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Find problems and solve them!

In this lesson, we will recall the concept of an angle bisector, formulate and prove direct and inverse theorems about the properties of an angle bisector, and generalize them. Let's solve a problem in which, in addition to the facts about the bisector, we apply other geometric facts.

Topic: Circle

Lesson: Properties of an angle bisector. Tasks

The triangle is the central figure of all geometry, and it is jokingly said that it is inexhaustible, like an atom. Its properties are numerous, interesting, entertaining. We look at some of these properties.

Any triangle is, first of all, three angles and three segments (see Fig. 1).

Rice. 1

Consider an angle with vertex A and sides B and C - angle .

In any angle, including the angle of a triangle, you can draw a bisector - that is, a straight line that divides the angle in half (see Fig. 2).

Rice. 2

Let's consider the properties of a point lying on the bisector of an angle (see Fig. 3).

Consider the point M lying on the bisector of the angle.

Recall that the distance from a point to a line is the length of the perpendicular drawn from this point to the line.

Rice. 3

Obviously, if we take a point that does not lie on the bisector, then the distances from this point to the sides of the angle will be different. The distance from point M to the sides of the angle is the same.

Theorem

Each point of the bisector of an undeveloped angle is equidistant from the sides of the angle, that is, the distances from point M to AC and to BC of the sides of the angle are equal.

The angle is given, its bisector is AL, point M lies on the bisector (see Fig. 4).

Prove that .

Rice. 4

Proof:

Consider triangles and . These are right triangles, and they are equal, because they have a common hypotenuse AM, and the angles are equal, since AL is the bisector of the angle. Thus, right triangles are equal in hypotenuse and acute angle, it follows that , which is what needed to be proved. Thus, a point on the bisector of an angle is equidistant from the sides of that angle.

The converse theorem is true.

Theorem

If a point is equidistant from the sides of an undeveloped angle, then it lies on its bisector.

An undeveloped angle is given, point M, such that the distance from it to the sides of the angle is the same.

Prove that point M lies on the bisector of the angle (see Fig. 5).

Rice. 5

Proof:

The distance from a point to a line is the length of the perpendicular. From point M we draw perpendiculars MK to side AB and MR to side AC.

Consider triangles and . These are right triangles, and they are equal, because they have a common hypotenuse AM, the legs MK and MR are equal by condition. Thus, right triangles are equal in hypotenuse and leg. From the equality of triangles follows the equality of the corresponding elements; equal angles lie opposite equal sides, thus, Therefore, point M lies on the bisector of the given angle.

Sometimes the direct and converse theorems are combined as follows:

Theorem

A point is equidistant from the sides of an angle if and only if it lies on the bisector of this angle.

The equidistance of bisector points from the sides of an angle is widely used in various problems.

Problem No. 674 from Atanasyan's textbook, geometry, grades 7-9:

From point M of the bisector of an undeveloped angle, perpendiculars MA and MB are drawn to the sides of this angle (see Fig. 6). Prove that .

Given: angle, bisector OM, perpendiculars MA and MB to the sides of the angle.

Rice. 6

Prove that:

Proof:

According to the direct theorem, point M is equidistant from the sides of the angle, since by condition it lies on its bisector. .

Consider right triangles and (see Fig. 7). They have a common hypotenuse OM, the legs MA and MB are equal, as we proved earlier. Thus, two rectangular

Rice. 7

triangles are equal in leg and hypotenuse. From the equality of triangles follows the equality of their corresponding elements, hence the equality of angles and equality of other legs.

From the equality of legs OA and OB it follows that the triangle is isosceles, and AB is its base. The straight line OM is the bisector of a triangle. According to the property of an isosceles triangle, this bisector is also an altitude, which means that lines OM and AB intersect at right angles, which is what needed to be proven.

So, we examined the direct and converse theorems about the property of a point lying on the bisector of an angle, generalized them and solved the problem using various geometric facts, including this theorem.

Bibliography

1. Alexandrov A.D. and others. Geometry, 8th grade. - M.: Education, 2006.
2. Butuzov V.F., Kadomtsev S.B., Prasolov V.V. Geometry, 8th grade. - M.: Education, 2011.
3. Merzlyak A.G., Polonsky V.B., Yakir S.M. Geometry, 8th grade. - M.: VENTANA-GRAF, 2009.

1. Bymath.net().
2. Oldskola1.narod.ru ().

Homework

1. Atanasyan L.S., Butuzov V.F., Kadomtsev S.B. and others. Geometry, 7-9, No. 676-678, Art. 180.

In this lesson we will look in detail at the properties of points lying on the bisector of an angle and points that lie on the perpendicular bisector to a segment.

Topic: Circle

Lesson: Properties of the bisector of an angle and the perpendicular bisector of a segment

Let's consider the properties of a point lying on the bisector of an angle (see Fig. 1).

