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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Committee on Education of the Administration of Saratov, Oktyabrsky District Municipal Autonomous Educational Institution Lyceum No. 3 named after. A. S. Pushkin.

Municipal Scientific and Practical

conference

"First Steps"

Topic: 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 that help solve various problems.

When considering this topic, you can see that geometry textbooks say very little 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 for the school curriculum themselves. 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 BUT, if two conditions are met:

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

Consequently,

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

Consequently,.

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.

Theorem. The bisector of the interior angle of a 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 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.

PROPERTIES OF THE BISSECTOR

Bisector property: In a triangle, the bisector divides the opposite side into segments proportional to the adjacent sides.

Bisector of an external angle The bisector of an external angle of a triangle intersects the extension of its side at a point, the distances from which to the ends of this side are proportional, respectively, to the adjacent sides of the triangle. C B A D

Bisector length formulas:

The formula for finding the lengths of the segments into which the bisector divides the opposite side of the triangle

The formula for finding the ratio of the lengths of the segments into which the bisector is divided by the intersection point of the bisectors

Problem 1. One of the bisectors of a triangle is divided by the intersection point of the bisectors in a ratio of 3:2, counting from the vertex. Find the perimeter of a triangle if the length of the side of the triangle to which this bisector is drawn is 12 cm.

Solution We use the formula to find the ratio of the lengths of the segments into which the bisector is divided by the intersection point of the bisectors in the triangle: 30. Answer: P = 30cm.

Task 2 . Bisectors BD and CE ∆ ABC intersect at point O. AB=14, BC=6, AC=10. Find O D .

Solution. Let's use the formula for finding the length of the bisector: We have: BD = BD = = According to the formula for the ratio of the segments into which the bisector is divided by the intersection point of the bisectors: l = . 2 + 1 = 3 parts of everything.

this is part 1  OD = Answer: OD =

Problems In ∆ ABC, the bisectors AL and BK are drawn. Find the length of the segment KLif AB \u003d 15, AK \u003d 7.5, BL \u003d 5. In ∆ ABC, the bisector AD is drawn, and through point D is a straight line parallel to AC and intersecting AB at point E. Find the ratio of areas ∆ ABC and ∆ BDE , if AB = 5, AC = 7. Find the bisectors of acute angles of a right triangle with legs 24 cm and 18 cm. In a right triangle, the bisector of an acute angle divides the opposite leg into segments 4 and 5 cm long. Determine the area of ​​the triangle.

5. In an isosceles triangle, the base and side are 5 and 20 cm, respectively. Find the bisector of the angle at the base of the triangle. 6. Find the bisector of the right angle of a triangle whose legs are equal a and b. 7. Calculate the length of the bisector of angle A of triangle ABC with side lengths a = 18 cm, b = 15 cm, c = 12 cm. Find the ratio in which the bisectors of the interior angles divide at the point of their intersection.

Answers: Answer: Answer: Answer: Answer: Answer: Answer: Answer: Answer: AP = 6 AP = 10 see KL = CP =

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:

1. Every point lying on the bisector of an angle is at the same distance from the sides of that 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, 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:

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 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.