Perpendicular bisector to a line segment

Definition 1. Perpendicular bisector to a segment called a straight line perpendicular to this segment and passing through its middle (Fig. 1).

Theorem 1. Each point of the perpendicular bisector to a segment is located at the same distance from the ends this segment.

Proof . Let's consider an arbitrary point D lying on the perpendicular bisector to the segment AB (Fig. 2), and prove that triangles ADC and BDC are equal.

Indeed, these triangles are right triangles in which legs AC and BC are equal, and leg DC is common. The equality of triangles ADC and BDC implies the equality of segments AD and DB. Theorem 1 is proven.

Theorem 2 (Converse to Theorem 1). If a point is at the same distance from the ends of a segment, then it lies on the perpendicular bisector to this segment.

Proof . Let us prove Theorem 2 by contradiction. For this purpose, assume that some point E is at the same distance from the ends of the segment, but does not lie on the perpendicular bisector to this segment. Let us bring this assumption to a contradiction. Let us first consider the case when points E and A lie on opposite sides of the perpendicular bisector (Fig. 3). In this case, the segment EA intersects the perpendicular bisector at some point, which we will denote by the letter D.

Let us prove that the segment AE is longer than the segment EB. Really,

Thus, in the case when points E and A lie on opposite sides of the perpendicular bisector, we have a contradiction.

Now consider the case when points E and A lie on the same side of the perpendicular bisector (Fig. 4). Let us prove that the segment EB is longer than the segment AE. Really,

The resulting contradiction completes the proof of Theorem 2

Circle circumscribed about a triangle

Definition 2. A circle circumscribed by a triangle, is called a circle passing through all three vertices of the triangle (Fig. 5). In this case the triangle is called triangle inscribed in a circle or inscribed triangle.

Properties of the circumscribed circle of a triangle. Theorem of sines

Figure	Drawing	Property
Perpendicular bisectors to the sides of the triangle		intersect at one point .
Center circle circumscribed about an acute triangle		Center described about acute-angled inside triangle.
Center circle circumscribed about a right triangle		The center described about rectangular middle of the hypotenuse .
Center circle circumscribed about an obtuse triangle		Center described about obtuse-angled triangle circle lies outside triangle.
		,
Square triangle		S= 2R 2 sin A sin B sin C ,
Circumradius		For any triangle the equality is true:

Perpendicular bisectors to the sides of a triangle

All perpendicular bisectors , drawn to the sides of an arbitrary triangle, intersect at one point .

Circle circumscribed about a triangle

Any triangle can be surrounded by a circle . The center of a circle circumscribed about a triangle is the point at which all the perpendicular bisectors drawn to the sides of the triangle intersect.

Center of the circumscribed circle of an acute triangle

Center described about acute-angled triangle circle lies inside triangle.

Center of the circumscribed circle of a right triangle

The center described about rectangular triangle circle is middle of the hypotenuse .

Center of the circumscribed circle of an obtuse triangle

Center described about obtuse-angled triangle circle lies outside triangle.

For any triangle the following equalities are true (sine theorem): , where a, b, c are the sides of the triangle, A, B, C are the angles of the triangle, R is the radius of the circumscribed circle.

Area of ​​a triangle

For any triangle the equality is true: S= 2R 2 sin A sin B sin C , where A, B, C are the angles of the triangle, S is the area of ​​the triangle, R is the radius of the circumscribed circle.

Circumradius

For any triangle the equality is true: where a, b, c are the sides of the triangle, S is the area of ​​the triangle, R is the radius of the circumscribed circle.

Proofs of theorems on the properties of the circumcircle of a triangle

Theorem 3. All perpendicular bisectors drawn to the sides of an arbitrary triangle intersect at one point.

Proof . Let's consider two perpendicular bisectors drawn to sides AC and AB of triangle ABC, and denote their intersection point with the letter O (Fig. 6).

