One of the most interesting topics in geometry from the school course is “Quadrilaterals” (8th grade). What types of such figures exist, what special properties do they have? What is unique about quadrilaterals with ninety-degree angles? Let's figure it all out.
What geometric figure is called a quadrilateral?
Polygons that consist of four sides and, accordingly, four vertices (angles) are called quadrilaterals in Euclidean geometry.
The history of the name of this type of figure is interesting. In the Russian language, the noun “quadrangle” is formed from the phrase “four corners” (just like “triangle” - three corners, “pentagon” - five corners, etc.).
However, in Latin (through which many geometric terms came to most languages of the world) it is called quadrilateral. This word is formed from the numeral quadri (four) and the noun latus (side). So we can conclude that the ancients called this polygon nothing more than a “quadrilateral”.
By the way, this name (with an emphasis on the presence of four sides, rather than corners, in figures of this type) was preserved in some modern languages. For example, in English - quadrilateral and in French - quadrilatère.
Moreover, in most Slavic languages, the type of figure in question is still identified by the number of angles, not sides. For example, in Slovak (štvoruholník), in Bulgarian ("chetirigalnik"), in Belarusian ("chatyrokhkutnik"), in Ukrainian ("chotirikutnik"), in Czech (čtyřúhelník), but in Polish the quadrilateral is called by the number of sides - czworoboczny.
What types of quadrilaterals are studied in the school curriculum?
In modern geometry, there are 4 types of polygons with four sides.
However, due to the overly complex properties of some of them, schoolchildren are introduced to only two types in geometry lessons.
- Parallelogram. The opposite sides of such a quadrilateral are parallel to each other in pairs and, accordingly, are also equal in pairs.
- Trapezium (trapezium or trapezoid). This quadrilateral consists of two opposite sides parallel to each other. However, the other pair of sides does not have this feature.
Types of quadrilaterals not studied in the school geometry course
In addition to the above, there are two more types of quadrilaterals that schoolchildren are not introduced to in geometry lessons because of their particular complexity.
- Deltoid (kite)- a figure in which each of two pairs of adjacent sides is equal in length. This quadrilateral got its name due to the fact that appearance it quite closely resembles the letter of the Greek alphabet - “delta”.
- Antiparallelogram- this figure is as complex as its name. In it, two opposite sides are equal, but at the same time they are not parallel to each other. In addition, the long opposite sides of this quadrilateral intersect each other, as do the extensions of the other two, shorter sides.
Types of parallelogram
Having dealt with the main types of quadrangles, it is worth paying attention to its subtypes. So, all parallelograms, in turn, are also divided into four groups.
- Classic parallelogram.
- Rhombus- a quadrangular figure with equal sides. Its diagonals intersect at right angles, dividing the rhombus into four equal right-angled triangles.
- Rectangle. The name speaks for itself. Since it is a quadrilateral with right angles (each of them is equal to ninety degrees). Its opposite sides are not only parallel to each other, but also equal.
- Square. Like a rectangle, it is a quadrilateral with right angles, but all its sides are equal. In this way, this figure is close to a rhombus. So we can say that a square is a cross between a rhombus and a rectangle.
Special properties of a rectangle
When considering figures in which each of the angles between the sides is equal to ninety degrees, it is worth taking a closer look at the rectangle. So, what special features does it have that distinguish it from other parallelograms?
To claim that the parallelogram in question is a rectangle, its diagonals must be equal to each other, and each of the angles must be right. In addition, the square of its diagonals must correspond to the sum of the squares of two adjacent sides of this figure. In other words, a classic rectangle consists of two right triangles, and in them, as is known, the role of the hypotenuse is played by the diagonal of the quadrilateral in question.
The last of the listed features of this figure is also its special property. Besides this, there are others. For example, the fact that all sides of the quadrilateral being studied with right angles are also its heights.
In addition, if a circle is drawn around any rectangle, its diameter will be equal to the diagonal of the inscribed figure.
Among other properties of this quadrilateral is that it is flat and does not exist in non-Euclidean geometry. This is due to the fact that in such a system there are no quadrangular figures, the sum of the angles of which is equal to three hundred and sixty degrees.
