Prove the test for the bisector. Triangle elements. Bisector. Basic property of the angle bisector

In this lesson, we will consider in detail what properties the points lying on the bisector of the angle and the points that lie on the perpendicular bisector to the segment have.

Theme: Circle

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

Consider the properties of a point lying on the angle bisector (see Fig. 1).

Rice. one

Given an angle , its bisector AL, point M lies on the bisector.

Theorem:

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

Proof:

Consider triangles and . These are right-angled triangles, and they are equal, because. have a common hypotenuse AM, and the angles and are equal, since AL is the bisector of angle . Thus, right-angled triangles are equal in hypotenuse and acute angle, hence it follows that , which was required 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 a non-expanded angle, then it lies on its bisector.

Rice. 2

An unfolded 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 the point M lies on the bisector of the angle.

Proof:

The distance from a point to a line is the length of the perpendicular. Draw from the point M perpendiculars MK to side AB and MP to side AC.

Consider triangles and . These are right-angled 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 against equal legs, thus, , therefore, the point M lies on the bisector of the given angle.

The direct and inverse theorems can be combined.

Theorem

The bisector of a non-expanded angle is the locus of points equidistant from the sides of the given angle.

Theorem

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

Rice. 3

Proof:

Consider first two bisectors BB 1 and СС 1 . They intersect, the intersection point O exists. To prove this, suppose the opposite - let the given bisectors do not intersect, in which case they are parallel. Then the line BC is a secant, and the sum of the angles , this contradicts the fact that in the whole triangle the sum of the angles is .

So, the point O of intersection of two bisectors exists. Consider its properties:

Point O lies on the bisector of angle , which means that 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 to -. Also, the 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 got the following equalities:

, that is, all three perpendiculars dropped from the 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 the point O is equidistant from the sides of the angle, hence it lies on its bisector AA 1.

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

Let us turn to the consideration of the segment, its perpendicular bisector and the properties of the point that lies on the perpendicular bisector.

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

Theorem

Rice. four

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 the legs of AO and OB are equal by condition, thus, we have two right-angled triangles equal in two legs. It follows that the hypotenuses of the triangles are also equal, that is, which was to be proved.

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

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

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

Rice. 5

The converse theorem is true.

Theorem

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

The segment AB is given, the median perpendicular to it p, the point M, equidistant from the ends of the segment (see Fig. 6).

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

Rice. 6

Proof:

Let's consider a triangle. It is isosceles, as by condition. Consider the median of the triangle: point O is the midpoint of the base AB, OM is the median. According to the property of an isosceles triangle, the median drawn to its base is both a height and a bisector. Hence it follows that . But the line p is also perpendicular to AB. We know that a single perpendicular to the segment AB can be drawn to the point O, which means that the lines OM and p coincide, hence it follows that the point M belongs to the line p, which was required to be proved.

The direct and inverse 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. Perpendicular 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 Р 1 , Р 2 and Р 3 intersect at the point O.

Do you know what the midpoint of a line is? Of course you do. And the center of the circle? Too.

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.

Do you remember the joke: the bisector is a rat that runs around corners and bisects the corner. 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.

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

By the way, do you remember all these terms? Do you remember how they differ from each other? Not? Not scary. Now let's figure it out.

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 a line drawn from the vertex of a triangle and bisects the opposite side (this again). 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.
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.

To better understand and remember forever what a bisector, median and height are, they need compare 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 in “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 (do not 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

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?

Go 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 meet the bisector in the problem? Try to apply one (and sometimes you can several) of the following amazing properties.

1. Bisector in an isosceles triangle.

Are you afraid of the word "theorem"? If you are afraid, then - in vain. Mathematicians are accustomed to call any statement that can be somehow deduced from other, simpler statements, a theorem of mathematics.

So, attention, the theorem!

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

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

- general.

And this means (rather, remember the first sign of the equality of triangles!), That.

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

And now let's see. Once, then absolutely exactly and even in addition,.

So it happened that

divided the side in half, that is, turned out to be the median
, which means they are both on, since (look again at the figure).

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

Hooray! We have proved the theorem. But guess what, that's not all. Faithful and converse theorem:

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

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

Why is it hard to remember? How can it help? Imagine that you have a task:

Given: .

Find: .

You immediately think, 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 conclude, which means, write the answer:. It's 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? Actually, it's nothing to worry about. Lazy mathematicians hid four in two lines. So, what does it mean, "Bisector - locus of points"? And this means that they are executed immediately twostatements:

If a point lies on a bisector, then the distances from it to the sides of the angle are equal.
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, then remember the Hatter from "Alice in Wonderland": "So you still have something good to 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: "the bisector is the locus of points equidistant from the sides of the angle" will be proved!

