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The bisector divides the opposite side. Angle bisector. Complete Lessons - Knowledge Hypermarket

Geometry is one of the most complex and intricate sciences. In it, what seems obvious at first glance, very rarely turns out to be correct. Bisectors, heights, medians, projections, tangents - a huge number of really difficult terms, which are very easy to get confused.

In fact, with due desire, you can understand the theory of any complexity. When it comes to bisector, median, and height, you need to understand that they are not unique to triangles. At first glance this simple lines, but each of them has its own properties and functions, the knowledge of which greatly simplifies the solution of geometric problems. So, what is the bisector of a triangle?

Definition

The term "bisector" itself comes from a combination of the Latin words "two" and "cut", "cut", which already indirectly indicates its properties. Usually, when children are introduced to this ray, they are offered a short phrase to memorize: "The bisector is a rat that runs around the corners and divides the corner in half." Naturally, such an explanation is not suitable for older students, besides, they are usually asked not about the angle, but about the geometric figure. So the bisector of a triangle is a ray that connects the vertex of the triangle to the opposite side, while dividing the angle into two equal parts. The point of the opposite side, to which the bisector comes, for an arbitrary triangle is chosen randomly.

Basic functions and properties

This ray has few basic properties. First, because the bisector of a triangle bisects the angle, any point lying on it will be at an equal distance from the sides that form the vertex. Secondly, in each triangle, three bisectors can be drawn, according to the number of available angles (hence, in the same quadrilateral there will already be four of them, and so on). The point at which all three rays intersect is the center of the circle inscribed in the triangle.

Properties get more complex

Let's complicate the theory a bit. Another interesting property: the angle bisector of a triangle divides the opposite side into segments whose ratio is equal to the ratio of the sides forming the vertex. At first glance, this is difficult, but in fact everything is simple: in the proposed figure, RL:LQ = PR:PK. By the way, this property is called "the bisector theorem" and first appeared in the works ancient Greek mathematician Euclid. They remembered him in one of the Russian textbooks only in the first quarter of the seventeenth century.

A little more difficult. In a quadrilateral, the bisector cuts off an isosceles triangle. In this figure, all equal angles for the median AF are labeled.

And also in quadrilaterals and trapezoids, the bisectors of one-sided angles are perpendicular to each other. In the drawing, the angle APB is 90 degrees.

In an isosceles triangle

Bisector isosceles triangle- much more useful beam. It is at the same time not only a divider of the angle in half, but also a median and a height.

The median is a segment that comes out of some angle and falls to the middle of the opposite side, thereby dividing it into equal parts. Height is a perpendicular dropped from the vertex to the opposite side, it is with its help that any problem can be reduced to a simple and primitive Pythagorean theorem. In this situation, the bisector of the triangle is equal to the root of the difference between the square of the hypotenuse and the other leg. By the way, it is this property that occurs most often in geometric problems.

To fix: in this triangle, the bisector FB is the median (AB=BC) and the height (angles FBC and FBA are 90 degrees).

In outline

So what do you need to remember? The bisector of a triangle is a ray that bisects its vertex. At the intersection of three rays is the center of the circle inscribed in this triangle (the only disadvantage of this property is that it has no practical value and serves only for the competent execution of the drawing). It also divides the opposite side into segments, the ratio of which is equal to the ratio of the sides between which this ray passed. In a quadrilateral, the properties are a little more complicated, but, to be honest, they practically do not occur in school-level tasks, so they are usually not affected in the program.

The bisector of an isosceles triangle is the ultimate dream of any student. It is both the median (that is, it divides the opposite side in half) and the height (perpendicular to this side). Solving problems with such a bisector is reduced to the Pythagorean theorem.

Knowledge of the basic functions of the bisector, as well as its main properties, is necessary for solving geometric problems of both mean and high level difficulties. In fact, this ray is found only in planimetry, so it cannot be said that memorizing information about it will allow you to cope with all types of tasks.

Sorokina Vika

Proofs of the properties of the bisector of a triangle are given and the application of the theory to solving problems is considered.

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Education Committee of the Administration of Saratov, Oktyabrsky District Municipal Autonomous educational institution Lyceum №3 im. A. S. Pushkin.

Municipal Scientific and Practical

conference

"First Steps"

Subject: Bisector and its properties.

