2 theorem on the bisector property of an angle. Bisector of a triangle. Detailed theory with examples (2019). Bisector and opposite side

Hello again! The first thing I want to show you in this video is what the bisector theorem is, the second thing is to give you its proof. So, we have an arbitrary triangle, triangle ABC. And I'm going to draw the bisector of this top corner. This can be done for any of the three angles, but I chose the top one (this will make the proof of the theorem a little easier). So, let's draw the bisector of this angle, ABC. And now this left corner is equal to this right corner. Let's call the point of intersection of the bisector with side AC D. The bisector theorem states that the ratio of the sides separated by this bisector... Well, you see: I drew the bisector - and from the large triangle ABC two smaller triangles were obtained. So, according to the bisector theorem, the ratios between the other two sides of these smaller triangles (i.e., not including the bisector side) will be equal. Those. this theorem states that the ratio AB/AD will be equal to the ratio BC/CD. I will mark this with different colors. The ratio of AB (this side) to AD (this side) will be equal to the ratio of BC (this side) to CD (this side). Interesting! The attitude of this side to this is equal to the attitude of this side to this... An excellent result, but you are unlikely to take my word for it and will definitely want us to prove it for ourselves. And maybe you guessed that since we now have some established aspect ratios, we will prove the theorem using the similarity of triangles. Unfortunately for us, these two triangles are not necessarily similar. We know that these two angles are equal, but we don’t know, for example, whether this angle (BAD) is equal to this one (BCD). We do not know and cannot make such assumptions. To establish this equality, we may need to construct another triangle, which will be similar to one of the triangles in this figure. And one way to do this is to draw another line. Frankly, this proof was not clear to me when I first studied this topic, so if it is not clear to you now, that’s okay. What if we extend this bisector of this angle here? Let's extend it... Let's say it goes on forever. Maybe we can construct a triangle similar to this triangle here, BDA, if we draw a line parallel to AB here below? Let's try to do this. According to the property of parallel lines, if point C does not belong to segment AB, then through point C it is always possible to draw a line parallel to segment AB. Then let's take another segment here. Let's call this point F. And suppose that this segment FC is parallel to segment AB. Segment FC is parallel to segment AB... Let me write this down: FC is parallel to AB. And now we have some interesting points here. By drawing a segment parallel to segment AB, we have constructed a triangle similar to triangle BDA. Let's see how it turned out. Before we talk about similarity, let's first think about what we know about some of the angles formed here. We know that there are internal crosswise angles here. If we take the same parallel lines... Well, one can imagine that AB continues indefinitely and FC continues indefinitely. And the segment BF in this case is a secant. Then, whatever this angle, ABD, this angle, CFD, will be equal to it (by the property of internal intersecting angles). We have come across such angles many times when we talk about the angles formed when parallel lines intersect with transversals. So these two angles will be equal. But this angle, DBC, and this one, CFD, will also be equal, because angles ABD and DBC are equal. After all, BD is a bisector, which means that angle ABD is equal to angle DBC. So, whatever these two angles are, the CFD angle will be equal to them. And this leads to an interesting result. Because it turns out that in this larger triangle BFC the angles at the base are equal. This, in turn, means that triangle BFC is isosceles. Then side BC must be equal to side FC. BC must be equal to FC. Great! We have used the property of interior cross-lying angles formed by a transversal to show that triangle BFC is isosceles and therefore sides BC and FC are equal. And this may be useful to us, because... we know that... Well, if we don’t know, then at least we feel that these two triangles will turn out to be similar. We haven't proven this yet. But how can what we just proved help us learn anything about the BC side? Well, we just proved that side BC is equal to side FC. If we can prove that the ratio AB/AD is equal to the ratio FC/CD, consider it done, because we just proved that BC = FC. But let's not turn to the theorem - let's come to it as a result of the proof. So, the fact that the segment FC is parallel to AB helped us find out that the triangle BFC is isosceles, and its lateral sides BC and FC are equal. Now let's look at other angles here. If we look at triangle ABD (this one) and triangle FDC, we have already found out that they have one pair of equal angles. But also this angle of triangle ABD is vertical in relation to this angle of triangle FDC - this means that these angles are equal. And we know that if two angles of one triangle are respectively equal to two angles of another (well, then the third corresponding angles will also be equal), then based on the similarity of the triangles at the two angles we can conclude that these two triangles are similar. I'll write this down. And you need to make sure that when recording, the vertices correspond to each other. So, based on the similarity between the two corners, we know... And I'll start with the corner marked in green. We know that triangle B... Then move to the corner marked in blue... Triangle BDA is similar to a triangle... And again we start with the corner marked in green: F (then move to the corner marked in blue)... Similar to a triangle FDC. Now let's return to the bisector theorem. We are interested in the aspect ratio AB/AD. The ratio of AB to AD... As we already know, the ratios of the corresponding sides of similar triangles are equal. Or one could find the ratio of two sides of one similar triangle and compare it with the ratio of the corresponding sides of another similar triangle. They must also be equal. So, since triangles BDA and FDC are similar, then the ratio AB... Well, by the way, the triangles are similar at two angles, so I’ll write it down here. Because triangles are similar, then we know that the ratio AB/AD will be equal... And we can look here at the similarity statement to find the corresponding sides. The side corresponding to AB is the CF side. Those. AB/AD equals CF divided by... Side AD corresponds to side CD. So CF/CD. So, we got the following ratio: AB/AD=CF/CD. But we have already proven that (since the triangle BFC is isosceles) CF is equal to BC. This means that here CF can be replaced by BC. This is what needed to be proven. We have proven that AB/AD=BC/CD. So, to prove this theorem, you need, firstly, to construct another triangle, this one. And assuming that the segments AB and CF are parallel, we can obtain two corresponding equal angles of two triangles - this, in turn, indicates the similarity of the triangles. After constructing another triangle, in addition to the fact that there are two similar triangles, we will also be able to prove that this larger triangle is isosceles. And then we can say: the ratio between this and this side of one similar triangle is equal to the ratio of the corresponding sides (this and this) of another similar triangle. And this means that we have proven that the ratio between this side and this side is equal to the ratio BC/CD. Q.E.D. See you!