Why is 1 correct?

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 forgot them, then look at the section.

So ... two right triangles: 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 this is great, because the 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 angles of the 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.

What for? 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:

Ouch! 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:

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

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

Now the most important thing.

You've figured out the theory on this topic. And, I repeat, it's ... it's 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 the successful passing of the exam, for admission to the institute on the budget and, MOST IMPORTANTLY, for life.

I will not convince you of anything, I will 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 much 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 exam and be ultimately ... happier?

FILL YOUR HAND, SOLVING PROBLEMS ON THIS TOPIC.

On the exam, you will not be asked theory.

You will need solve problems on time.

And, if you haven’t solved them (LOTS!), you will definitely make a stupid mistake somewhere or simply won’t make it in time.

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

Find a collection anywhere you want necessarily with solutions, detailed analysis and decide, decide, decide!

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“Understood” and “I know how to solve” are completely different skills. You need both.

Find problems and solve!

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

Theme: 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 consider some of these properties.

Any triangle is primarily three angles and three segments (see Fig. 1).

Rice. one

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

Consider the properties of a point lying on the bisector of an angle (see Fig. 3).

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

Recall that the distance from a point to a line is the length of the perpendicular dropped 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 the point M to the sides of the corner is the same.

Theorem

Each point of the bisector of a non-expanded angle is equidistant from the sides of the angle, that is, the distances from the point M to AC and to the BC of the sides of the angle are equal.

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

Prove that .

Rice. four

Proof:

Consider triangles and . These are right-angled triangles, and they are equal, because they have a common hypotenuse AM, and the angles and are equal, since AL is the angle bisector. Thus, right-angled triangles are equal in hypotenuse and acute angle, hence it follows that , which was required 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 a non-expanded 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 the 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. Draw from the point M perpendiculars MK to side AB and MP to side AC.

Consider triangles and . These are right-angled triangles, and they are equal, because they have a common hypotenuse AM, the legs of 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 against equal legs, thus, , therefore, the point M lies on the bisector of the given angle.

Sometimes the direct and inverse 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 that angle.

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

Problem #674 from Atanasyan's textbook, geometry, grades 7-9:

From the point M of the bisector of a non-expanded 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, the 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, legs MA and MB are equal, as we proved earlier. So 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 the angles and equality of other legs.

From the equality of the legs OA and OB it follows that the triangle is isosceles, and AB is its base. The line OM is the bisector of a triangle. According to the property of an isosceles triangle, this bisector is also a height, which means that the lines OM and AB intersect at a right angle, which was to be proved.

So, we have considered the direct and inverse theorems on the property of a point lying on the bisector of an angle, generalized them and solved the problem by applying various geometric facts, including this theorem.

Bibliography

Aleksandrov A.D. etc. Geometry, grade 8. - M.: Education, 2006.
Butuzov V.F., Kadomtsev S.B., Prasolov V.V. Geometry, 8th grade. - M.: Education, 2011.
Merzlyak A.G., Polonsky V.B., Yakir S.M. Geometry, 8th grade. - M.: VENTANA-GRAF, 2009.

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

Homework

Atanasyan L.S., Butuzov V.F., Kadomtsev S.B. et al. Geometry, 7-9, no. 676-678, art. 180.

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

Bisector Properties

1. The bisector of a triangle bisects the angle.

2. The angle bisector of a triangle divides the opposite side in a ratio equal to the ratio of 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 a circle inscribed in this triangle.

Some formulas related to the bisector of a triangle

(proof of the formula - )
, where
- the length of the bisector drawn to the side,
- the sides of the triangle against the vertices, respectively,
- the length of the segments into which the bisector divides the side,

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

Tasks covered in the video:
1. In the triangle ABC with sides AB=2 cm, BC=3 cm, AC=3 cm, the bisector BM is drawn. Find the lengths of segments AM and MC
2. The bisector of the inner angle at vertex A and the bisector of the outer angle at vertex C of triangle ABC intersect at point M. Find the angle BMC, if angle B is 40, 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 may also be interested in a short video tutorial where one of the properties of the bisector is applied

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, we have an 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:

Every point lying on the bisector of an angle is at the same distance from the sides of that angle.
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 is easy to determine the distance from a point to the sides of the 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:

$\angle MO((H)_(1))=\angle MO((H)_(2))$ by assumption (since $OM$ is a bisector);

$\angle M((H)_(1))O=\angle M((H)_(2))O=90()^\circ $ by construction;

$\angle OM((H)_(1))=\angle OM((H)_(2))=90()^\circ -\angle MO((H)_(1))$ 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:

The hypotenuse $OM$ is common;

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

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 to a new level. In the next lesson, we will analyze more complex properties of the bisector and learn how to apply them to solve real problems.