Square and its features
Having understood the signs and properties of a rectangle, it is worth paying attention to the second known to science a quadrilateral with right angles (this is a square).
Being in fact the same rectangle, but with equal sides, this figure has all its properties. But unlike it, the square is present in non-Euclidean geometry.
In addition, this figure has other distinctive features of its own. For example, the fact that the diagonals of a square are not only equal to each other, but also intersect at right angles. Thus, like a rhombus, a square consists of four right triangles into which diagonals divide it.
In addition, this figure is the most symmetrical among all quadrilaterals.
What is the sum of the angles of a quadrilateral?
When considering the features of quadrilaterals of Euclidean geometry, it is worth paying attention to their angles.
So, in each of the above figures, regardless of whether it has right angles or not, their total sum is always the same - three hundred and sixty degrees. This is a unique distinguishing feature of this type of figure.
Perimeter of quadrilaterals
Having figured out what the sum of the angles of a quadrilateral is equal to and other special properties of figures of this type, it is worth finding out what formulas are best used to calculate their perimeter and area.
To determine the perimeter of any quadrilateral, you just need to add the lengths of all its sides together.
For example, in the KLMN figure, its perimeter can be calculated using the formula: P = KL + LM + MN + KN. If you substitute the numbers here, you get: 6 + 8 + 6 + 8 = 28 (cm).
In the case when the figure in question is a rhombus or a square, to find the perimeter, you can simplify the formula by simply multiplying the length of one of its sides by four: P = KL x 4. For example: 6 x 4 = 24 (cm).
Formulas for quadrilaterals with area
Having figured out how to find the perimeter of any figure with four corners and sides, it’s worth considering the most popular and simple ways finding its area.
Other properties of quadrilaterals: incircles and circumcircles
Having considered the features and properties of a quadrilateral as a figure of Euclidean geometry, it is worth paying attention to the ability to describe circles around or inscribe circles inside it:
- If the sums of the opposite angles of a figure are one hundred and eighty degrees and are equal in pairs, then a circle can be freely described around such a quadrilateral.
- According to Ptolemy's theorem, if a circle is circumscribed outside a polygon with four sides, then the product of its diagonals is equal to the sum of the products of the opposite sides of the given figure. Thus, the formula will look like this: KM x LN = KL x MN + LM x KN.
- If you build a quadrilateral in which the sums of opposite sides are equal to each other, then you can inscribe a circle in it.
Having figured out what a quadrilateral is, what types of it exist, which of them have only right angles between the sides and what properties they have, it’s worth remembering all this material. In particular, formulas for finding the perimeter and area of the polygons considered. After all, figures of this shape are among the most common, and this knowledge can be useful for calculations in real life.
"Circle" We have seen that a circle can be circumscribed around any triangle. That is, for every triangle there is a circle such that all three vertices of the triangle “sit” on it. Like this:
Question: can the same be said about a quadrilateral? Is it true that there will always be a circle on which all four vertices of the quadrilateral will “sit”?
It turns out that this is NOT TRUE! A quadrilateral can NOT ALWAYS be inscribed in a circle. There is a very important condition:
In our picture:
| . |
Look, the angles and lie opposite each other, which means they are opposite. What then about the angles and? They seem to be opposites too? Is it possible to take angles and instead of angles and?
Of course you can! The main thing is that the quadrilateral has some two opposite angles, the sum of which will be. The remaining two angles will then also add up by themselves. Do not believe? Let's make sure. Look:
Let be. Do you remember what the sum of all four angles of any quadrilateral is? Certainly, . That is - always! . But, → .
Magic right there!
So remember this very firmly:
If a quadrilateral is inscribed in a circle, then the sum of any two of its opposite angles is equal to
and vice versa:
If a quadrilateral has two opposite angles whose sum is equal, then the quadrilateral is cyclic.
We will not prove all this here (if you are interested, look into the next levels of theory). But let's see what this remarkable fact leads to: that in an inscribed quadrilateral the sum of the opposite angles is equal.
For example, the question comes to mind: is it possible to describe a circle around a parallelogram? Let's try the “poke method” first.
Somehow it doesn't work out.
Now let's apply the knowledge:
Let's assume that we somehow managed to fit a circle onto a parallelogram. Then there must certainly be: , that is.