Theorem. The bisector of an interior angle of a triangle divides the opposite side into parts proportional to the adjacent sides.

Proof. Consider triangle ABC (Fig. 259) and the bisector of its angle B. Draw through vertex C a straight line CM, parallel to the bisector BC, until it intersects at point M with the continuation of side AB. Since BK is the bisector of angle ABC, then . Further, as corresponding angles for parallel lines, and as crosswise angles for parallel lines. Hence and therefore - isosceles, whence . By the theorem about parallel lines intersecting the sides of an angle, we have and in view we get , which is what we needed to prove.

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

This property is proven 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 VMS and VSM, and therefore the sides VM and BC of the triangle VMS, after which the required proportion will be obtained immediately.

We can say that the bisector of an external angle divides the opposite side into parts proportional to the adjacent sides; you just need to agree to allow “external division” of the segment.

Point L, lying outside the segment AC (on its continuation), divides it externally in the relation if Thus, the bisectors of the angle of a 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 equal to 12 and 15, the bases are equal to 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 a proportion for the segment that serves as a continuation of the lateral side, from which we easily find. In a similar way, we determine the second lateral 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 the ratio 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 side AB, cutting off the 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 lateral sides of the trapezoid.

Problem 3. The bisector of the internal angle B of triangle ABC cuts side AC into segments at what distance from 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. Let us denote by L the point of intersection of the continuation AC and the bisector of the external angle B. Since AK Let us denote the unknown distance AL by then and we will have a proportion The solution of which gives us the required distance

Complete 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 straight segments dividing the trapezoid into strips.

2. The perimeter of the triangle is 32. The bisector of angle A divides 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 intersection points of the bisectors of the corners of the base with the sides.

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.

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, 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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Ministry of Education and Science of the Republic of Tatarstan

Executive Committee Education Department

Bugulma municipal district of the Republic of Tatarstan

Bugulma

MBOU secondary school No. 1 with in-depth study of individual subjects

Class: 9 A

Research work

Subject:Angle bisector of a triangle

Student: Alexandrov A.A.

Head: Chukanova I.I.

Bugulma, 2012

Content.

1. Introduction …………………………………………………………………………3

2.Main part:

2.1. Formulation of the theorem on the bisector of a triangle angle……………...4

2.2. Various ways to prove the theorem on the angle bisector of a triangle…………………………………………………………………………………………..

2.21. Similarity method………………………………………………………………………………

2.22. Area method………………………………………………………5

2.23. Circumscribed circle………………………………………………………………..

2.24 Theorem of sines. ………………………………………………………...6

2.25.Vector method………………………………………………………7

2.26. Proof using axial symmetry……………………

2.3. Solving application problems……………………………………………..8

2.31. Solving problems from the textbook……………………………………………………………....

2.32. Solving Olympiad problems…………………………………………….

3.Conclusion ………………………………………………………………...........10

4.Literature used …………………………………………………….11

1. Introduction.

The first geometric concepts arose in prehistoric times. Man not only passively observed nature, but practically mastered and used its wealth. Material needs prompted people to make tools, cut stones and build houses, sculpt pottery and string a bow. People pulled their bows, made various objects with straight edges, and gradually reached the abstract concept of a straight line.

Practical human activity served as the basis for a long process of developing abstract concepts and discovering the simplest geometric dependencies and relationships.

Over time, when a large number of geometric facts accumulated, people began to need generalization, to understand the dependence of some elements on others, to establish logical connections and proofs. Geometry has becomescience only after the appearance of theorems and proofs in it.

Among the basic geometric facts is the theorem on the bisector of the angle of a triangle.

The triangle bisector theorem is often used in solving geometric problems. The theorem is interesting because there are many methods for proving it (method of similarity, method of areas, method of motion, etc.). This paper offers only 4 ways to prove this remarkable theorem.