The work was completed by: a student of the 8th grade

Sorokina VictoriaSupervisor: Mathematics teacher of the highest categoryPopova Nina Fyodorovna

Saratov 2011

1. Title page…………………………………………………………...1
2. Contents …………………………………………………………………2
3. Introduction and objectives………………………………………………………... ..3
4. Consideration of the properties of the bisector

• Third locus of points………………………………….3
• Theorem 1……………………………………………………………....4
• Theorem 2…………………………………………………………………4
• The main property of the bisector of a triangle:

1. Theorem 3……………………………………………………………...4
2. Task 1…………………………………………………………… ….7
3. Task 2……………………………………………………………….8
4. Task 3…………………………………………………………….....9
5. Task 4…………………………………………………………….9-10

• Theorem 4…………………………………………………………10-11
• Formulas for finding the bisector:

1. Theorem 5…………………………………………………………….11
2. Theorem 6…………………………………………………………….11
3. Theorem 7…………………………………………………………….12
4. Task 5…………………………………………………………...12-13

• Theorem 8…………………………………………………………….13
• Task 6………………………………………………………...…….14
• Task 7……………………………………………………………14-15
• Determination using the bisector of the cardinal points………………15

1. Conclusion and conclusion……………………………………………………..15
2. List of used literature ……………………………………..16

Bisector

In a geometry lesson, studying the topic of similar triangles, I met with a problem on the theorem on the ratio of the bisector to opposite sides. It would seem that there could be something interesting in the topic of the bisector, but this topic interested me, and I wanted to study it more deeply. After all, the bisector is very rich in its amazing properties to help solve various problems.

When considering this topic, you can notice that in geometry textbooks very little is said about the properties of the bisector, and in exams, knowing them, you can solve problems much easier and faster. In addition, in order to pass the GIA and the Unified State Examination, modern students need to study Additional materials to the school curriculum. That is why I decided to study the topic of the bisector in more detail.

Bisector (from Latin bi- “double”, and sectio “cutting”) of an angle - a ray with the beginning at the apex of the angle, dividing the angle into two equal parts. The bisector of an angle (together with its extension) is the locus of points equidistant from the sides of the angle (or their extensions)

Third locus of points

Figure F is the locus of points (the set of points) that have some property A, if two conditions are met:

1. from the fact that the point belongs to the figure F, it follows that it has the property A;
2. from the fact that the point satisfies the property A, it follows that it belongs to the figure F.

The first locus of points considered in geometry is a circle, i.e. locus of points equidistant from one fixed point. The second is the perpendicular bisector of the segment, i.e. locus of points equidistant from the end of a segment. And finally, the third - the bisector - the locus of points equidistant from the sides of the angle

Theorem 1:

The points of the bisector are equally distant from the sides he's a corner.

Proof:

Let P - bisector point A. Drop from pointR perpendiculars RV and PC per side corner. Then VAR = SAR hypotenuse and acute angle. Hence, RV = PC

Theorem 2:

If point P is equidistant from the sides of angle A, then it lies on the bisector.

Proof: РВ = PC => ВАР = СAP => BAP= CAP => АР is a bisector.

Among the basic geometric facts should be attributed the theorem that the bisector divides the opposite side in relation to opposite sides. This fact has long remained in the shadows, but everywhere there are problems that are much easier to solve if you know this and other facts about the bisector. I became interested, and I decided to explore this property of the bisector more deeply.

Basic property of the angle bisector of a triangle

Theorem 3. The bisector divides the opposite side of the triangle in relation to the adjacent sides.

Proof 1:

Given: AL- bisector of triangle ABC

Prove:

Proof: Let F - point of intersection of a line AL and a line passing through a point IN parallel to side AC.

Then BFA = FAC = BAF. Therefore BAF isosceles and AB = BF. From the similarity of triangles ALC and FLB we have

ratio

where

Proof 2

Let F be the point intersected by the line AL and the line passing through the point C parallel to the base AB. Then you can repeat the reasoning.

Proof 3

Let K and M be the bases of the perpendiculars dropped onto the line AL from points B and C respectively. Triangles ABL and ACL are similar in two angles. That's why
. And from the similarity of BKL and CML we have

From here

Proof 4

Let's use the area method. Calculate the areas of triangles ABL and ACL two ways.

From here.

Proof 5

Let α= BAC,φ= BLA. By the sine theorem in triangle ABL

And in triangle ACL.

Because ,

Then, dividing both parts of the equality by the corresponding parts of the other, we get.

Task 1

Given: In the triangle ABC, VC is the bisector, BC=2, KS=1,

Solution:

Task 2

Given:

Find the bisectors of the acute angles of a right triangle with legs 24 and 18

Solution:

Let leg AC = 18, leg BC = 24,

AM is the bisector of the triangle.

By the Pythagorean theorem, we find

that AB = 30.

Since , then

Similarly, we find the second bisector.

Answer:

Task 3

In a right triangle ABC with right angle B angle bisector A crosses side BC

At point D. It is known that BD = 4, DC = 6.