Now let's remember the properties of a parallelogram:
Every parallelogram has equal opposite angles.
It turned out that
What about the angles and? Well, the same thing of course.
Inscribed → →
Parallelogram→ →
Amazing, right?
It turns out that if a parallelogram is inscribed in a circle, then all its angles are equal, that is, it is a rectangle!
And at the same time - the center of the circle coincides with the intersection point of the diagonals of this rectangle. This is included as a bonus, so to speak.
Well, that means we found out that a parallelogram inscribed in a circle is rectangle.
Now let's talk about the trapezoid. What happens if a trapezoid is inscribed in a circle? But it turns out there will be isosceles trapezoid. Why?
Let the trapezoid be inscribed in a circle. Then again, but due to the parallelism of the lines and.
This means we have: → → isosceles trapezoid.
Even easier than with a rectangle, right? But you need to firmly remember - it will come in handy:
Let's list the most important ones again main statements tangent to a quadrilateral inscribed in a circle:
- A quadrilateral is inscribed in a circle if and only if the sum of its two opposite angles is equal to
- A parallelogram inscribed in a circle - certainly rectangle and the center of the circle coincides with the intersection point of the diagonals
- A trapezoid inscribed in a circle is equilateral.
Inscribed quadrilateral. Average level
It is known that for every triangle there is a circumscribed circle (we proved this in the topic “The Circumscribed Circle”). What can be said about the quadrilateral? It turns out that NOT EVERY quadrilateral can be inscribed in a circle, and there is such a theorem:
A quadrilateral is inscribed in a circle if and only if the sum of its opposite angles is equal to.
In our drawing -
Let's try to understand why this is so? In other words, we will now prove this theorem. But before you prove it, you need to understand how the statement itself works. Did you notice the words “then and only then” in the statement? Such words mean that harmful mathematicians have crammed two statements into one.
Let's decipher:
- “Then” means: If a quadrilateral is inscribed in a circle, then the sum of any two of its opposite angles is equal.
- “Only then” means: If a quadrilateral has two opposite angles whose sum is equal, then such a quadrilateral can be inscribed in a circle.
Just like Alice: “I think what I say” and “I say what I think.”
Now let’s figure out why both 1 and 2 are true?
First 1.
Let a quadrilateral be inscribed in a circle. Let's mark its center and draw radii and. What will happen? Do you remember that an inscribed angle is half the size of the corresponding central angle? If you remember, we’ll use it now, and if not, take a look at the topic "Circle. Inscribed angle".
Inscribed
Inscribed
But look: .
We get that if - is inscribed, then
Well, it’s clear that it also adds up. (we also need to consider).
Now “vice versa”, that is, 2.
Let it turn out that in a quadrilateral the sum of some two opposite angles is equal. Let's say let
We don't know yet whether we can describe a circle around it. But we know for sure that we are guaranteed to be able to describe a circle around a triangle. So let's do it.
If a point does not “sit” on the circle, then it inevitably ends up either outside or inside.
Let's consider both cases.
Let the point be outside first. Then the segment intersects the circle at some point. Let's connect and. The result is an inscribed (!) quadrilateral.
We already know about it that the sum of its opposite angles is equal, that is, and according to our condition.
It turns out that it should be so that.
But this cannot possibly be the case since - is an external angle for and means .
What about inside? Let's do similar things. Let the point be inside.
Then the continuation of the segment intersects the circle at a point. Again - an inscribed quadrilateral, and according to the condition it must be satisfied, but - an external angle for and means, that is, again it cannot be that.
That is, a point cannot be either outside or inside the circle - that means it is on the circle!
The whole theorem has been proven!
Now let's see what good consequences this theorem gives.
Corollary 1
A parallelogram inscribed in a circle can only be a rectangle.
Let's understand why this is so. Let a parallelogram be inscribed in a circle. Then it should be done.
But from the properties of a parallelogram we know that.
And the same, naturally, regarding the angles and.
So it turns out to be a rectangle - all the corners are along.
But, in addition, there is an additional pleasant fact: the center of the circle circumscribed about the rectangle coincides with the point of intersection of the diagonals.
Let's understand why. I hope you remember very well that the angle subtended by the diameter is a straight line.