Since the point O lies on the perpendicular bisector to the segment AC, then by virtue of Theorem 1 the equality holds:

Since the point O lies on the perpendicular bisector to the segment AB, then by virtue of Theorem 1 the following equality holds:

Therefore, the equality is true:

whence, using Theorem 2, we conclude that point O lies on the perpendicular bisector to the segment BC.

Thus, all three perpendicular bisectors pass through the same point, as required to be proved. Any triangle can be surrounded by a circle . The center of a circle circumscribed about a triangle is the point at which all the perpendicular bisectors drawn to the sides of the triangle intersect.

Proof . Let's consider point O, at which all the bisectors drawn to the sides of triangle ABC intersect (Fig. 6).

When proving Theorem 3, the following equality was obtained:

from which it follows that a circle with a center at point O and radii OA, OB, OC passes through all three vertices of triangle ABC, which was what needed to be proven.

First level

Circumscribed circle. Visual Guide (2019)

The first question that may arise is: what is described - around what?

Well, actually, sometimes it happens around anything, but we will talk about a circle circumscribed around (sometimes they also say “about”) a triangle. What is it?

And just imagine, an amazing fact takes place:

Why is this fact surprising?

But triangles are different!

And for everyone there is a circle that will go through through all three peaks, that is, the circumscribed circle.

The proof of this amazing fact can be found in the following levels of the theory, but here we only note that if we take, for example, a quadrilateral, then not for everyone there will be a circle passing through the four vertices. For example, a parallelogram is an excellent quadrilateral, but there is no circle passing through all its four vertices!

And there is only for a rectangle:

Here you go, and every triangle always has its own circumscribed circle! And it’s even always quite easy to find the center of this circle.

Do you know what it is perpendicular bisector?

Now let's see what happens if we consider as many as three perpendicular bisectors to the sides of the triangle.

It turns out (and this is precisely what needs to be proven, although we will not) that all three perpendiculars intersect at one point. Look at the picture - all three perpendicular bisectors intersect at one point.

Do you think the center of the circumscribed circle always lies inside the triangle? Imagine - not always!

But if acute-angled, then - inside:

What to do with a right triangle?

And with an additional bonus:

Since we are talking about the radius of the circumscribed circle: what is it equal to for an arbitrary triangle? And there is an answer to this question: the so-called .

Namely:

And, of course,

1. Existence and circumcircle center

Here the question arises: does such a circle exist for every triangle? It turns out that yes, for everyone. And moreover, we will now formulate a theorem that also answers the question of where the center of the circumscribed circle is located.

Look, like this:

Let's be brave and prove this theorem. If you have already read the topic “” and understood why three bisectors intersect at one point, then it will be easier for you, but if you haven’t read it, don’t worry: now we’ll figure it out.

We will carry out the proof using the concept of locus of points (GLP).

Well, for example, is the set of balls the “geometric locus” of round objects? No, of course, because there are round... watermelons. Is it a set of people, a “geometric place”, who can speak? No, either, because there are babies who cannot speak. In life, it is generally difficult to find an example of a real “geometric location of points.” It's easier in geometry. Here, for example, is exactly what we need:

Here the set is the perpendicular bisector, and the property “ ” is “to be equidistant (a point) from the ends of the segment.”

Shall we check? So, you need to make sure of two things:

1. Any point that is equidistant from the ends of a segment is located on the perpendicular bisector to it.

Let's connect c and c.Then the line is the median and height b. This means - isosceles - we made sure that any point lying on the perpendicular bisector is equally distant from the points and.

Let's take the middle and connect and. The result is the median. But according to the condition, not only the median is isosceles, but also the height, that is, the perpendicular bisector. This means that the point exactly lies on the perpendicular bisector.

All! We have fully verified the fact that The perpendicular bisector of a segment is the locus of points equidistant from the ends of the segment.

This is all well and good, but have we forgotten about the circumscribed circle? Not at all, we have just prepared ourselves a “springboard for attack.”

Consider a triangle. Let's draw two bisectoral perpendiculars and, say, to the segments and. They will intersect at some point, which we will name.

