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 points of intersection 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 really doing 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:
- Any point lying on the bisector of a certain angle is at the same distance from the sides of this 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, 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
Let's consider 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
Got two right triangle: $\vartriangle OM((H)_(1))$ and $\vartriangle OM((H)_(2))$. They have a common hypotenuse $OM$ and equal angles:
- $\angle MO((H)_(1))=\angle MO((H)_(2))$ by condition (since $OM$ is a bisector);
- $\angle M((H)_(1))O=\angle M((H)_(2))O=90()^\circ $ by construction;
- $\angle OM((H)_(1))=\angle OM((H)_(2))=90()^\circ -\angle MO((H)_(1))$, 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:
- Hypotenuse $OM$ - general;
- Legs $M((H)_(1))=M((H)_(2))$ by condition (after all, the point $M$ is equidistant from the sides of the angle);
- 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 bisectors and learn how to use them to solve real problems.
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. it turned out to be two smaller triangles. 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. And the segment BF in this case is a secant. .. Triangle BDA is similar to triangle... And again we start with the corner marked in green: F (then move to the corner marked in blue)... Similar to triangle FDC. 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).