Diameter,
Diameter
which means it is the center. That's all.
Corollary 2
A trapezoid inscribed in a circle is isosceles.
Let the trapezoid be inscribed in a circle. Then.
And also.
Have we discussed everything? Not really. In fact, there is another, “secret” way to recognize an inscribed quadrilateral. We will not formulate this method very strictly (but clearly), but will prove it only at the last level of the theory.
If in a quadrilateral one can observe such a picture as here in the figure (here the angles “looking” at the side of the points and are equal), then such a quadrilateral is inscribed.
This is a very important drawing - in problems it is often easier to find equal angles than the sum of the angles and.
Despite the complete lack of rigor in our formulation, it is correct, and moreover, it is always accepted by the Unified State Exam examiners. You should write something like this:
“- inscribed” - and everything will be fine!
Don’t forget this important sign - remember the picture, and perhaps it will catch your eye in time when solving the problem.
Inscribed quadrilateral. Brief description and basic formulas
If a quadrilateral is inscribed in a circle, then the sum of any two of its opposite angles is equal to
and vice versa:
If a quadrilateral has two opposite angles whose sum is equal, then the quadrilateral is cyclic. 
A quadrilateral is inscribed in a circle if and only if the sum of its two opposite angles is equal.
Parallelogram inscribed in a circle- certainly a rectangle, and the center of the circle coincides with the intersection point of the diagonals.
A trapezoid inscribed in a circle is isosceles.
Definition 1. A quadrilateral is a figure consisting of four points (vertices), no three of which lie on the same straight line, and four consecutive non-intersecting segments (sides) connecting them.Definition 2. Neighboring vertices are those that are the ends of one side.
Definition 3. Vertices that are not adjacent are called opposite.
Definition 4. The segments connecting opposite vertices of a quadrilateral are called its diagonals.
Theorem 1. The sum of the angles of a quadrilateral is 360 degrees.
Indeed, dividing a quadrilateral by a diagonal into two triangles, we find that the sum of its angles is equal to the sum of the angles of these two triangles. Knowing that the sum of the angles of a triangle is 180 o, we get the desired: 2 * 180 o = 360 o
Definition d1. A circumscribed quadrilateral is a quadrilateral whose sides all touch a certain circle. Recall that the concept of a side tangent to a circle: a circle is considered to be tangent to a given side if it touches the line containing that side and the point of tangency lies on that side.
Definition d2. An inscribed quadrilateral is a quadrilateral whose vertices all belong to a certain circle.
Theorem 2. For any quadrilateral inscribed in a circle, the sum of pairs of opposite angles is equal to 180 degrees.
Angles A and C both rest on the arc BD only from different sides, that is, they cover the entire circle, and the circle itself is an arc measuring 360 o, but we know the theorem that states that the magnitude of the inscribed angle is equal to half the angular magnitude of the arc, on which it rests on, therefore we can state that the sum of these angles (A and C in particular) is equal to 180 o. In the same way, you can prove this theorem for another pair of angles.
Theorem 3.
If a circle can be inscribed in a quadrilateral, then the sums of the lengths of its opposite sides are equal. To prove this theorem, we will use the theorem from the topic circle and circle, which reads: Tangent segments drawn from one point to a circle are equal, i.e. VC=BP, CP=CH, DH=DT and AT=AK. Let's sum up the sides AB and CD: AB+CD=(AK+KB)+(DH+HC)=AT+BP+DT+CP=(AT+TD)+(BP+PC)=AD+BC, i.e. d.
There are converses for Theorems 2 and 3. Let's write them accordingly:
Theorem 4.
A circle can be described around a quadrilateral if and only if the sum of the opposite angles is equal to 180 degrees
Theorem 5.
A circle can be inscribed in a quadrilateral if and only if the sums of the lengths of opposite sides are equal. 