Now, pay attention!

The point lies on the perpendicular bisector;
the point lies on the perpendicular bisector.
And that means, and.

Several things follow from this:

Firstly, the point must lie on the third bisector perpendicular to the segment.

That is, the perpendicular bisector must also pass through the point, and all three perpendicular bisectors intersect at one point.

Secondly: if we draw a circle with a center at a point and a radius, then this circle will also pass through both the point and the point, that is, it will be a circumscribed circle. This means that it already exists that the intersection of three perpendicular bisectors is the center of the circumscribed circle for any triangle.

And the last thing: about uniqueness. It is clear (almost) that the point can be obtained in a unique way, therefore the circle is unique. Well, we’ll leave “almost” for your reflection. So we proved the theorem. You can shout “Hurray!”

What if the problem asks “find the radius of the circumscribed circle”? Or vice versa, the radius is given, but you need to find something else? Is there a formula that relates the radius of the circumcircle to the other elements of the triangle?

Please note: the sine theorem states that in order to find the radius of the circumscribed circle, you need one side (any!) and the angle opposite to it. That's all!

3. Center of the circle - inside or outside

Now the question is: can the center of the circumscribed circle lie outside the triangle?
Answer: as much as possible. Moreover, this always happens in an obtuse triangle.

And generally speaking:

CIRCULAR CIRCLE. BRIEFLY ABOUT THE MAIN THINGS

1. Circle circumscribed about a triangle

This is the circle that passes through all three vertices of this triangle.

2. Existence and circumcircle center

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.

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

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A triangle is the simplest of flat polygonal figures. If the value of any angle at its vertices is 90°, then the triangle is called a right triangle. It is possible to draw a circle around such a polygon in such a way that each of the 3 vertices has one common point with its boundary (circle). This circle will be called a circumscribed circle, and the presence of a right angle greatly simplifies the task of constructing it.

You will need

• Ruler, compass, calculator.

Instructions

1. Start by determining the radius of the circle you will need to construct. If it is possible to measure the lengths of the sides of a triangle, then pay attention to its hypotenuse - the side lying opposite the right angle. Measure it and divide the resulting value in half - this will be the radius of the circle described around the right triangle.

2. If the length of the hypotenuse is unknown, but there are lengths (a and b) of the legs (2 sides adjacent to the right angle), then find the radius (R) using the Pythagorean theorem. It follows from it that this parameter will be equal to half the square root extracted from the sum of the squared lengths of the legs: R=?*?(a?+b?).

3. If the length of only one of the legs (a) and the value of the adjacent acute angle (?) are known, then to determine the radius of the circumscribed circle (R), use the trigonometric function - cosine. In a right triangle, it determines the ratio of the lengths of the hypotenuse and this leg. Calculate half of the quotient of the leg length divided by the cosine of the famous angle: R=?*a/cos(?).

4. If, in addition to the length of one of the legs (a), the value of the acute angle (?) lying opposite it is known, then to calculate the radius (R), use another trigonometric function - the sine. Apart from replacing the function and the side, nothing will change in the formula - divide the length of the leg by the sine of the known acute angle, and divide the result in half: R=?*b/sin(?).

5. After finding the radius using any of the listed methods, determine the center of the circumscribed circle. To do this, put the resulting value on a compass and set it to each vertex of the triangle. There is no need to describe a complete circle; easily mark the place where it intersects with the hypotenuse - this point will be the center of the circle. This is the quality of a right triangle - the center of the circle circumscribed around it is invariably in the middle of its longest side. Draw a circle of the radius laid out on the compass with the center at the detected point. This will complete the construction.

Occasionally, it is possible to draw a circle around a convex polygon in such a way that the vertices of all angles lie on it. Such a circle in relation to the polygon should be called circumscribed. Her center must not necessarily be located inside the perimeter of the inscribed figure, but using the properties of the described circle, discovering this point, as usual, is not very difficult.

You will need

• Ruler, pencil, protractor or square, compass.