Purpose and objectives of the study:

Study the proofs of the triangle angle bisector theorem.

Learn to work with drawings.

Solve problems using the theorem.

Compose and solve practical problems.

Main part.

2.1. Statement of the theorem on the angle bisector of a triangle.

Theorem: The bisector of an interior angle of a triangle divides

the opposite side into parts proportional to

adjacent sides of the triangle.

IfBD– bisector ∆ABC, then the equality.

2.2. Different ways to prove the bisector theorem

angle of a triangle.

2.21. Similarity method

Let's make a directmparallel to the bisectorBD.

ABD =  DBC(becauseBD– bisector).

DBC = BCD₁ (becausemǁ BDAndB.C.– secant).

BD₁ C = ABD(becausemǁ BDAndBD₁ - secant).

BCD₁ = BD₁ C.

So ∆BCD₁ - isosceles=> B.C.= BD₁ .

∆ABD ∆ AD₁ C(at two corners).

Hence:

Q.E.D.

2.22. Area method.

Consider ∆ABDand ∆CBD.

S ∆ ABD : S ∆ CBD = AD : DC (sinceh– total height).

BD– bisector ∆ABC. Each point of the bisectorBDequidistant from

parties ∆ ABC. MeansD.H. = DM- heights ∆ ABDand ∆CBD.

S ∆ ABD : S ∆ CBD = AB : B.C..

So: AB: BC = AD: DWITH=> AB: AD = BC: DWITH.

Q.E.D.

2.23. Circumscribed circle.

Let's describe around ∆ABCcircle. Let's continueBDto the intersection with

circle at point E.

∆B.A.E.∆ BDC(at two corners). Means: (1).

∆B.C.E.∆ BAD(at two corners). Means: (2).

Since ∆ACE– isosceles, thenA.E. = C.E.. ThenAB ∙ DC = BC ∙ AD=>

Q.E.D.

2.24. According to the theorem of sines.

In a triangleABCABD =  DBC = β (becauseBD– bisector ∆ABC).

Consider ∆ABD. According to the law of sines: (1).

Consider ∆BCD. According to the law of sines:

(2).

Hence:.

Q.E.D.

2.25.Vector method.

For any point D of the segment AC the vector ,

Wherek = and 1-k = .

Really,

In our case, the vector parallel to the vector ∙ + ∙ ,

and therefore = : , Then = , where = .

Q.E.D.

2.26. Axial symmetry.

Let's perform axial symmetryStriangleABCrelativelyBD,

we getS BD (A) = A 1 , S BD (C) = C 1 AndS BD (B) = B.

Then ∆CDC 1 ∆ ADA 1 (at two corners) and ∆ СС 1 B ∆ A.A. 1 B(at two corners).

AB = A 1 B(since ∆ABA 1 – isosceles).

Then And . Hence, .

Q.E.D.

2.3 Solving application problems.

2.31.Task from the textbook.

The median and altitude divide the triangle into three equal parts. Find the angles of the triangle.

∆ ACH=∆ MCHalong the leg and acute angle.

Therefore ∆A.C.M - isosceles, AN=HM. Let AN = NM = a, MV = 2a.

According to the property of the bisector SM ∆HSun we have: . Then CB=2CH,

RSVN=30⁰ , РВСН = 60⁰ , β =30 ⁰ , РС=90⁰

Answer: 30⁰ , 60 ⁰ , 90 ⁰ .

2.32. Olympiad task.

In triangle ABC, sides AB and BC are marked with points M and N, respectively, with BM = BN. A line perpendicular to BC is drawn through point M, and a line perpendicular to AB is drawn through point N. These lines intersect at point O. The extension of segment BO intersects side AC at point P and divides it into segments AP = 5 and PC = 4. Find the length of segment BP if it is known that BC = 6.

Given:

ВС=6cm, ВК=4cm, ВК- bisector ∆ АВС.

KS=3cm,R ∆ BKC=1cm.S ∆ ABC=60cm².

Find: AB.

Solution:

1. 3. The areas of triangles having equal heights are related as

In this work, by citing various methods of this proof, I showed how universal the theorem is.

It is easy to understand, but at the same time helps me in solving very complex and confusing problems.

Having studied this theorem, I discovered a lot of new things for myself, expanded my knowledge and I think that I paved the way for further study of geometry.

4. Literature used.

Supplement to the journal KVANT No. 1/1995.

Articles: L.N.Smolyakov. 13 more proofs of the theorem about

bisector.//Kvant, No. 2, 1985.

S.R. Sefibekov. Four proofs of the theorem about

bisector.//Kvant, No. 8, 1983.

L. S. Atanasyan, V. F. Butuzov, S. B. Kadomtsev, E. G. Poznyak, I. I.

Yudina. Textbook for general education institutions.

Enlightenment, 2003.

I.F. Sharygin. Geometry grades 7-9. Moscow, Publishing House

"Bustard", 1997.

Unified collection of TsOR.

G.K.Pak. "Bisector". Series: Getting ready for math

Olympics Vladivostok, 2003.