Rice. 1

The angle is given, its bisector is AL, point M lies on the bisector.

Theorem:

If point M lies on the bisector of an angle, then it is equidistant from the sides of the angle, that is, the distances from point M to AC and to BC of the sides of the angle are equal.

Proof:

Consider triangles and . These are right triangles and they are equal because... have a common hypotenuse AM, and the angles are equal, since AL is the bisector of the angle. Thus, right triangles are equal in hypotenuse and acute angle, it follows that , which is what needed to be proved. Thus, a point on the bisector of an angle is equidistant from the sides of that angle.

The converse theorem is true.

If a point is equidistant from the sides of an undeveloped angle, then it lies on its bisector.

Rice. 2

An undeveloped angle is given, point M, such that the distance from it to the sides of the angle is the same (see Fig. 2).

Prove that point M lies on the bisector of the angle.

Proof:

The distance from a point to a line is the length of the perpendicular. From point M we draw perpendiculars MK to side AB and MR to side AC.

Consider triangles and . These are right triangles and they are equal because... have a common hypotenuse AM, legs MK and MR are equal by condition. Thus, right triangles are equal in hypotenuse and leg. From the equality of triangles follows the equality of the corresponding elements; equal angles lie opposite equal sides, thus, Therefore, point M lies on the bisector of the given angle.

The direct and converse theorems can be combined.

Theorem

The bisector of an undeveloped angle is the locus of points equidistant from the sides of a given angle.

Theorem

The bisectors AA 1, BB 1, СС 1 of the triangle intersect at one point O (see Fig. 3).

Rice. 3

Proof:

Let us first consider two bisectors BB 1 and CC 1. They intersect, the intersection point O exists. To prove this, let us assume the opposite - even if these bisectors do not intersect, in which case they are parallel. Then straight line BC is a secant, and the sum of the angles , this contradicts the fact that in the entire triangle the sum of the angles is .

So, point O of the intersection of two bisectors exists. Let's consider its properties:

Point O lies on the bisector of the angle, which means it is equidistant from its sides BA and BC. If OK is perpendicular to BC, OL is perpendicular to BA, then the lengths of these perpendiculars are equal - . Also, point O lies on the bisector of the angle and is equidistant from its sides CB and CA, the perpendiculars OM and OK are equal.

We obtained the following equalities:

, that is, all three perpendiculars dropped from point O to the sides of the triangle are equal to each other.

We are interested in the equality of perpendiculars OL and OM. This equality says that point O is equidistant from the sides of the angle, it follows that it lies on its bisector AA 1.

Thus, we have proven that all three bisectors of a triangle intersect at one point.

Let's move on to consider the segment, its perpendicular bisector and the properties of the point that lies on the perpendicular bisector.

A segment AB is given, p is the perpendicular bisector. This means that straight line p passes through the middle of segment AB and is perpendicular to it.

Theorem

Rice. 4

Any point lying on the perpendicular bisector is equidistant from the ends of the segment (see Fig. 4).

Prove that

Proof:

Consider triangles and . They are rectangular and equal, because. have a common leg OM, and legs AO and OB are equal by condition, thus, we have two right triangles, equal in two legs. It follows that the hypotenuses of the triangles are also equal, that is, what was required to be proved.

Note that the segment AB is a common chord for many circles.

For example, the first circle with a center at point M and radius MA and MB; second circle with center at point N, radius NA and NB.

Thus, we have proven that if a point lies on the perpendicular bisector of a segment, it is equidistant from the ends of the segment (see Fig. 5).

Rice. 5

The converse theorem is true.

Theorem

If a certain point M is equidistant from the ends of a segment, then it lies on the perpendicular bisector to this segment.

Given a segment AB, a perpendicular bisector to it p, a point M equidistant from the ends of the segment (see Fig. 6).

Prove that point M lies on the perpendicular bisector of the segment.

Rice. 6

Proof:

Consider a triangle. It is isosceles, as per the condition. Consider the median of a triangle: point O is the middle of the base AB, OM is the median. According to the property of an isosceles triangle, the median drawn to its base is both an altitude and a bisector. It follows that . But line p is also perpendicular to AB. We know that at point O it is possible to draw a single perpendicular to the segment AB, which means that the lines OM and p coincide, it follows that the point M belongs to the straight line p, which is what we needed to prove.

The direct and converse theorems can be generalized.

Theorem

The perpendicular bisector of a segment is the locus of points equidistant from its ends.

A triangle, as you know, consists of three segments, which means that three perpendicular bisectors can be drawn in it. It turns out that they intersect at one point.

The perpendicular bisectors of a triangle intersect at one point.

A triangle is given. Perpendiculars to its sides: P 1 to side BC, P 2 to side AC, P 3 to side AB (see Fig. 7).

Prove that the perpendiculars P 1, P 2 and P 3 intersect at point O.