Instructions

1. If the polygon around which it is necessary to describe a circle is drawn on paper, to find center and a circle is enough with a ruler, pencil and protractor or square. Measure the length of each side of the figure, determine its middle and place an auxiliary point in this place in the drawing. With the support of a square or protractor, draw a segment inside the polygon perpendicular to this side until it intersects with the opposite side.

2. Do the same operation with every other side of the polygon. The intersection of 2 constructed segments will be the desired point. This follows from the main property of the described circle- her center in a convex polygon with any number of sides invariably lies at the point of intersection of the perpendicular bisectors drawn to these sides.

3. For regular polygons, the definition center and inscribed circle it could be much simpler. Let's say, if this is a square, then draw two diagonals - their intersection will be center ohm inscribed circle. In a positive polygon with any even number of sides, it is enough to combine two pairs of opposite angles with auxiliary segments - center described circle must coincide with the point of their intersection. In a right triangle, to solve the problem, easily determine the middle of the longest side of the figure - the hypotenuse.

4. If it is not clear from the conditions whether it is allowed in the thesis to draw a circumscribed circle for a given polygon, after determining the position of the point center and you can find out using any of the methods described. Mark on the compass the distance between the detected point and each of the vertices, set the compass to the required center circle and draw a circle - the entire vertex should lie on this circle. If this is not the case, then one of the basic properties is not satisfied and it is impossible to describe a circle around this polygon.

According to the definition, described circle must pass through all the vertices of the corners of a given polygon. In this case, it ideally does not matter what kind of polygon it is - a triangle, square, rectangle, trapezoid or something else. It also doesn’t matter whether the polygon is true or false. You just need to consider that there are polygons around which circle impossible to describe. It is invariably allowed to describe circle around the triangle. As for quadrilaterals, then circle You can describe a square or rectangle or an isosceles trapezoid.

You will need

• Specified polygon
• Ruler
• Square
• Pencil
• Compass
• Protractor
• Sine and cosine tables
• Mathematical representations and formulas
• Pythagorean theorem
• Theorem of sines
• Cosine theorem
• Signs of similarity of triangles

Instructions

1. Construct a polygon with the given parameters and determine whether it is possible to describe around it circle. If you are given a quadrilateral, calculate the sum of its opposite angles. Each of them must be equal to 180°.

2. In order to describe circle, you need to calculate its radius. Remember where the center of the circumcircle lies in various polygons. In a triangle, it is located at the intersection point of all the altitudes of a given triangle. In a square and rectangles - at the point of intersection of the diagonals, for a trapezoid - at the point of intersection of the axis of symmetry to the line connecting the midpoints of the lateral sides, and for any other convex polygon - at the point of intersection of the median perpendiculars to the sides.

3. Calculate the diameter of a circle circumscribed around a square and a rectangle using the Pythagorean theorem. It will be equal to the square root of the sum of the squares of the sides of the rectangle. For a square with all sides equal, the diagonal is equal to the square root of twice the square of the side. Dividing the diameter by 2 gives you the radius.

4. Calculate the circumradius of the triangle. Since the parameters of the triangle are given in the conditions, calculate the radius using the formula R = a/(2·sinA), where a is one of the sides of the triangle, ? - the angle opposite to it. Instead of this side, you can take any other side and the angle opposite it.

5. Calculate the radius of the circle circumscribed around the trapezoid. R = a*d*c / 4 v(p*(p-a)*(p-d)*(p-c)) In this formula, a and b are the bases of the trapezoid, h is the height, d is the diagonal, p = 1 /2*(a+d+c) . Compute missing values. The height can be calculated using the theorem of sines or cosines, since the lengths of the sides of the trapezoid and the angles are specified in the conditions of the problem. Knowing the height and considering the similarity signs of triangles, calculate the diagonal. After this, all that remains is to calculate the radius using the above formula.

Video on the topic

Helpful advice
To calculate the radius of a circle circumscribed around another polygon, perform a number of additional constructions. Get more primitive figures whose parameters you know.