Find the area of ​​a triangle ADC

Solution:

By the property of the bisector of a triangle

Denote AB = 2 x , AC = 3 x . By theorem

Pythagorean BC 2 + AB 2 = AC 2, or 100 + 4 x 2 = 9 x 2

From here we find that x = Then AB = , S ABC=

Hence,

Task 4

Given:

In an isosceles triangle ABC side AB equals 10, base AC is 12.

Angle bisectors A and C intersect at a point D. Find BD.

Solution:

Since the bisectors of a triangle intersect at

One point, then BD is the bisector of B. Let's continue BD to the intersection with AC at point M . Then M is the midpoint of AC , BM AC . That's why

Because CD - triangle bisector BMC then

Hence,.

Answer:

Theorem 4 . The three bisectors of a triangle intersect at one point.

Indeed, consider first the point Р of the intersection of two bisectors, for example, AK 1 and VC 2 . This point is equally distant from the sides AB and AC, since it lies on the bisectorA, and equally removed from the sides AB and BC, as belonging to the bisectorB. Hence, it is equally removed from the sides AC and BC and thus belongs to the third bisector of SC 3 , that is, at the point P, all three bisectors intersect.

Formulas for finding the bisector
Theorem5: (the first formula for the bisector): If in triangle ABC segment AL is a bisector A, then AL² = AB AC - LB LC.

Proof: Let M be the point of intersection of the line AL with the circle circumscribed about the triangle ABC (Fig. 41). The BAM angle is equal to the MAC angle by convention. Angles BMA and BCA are equal as inscribed angles based on the same chord. Hence, triangles BAM and LAC are similar in two angles. Therefore, AL: AC = AB: AM. So AL AM = AB AC AL (AL + LM) = AB AC AL² = AB AC - AL LM = AB AC - BL LC. Q.E.D.

Theorem6: . (second formula for the bisector): In triangle ABC with sides AB=a, AC=b andA, equal to 2α and the bisector of l, the equality takes place:
l = (2ab / (a+b)) cosα.

Proof : Let ABC be a given triangle, AL its bisector, a=AB, b=AC, l=AL. Then S ABC = S ALB + S ALC . Therefore, ab sin2α = a l sinα + b l sinα 2ab sinα cosα = (a + b) l sinα l = 2 (ab / (a+b)) cosα. The theorem has been proven.

Theorem 7: If a, b are the sides of the triangle, Y is the angle between them,is the bisector of this angle. Then.

Average level

Bisector of a triangle. Detailed theory with examples (2019)

The bisector of a triangle and its properties

Do you know what the midpoint of a line is? Of course you do. And the center of the circle? Same. What is the midpoint of an angle? You can say that this doesn't happen. But why, the segment can be divided in half, but the angle cannot? It is quite possible - just not a dot, but .... line.

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

Bisector of a triangle is a segment of the bisector of the 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 a lot of interesting properties of the bisector. This knowledge has greatly simplified the lives of people. It has become easier to build, calculate distances, even correct the firing of cannons ... But knowledge of these properties will help us solve some tasks of the GIA and the Unified State Examination!

The first knowledge that will help in this - bisector of an isosceles triangle.

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

So, 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 - it's a side.

The median is the line drawn from the vertex of the triangle and dividing opposite side(it again) in half.

Notice we don't say, "The median of an isosceles triangle." Do you know why? Because the median drawn from the vertex of a triangle bisects the opposite side in ANY triangle.

Well, the 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. In order to better understand and remember forever what a bisector, median and height are, they need to be compared with each other and understand how they are similar and how they differ from each other. At the same time, in order to better remember, it is better to describe everything " human language". Then you will easily operate with the language of mathematics, but at first you do not understand this language and you need to comprehend everything in your own language.

So how are they similar? The bisector, median and height - they all "go out" from the vertex of the triangle and abut in the opposite direction and "do something" either with the angle from which they come out, or with the opposite side. I think it's simple, no?

And how do they differ?

• The bisector bisects the angle from which it exits.
• The median bisects the opposite side.
• The height is always perpendicular to the opposite side.

That's it. To understand is easy. Once you understand, you can remember.

Now the next question. Why, then, in the case of an isosceles triangle, the bisector turns out to be both the median and the height at the same time?

You can just look at the figure and make sure that the median splits 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 it - everything is simple! Look: and have equal sides and, they have a common side and. (- bisector!) And so, it turned out that two triangles have two equal sides and an angle between them. We recall the first sign of the equality of triangles (you don’t remember, look at the topic) and conclude that, which means = and.

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

But what is it?