Tip 4: How to draw a right triangle using an acute angle and hypotenuse

A triangle is called a right triangle if the angle at one of its vertices is 90°. The side opposite this angle is called the hypotenuse, and the sides opposite the two acute angles of the triangle are called legs. If the length of the hypotenuse and the size of one of the acute angles are known, then these data are enough to construct a triangle using at least two methods.

You will need

• A sheet of paper, a pencil, a ruler, a compass, a calculator.

Instructions

1. The 1st method requires, in addition to a pencil and paper, a ruler, a protractor and a square. First, draw the side that is the hypotenuse - put point A, set aside the known length of the hypotenuse from it, put point C and combine the points.

2. Attach the protractor to the drawn segment in such a way that the zero mark coincides with point A, measure the value of the known acute angle and place an auxiliary point. Draw a line that will start at point A and pass through the auxiliary point.

3. Attach the square to the segment AC in such a way that the right angle starts from point C. Mark the point where the square intersects the line drawn in the previous step with the letter B and combine it with point C. This completes the construction of a right triangle with the famous side length AC (hypotenuse) and an acute angle at vertex A will be completed.

4. Another method, in addition to pencil and paper, will require a ruler, compass and calculator. Start by calculating the lengths of the legs - knowing the size of one acute angle and the length of the hypotenuse is absolutely enough for this.

5. Calculate the length of that leg (AB), the one that lies opposite the angle of the known value (β) - it will be equal to the product of the length of the hypotenuse (AC) by the sine of the famous angle AB=AC*sin(β).

6. Determine the length of the other leg (BC) - it will be equal to the product of the length of the hypotenuse and the cosine of the given angle BC=AC*cos(β).

7. Place point A, measure the length of the hypotenuse from it, place point C and draw a line between them.

8. Set aside on the compass the length of leg AB, calculated in the fifth step, and draw an auxiliary semicircle with the center at point A.

9. Set aside the length of leg BC, calculated in the sixth step, on the compass and draw an auxiliary semicircle with the center at point C.

10. Mark the intersection point of the 2 semicircles with the letter B and draw segments between points A and B, C and B. This will complete the construction of the right triangle.

Tip 5: What are the names of the sides of a right triangle

People became interested in the stunning properties of right triangles back in ancient times. Many of these properties were described by the ancient Greek scientist Pythagoras. In Ancient Greece, the names of the sides of a right triangle also appeared.

Which triangle is called a right triangle?

There are several types of triangles. Some have all acute angles, others have one obtuse and two acute, and others have two acute and one straight. According to this sign, each type of these geometric figures received their name: acute-angled, obtuse-angled and rectangular. That is, a triangle in which one of the angles is 90° is called a right triangle. There is another definition similar to the first. A triangle whose two sides are perpendicular is called a right triangle.

Hypotenuse and legs

In acute and obtuse triangles, the segments connecting the vertices of the angles are called primitively sides. A right-angled triangle has other names for its sides. Those adjacent to the right angle are called legs. The side opposite the right angle is called the hypotenuse. Translated from Greek, the word “hypotenuse” means “tight”, and “cathetus” means “perpendicular”.

Relationships between the hypotenuse and legs

The sides of a right triangle are connected by certain relationships, which make calculations much easier. For example, knowing the dimensions of the legs, you can calculate the length of the hypotenuse. This relationship, named after the mathematician who discovered it, was called the Pythagorean theorem and it looks like this: c2 = a2 + b2, where c is the hypotenuse, a and b are the legs. That is, the hypotenuse will be equal to the square root of the sum of the squares of the legs. In order to discover each of the legs, it is enough to subtract the square of the other leg from the square of the hypotenuse and extract the square root from the resulting difference.