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

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

In any case, please remember:

Now it's harder: we'll count angle between bisectors in any triangle! Don't be afraid, it's not all 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

\angle AOC=90()^\circ +\frac(\angle B)(2)

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??!! Will they all intersect at the same point?

Or will it be?

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

Really, great?

Do you want to know why this happens?

So...two right triangle: And. They have:

• common hypotenuse.
• (because - the bisector!)

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

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

Why is 2 correct?

And connect the dots.

So, that is, lies on the bisector!

That's all!

How can all this be applied to problem solving? For example, in tasks there is often such a phrase: "The circle touches the sides of the angle ...". Well, you need to find something.

You quickly realize that

And you can use equality.

3. Three bisectors in a triangle intersect at one point

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

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

And the third bisector could go like this:

But in fact, everything is much better!

Let's consider the intersection point of two bisectors. Let's call her .

What did we use here both times? Yes paragraph 1, of course! If a point lies on the 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's going to work point 2: if the distances to the sides of the angle are equal, then the point lies on the bisector ... of what angle? Look at the picture again:

and are the distances to the sides of the angle, and they are equal, which means that 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.

(For fidelity, look at another topic).

Well, now you will 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, and a bisector has a lot of properties, right? And that's great, because more properties, the more tools for solving problems about the bisector.

4. Bisector and parallelism, bisectors of adjacent angles

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

Case 1

It's great, right? Let's understand why.

On the one hand, we are drawing a bisector!

But, on the other hand, - like crosswise lying corners (remember the topic).

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

Case 2

Imagine a triangle (or look at a picture)

Let's continue side by point. Now there are two corners:

• - inner corner
• - outer corner - it's outside, right?

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

And it will turn out rectangular!

Surprisingly, that's exactly what it is.

We understand.

What do you think the amount is?

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

And now we recall that and are bisectors and we will see that inside the angle is exactly half from the sum of all four angles: and - - that is, exactly. It can also be written as an equation:

So, unbelievable but true:

The angle between the bisectors of the inner and outer corner triangle is equal.

Case 3

See that everything is the same here as for the inner and outer corners?

Or do we think again why this is so?

Again, as for adjacent corners,

(as corresponding with parallel bases).

And again, make up exactly half from the sum

Conclusion: If there are bisectors in the problem related angles or bisectors respective angles of a parallelogram or trapezoid, then in this problem certainly a right triangle is involved, and maybe even a whole rectangle.

5. Bisector and opposite side

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

That is:

Amazing fact, isn't it?

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

Again - an exit to the "space" - an additional building!

Let's go straight.

For what? Now we'll see.

We continue the bisector to the intersection with the line.

A familiar picture? Yes, yes, yes, exactly the same as in paragraph 4, case 1 - it turns out that (- bisector)

Like lying crosswise

So, this is also.

Now let's look at the triangles and.

What can be said about them?

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

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

And now in short notation:

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

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

Now you can safely use it! Let's analyze one more property of the bisectors of the angles of a triangle - don't be scared, now the most difficult thing is over - it will be easier.

We get that

Theorem 1:

Theorem 2:

Theorem 3:

Theorem 4:

Theorem 5:

Theorem 6:

Theorem. Bisector inner corner triangle divides the opposite side into parts proportional to the adjacent sides.

Proof. Consider the triangle ABC (Fig. 259) and the bisector of its angle B. Let us draw a straight line CM through the vertex C, parallel to the bisector VC, until it intersects at the point M with the continuation of the side AB. Since VC is the bisector of angle ABC, then . Further, as corresponding angles at parallel lines, and as crosswise lying angles at parallel lines. From here and therefore - isosceles, from where. According to the theorem on parallel lines intersecting the sides of the angle, we have and in view of this we get, which was required to be proved.

The bisector of the external angle B of the triangle ABC (Fig. 260) has a similar property: the segments AL and CL from the vertices A and C to the point L of the intersection of the bisector with the continuation of the side AC are proportional to the sides of the triangle:

This property is proved in the same way as the previous one: in Fig. 260 an auxiliary straight line SM is drawn, parallel to the bisector BL. The reader himself will be convinced of the equality of the angles BMC and BCM, and hence the sides BM and BC of the triangle BMC, after which the required proportion will be obtained immediately.

We can say that the bisector of the external angle also divides the opposite side into parts proportional to the adjacent sides; it is only necessary to agree to allow "external division" of the segment.

The point L, which lies outside the segment AC (on its continuation), divides it externally with respect to if So, the bisectors of the angle of the triangle (internal and external) divide the opposite side (internal and external) into parts proportional to the adjacent sides.