Adjacent and opposite leg

Draw a right triangle DIA. The letter C usually denotes the vertex of a right angle, A and B - the vertices of acute angles. It is convenient to call the sides opposite the entire angle a, b and c, according to the names of the angles lying opposite them. Look at angle A. Leg a will be opposite for it, leg b will be adjacent. The ratio of the opposite side to the hypotenuse is called the sine. This trigonometric function can be calculated using the formula: sinA=a/c. The ratio of the adjacent leg to the hypotenuse is called cosine. It is calculated using the formula: cosA=b/c. Thus, knowing the angle and one of the sides, it is possible to calculate the other side using these formulas. Both sides are also connected by trigonometric relations. The ratio of the opposite to the adjacent is called the tangent, and the ratio of the adjacent to the opposite is called cotangent. These relationships can be expressed by the formulas tgA=a/b or ctgA=b/a.

A circle circumscribed by a right triangle. In this publication, we will look at the proof of one “mathematical fact” that is widely used in solving geometry problems. In some sources this fact is designated as a theorem, in others as a property, there are different formulations, but their essence is the same:

Any triangle constructed on the diameter of a circle whose third vertex lies on this circle is rectangular!

That is, the pattern in this geometric pattern is that wherever you place the vertex of the triangle, the angle at this vertex will always be right:

There are quite a lot of tasks in the mathematics exam, during the solutions of which this property is used.

I consider the standard proof to be very confusing and overloaded with mathematical symbols; you will find it in the textbook. We will consider the simple and intuitive. I discovered it in a wonderful essay called " Mathematician's cry", I recommend reading for teachers and students.

First, let's remember some theoretical points:

Parallelogram sign. A parallelogram has opposite sides that are equal. That is, if a quadrilateral has both pairs of opposite sides equal, then this quadrilateral is a parallelogram.

Rectangle sign. A rectangle is a parallelogram and its diagonals are equal. That is, if a parallelogram has equal diagonals, then it is a rectangle.

*A rectangle is a parallelogram; this is its special case.

So let's get started:

Let's take a triangle and rotate it 180 0 relative to the center of the circle (turn it over). We get a quadrilateral inscribed in a circle:

Since we simply rotated the triangle, the opposite sides of the quadrilateral are equal, which means it is a parallelogram. Since the triangle is rotated exactly 180 degrees, its vertex is diametrically opposite to the vertex of the “original” triangle.

It turns out that the diagonals of the quadrilateral are equal, so they are diameters. We have a quadrilateral whose opposite sides are equal and the diagonals are equal, therefore it is a rectangle, and all its angles are right angles.

That's all the proof!

You can also consider this, also simple and understandable:

View another proof =>>

From point C we will construct a segment passing through the center of the circle, the other end of which will lie on the opposite point of the circle (point D). Connect point D to vertices A and B:We got a quadrilateral. Triangle AOD is equal to triangle COB on two sides and the angle between them:

From the equality of triangles it follows that AD = CB.

Likewise, AC = DB.

We can conclude that the quadrilateral is a parallelogram. In addition, its diagonals are equal - AB is initially given as a diameter, CD is also a diameter (passes through point O).

Thus, ACBD is a rectangle, which means all its angles are right angles. Proven!

Another remarkable approach, which clearly and “beautifully” tells us that the angle in question is always right.

Look and remember the information about. Now look at the sketch:

Angle AOB is nothing more than the central angle based on the arc ADB, and it is equal to 180 degrees. Yes, AB is the diameter of a circle, but nothing prevents us from considering AOB a central angle (this is a straight angle). The angle ACB is inscribed for it; it also rests on the same arc on ADB.

And we know that the inscribed angle is equal to half the central one, that is, no matter how we place point C on the circle, the angle ACB will always be equal to 90 degrees, which means it is straight.

What conclusions can be drawn in relation to solving problems, in particular those included in the exam?

If the condition is about a triangle inscribed in a circle and built on the diameter of this circle, then this triangle is definitely a right triangle.

If it is said that a right triangle is inscribed in a circle, then this means that its hypotenuse coincides with its diameter (equal to it) and the center of the hypotenuse coincides with the center of the circle.

That's all. Good luck to you!

Sincerely, Alexander Krutitskikh.