Problem 1. The sides of the trapezoid are 12 and 15, the bases are 24 and 16. Find the sides of the triangle formed by the large base of the trapezoid and its extended sides.

Solution. In the notation of Fig. 261 we have for the segment serving as a continuation of the lateral side the proportion from which we easily find In a similar way we determine the second side of the triangle The third side coincides with the large base: .

Problem 2. The bases of the trapezoid are 6 and 15. What is the length of the segment parallel to the bases and dividing the sides in a ratio of 1:2, counting from the vertices of the small base?

Solution. Let's turn to Fig. 262 depicting a trapezoid. Through the vertex C of the small base we draw a line parallel to the lateral side AB, cutting off a parallelogram from the trapezoid. Since , then from here we find . Therefore, the entire unknown segment KL is equal to Note that to solve this problem, we do not need to know the sides of the trapezoid.

Problem 3. The bisector of the internal angle B of triangle ABC cuts the side AC into segments at what distance from the vertices A and C will the bisector of the external angle B intersect the extension AC?

Solution. Each of the bisectors of angle B divides AC in the same ratio, but one internally and the other externally. We denote by L the point of intersection of the continuation of AC and the bisector of the external angle B. Since AK We denote the unknown distance AL by then and we will have the proportion The solution of which gives us the desired distance

Do the drawing yourself.

Exercises

1. A trapezoid with bases 8 and 18 is divided by straight lines, parallel to the bases, into six strips of equal width. Find the lengths of the line segments dividing the trapezoid into strips.

2. The perimeter of the triangle is 32. The bisector of angle A divides the side BC into parts equal to 5 and 3. Find the lengths of the sides of the triangle.

3. The base of an isosceles triangle is a, the side is b. Find the length of the segment connecting the points of intersection of the bisectors of the corners of the base with the sides.

Today is going to be a very easy lesson. We will consider only 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 OGE or USE, in the first lesson, cannot even formulate the exact definition of the bisector.

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

To begin with, a slightly strange question: what is an angle? That's right: an angle is just two rays coming out of the same point. For example:

Examples of angles: acute, obtuse and right

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

The captain seems to hint that in addition to the rays $OA$ and $OB$, one can always draw a bunch of rays from the point $O$. But among them there will be one special one - it is called the bisector.

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

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

Examples of bisectors for acute, obtuse and right angles

Since in real drawings it is far from always obvious that a certain ray (in our case, this is the $OM$ ray) splits the initial angle into two equal ones, it is customary in geometry to mark equal angles with the same number of arcs (in our drawing this is 1 arc for an acute angle, two for blunt, three for straight).

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

Basic property of the angle bisector

In fact, the bisector has a lot of properties. And we will definitely consider 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 the given angle.

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

1. Any point lying on the bisector of an angle is on the same distance from the sides of this corner.
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, in fact, is called the distance from a point to a side of an angle? The good old definition of the distance from a point to a line will help us here:

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

For example, consider a line $l$ and a point $A$ not lying on this line. Draw a perpendicular $AH$, where $H\in l$. Then the length of this perpendicular will be the distance from the point $A$ to the line $l$.

Graphical representation of the distance from a point to a line

Since an angle is just two rays, and each ray is a piece of a line, it's easy to determine the distance from a point to the sides of an angle. It's just two perpendiculars:

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

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

As promised, we break the proof into two parts:

1. The distances from a 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 same point $M$ is at the same distance from the sides of the angle.

Proof. Let's draw perpendiculars from the 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 corner

We got 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 assumption (since $OM$ is a bisector);
2. $\angle M((H)_(1))O=\angle M((H)_(2))O=90()^\circ $ by construction;
3. $\angle OM((H)_(1))=\angle OM((H)_(2))=90()^\circ -\angle MO((H)_(1))$ because the sum acute angles of a right triangle is always equal to 90 degrees.

Therefore, 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 the 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 $O$ and a point $M$ equidistant from the sides of this angle be given:

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

Proof. To begin with, let's draw this very ray $OM$, otherwise there will be nothing to prove:

Spent the beam $OM$ inside the corner

We got two right triangles again: $\vartriangle OM((H)_(1))$ and $\vartriangle OM((H)_(2))$. Obviously they are equal because:

1. The hypotenuse $OM$ is common;
2. The legs $M((H)_(1))=M((H)_(2))$ by condition (because the point $M$ is equidistant from the sides of the corner);
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, 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.

In conclusion of the proof, we mark the formed equal angles with red arcs:

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

As you can see, nothing complicated. We have proved 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 on to new level. In the next lesson, we'll go over more complex properties bisectors and learn how to apply them to solve real problems.