Proof: Let ABCD be the given quadrilateral and let AB + CD = AD + BC. Let us draw the bisectors of its angles A and D. These bisectors are not parallel, which means they intersect at some point O. Let us drop the perpendiculars OK, OL and OM from point O to the sides AB, AD and CD. Then OK=OL, and OL=OM, which means that a circle with center at point O and radius OK touches sides AB, AD and CD of this quadrilateral. Let's draw a tangent to this circle from point B. Let this tangent intersect the line CD at point P. Then ABPD is a circumscribed quadrilateral. Therefore, by the property of the circumscribed quadrilateral, AB + DP = AD + BP. Also, by condition, AB+ CD = AD + BC. Therefore, BP + PC = BC, which means, by the triangle inequality, point P lies on the segment BC. Consequently, lines BP and BC coincide, which means line BC touches a circle with center at point O, that is, ABCD is a circumscribed quadrilateral by definition. The theorem has been proven. 
Theorem 6.
The area of a quadrilateral is equal to half the product of its diagonals and the sine of the angle between them.
Proof: Let ABCD be the given quadrilateral. Let also O be the point of intersection of the diagonals. Then
S ABCD = S ABO + S BCO +S CDO + S DAO =
= 1/2(AO·BO·sin∠ AOB + BO·CO·sin∠ BOC +
+ CO·DO·sin∠ COD + DO·AO·sin∠ AOD) =
= 1/2 sin∠ BOC (AO + CO) (BO + DO) =
= 1/2·sin∠ BOC·AC·BD.
The theorem has been proven.
Theorem d1.
(Varignon) A quadrilateral with vertices at the midpoints of the sides of any quadrilateral is a parallelogram, and the area of this parallelogram is equal to half the area of the original quadrilateral. 
Proof: Let ABCD be a given quadrilateral and K, L, M and N the midpoints of its sides. Then KL is the midline of triangle ABC, which means KL is parallel to AC. Also LM is parallel to BD, MN is parallel to AC, and NK is parallel to BD. Therefore, KL is parallel to MN, LM is parallel to KN. So KLMN is a parallelogram. The area of this parallelogram is KL·KN·sin∠ NKL =
1/2 AC BD sin DOC = 1/2S ABCD .
The theorem has been proven.
INSCRIBED AND CIRCULAR POLYGONS,
§ 106. PROPERTIES OF INSCRIBED AND DESCRIBED QUADRIAGONS.
Theorem 1. The sum of the opposite angles of a cyclic quadrilateral is 180°.
Let a quadrilateral ABCD be inscribed in a circle with center O (Fig. 412). It is required to prove that / A+ / C = 180° and / B + / D = 180°.
/
A, as inscribed in circle O, measures 1/2 BCD.
/
C, as inscribed in the same circle, measures 1/2 BAD.
Consequently, the sum of angles A and C is measured by the half-sum of arcs BCD and BAD; in sum, these arcs make up a circle, i.e. they have 360°.
From here /
A+ /
C = 360°: 2 = 180°.
Similarly, it is proved that / B + / D = 180°. However, this can be deduced in another way. We know that the sum of the interior angles of a convex quadrilateral is 360°. The sum of angles A and C is equal to 180°, which means that the sum of the other two angles of the quadrilateral also remains 180°.
Theorem 2(reverse). If in a quadrilateral the sum of two opposite angles is equal 180° , then a circle can be described around such a quadrilateral.
Let the sum of the opposite angles of the quadrilateral ABCD be equal to 180°, namely
/
A+ /
C = 180° and /
B + /
D = 180° (drawing 412).
Let us prove that a circle can be described around such a quadrilateral.
Proof. Through any 3 vertices of this quadrilateral you can draw a circle, for example through points A, B and C. Where will point D be located?
Point D can only take one of the following three positions: be inside the circle, be outside the circle, be on the circumference of the circle.
Let's assume that the vertex is inside the circle and takes position D" (Fig. 413). Then in the quadrilateral ABCD" we will have:
/ B + / D" = 2 d.
Continuing side AD" to the intersection with the circle at point E and connecting points E and C, we obtain the cyclic quadrilateral ABCE, in which, by the direct theorem
/ B+ / E = 2 d.
From these two equalities it follows:
/
D" = 2 d - /
B;
/
E=2 d - /
B;
/ D" = / E,
but this cannot be, because / D", being external relative to the triangle CD"E, must be greater than angle E. Therefore, point D cannot be inside the circle.
It is also proved that vertex D cannot take position D" outside the circle (Fig. 414).
It remains to recognize that vertex D must lie on the circumference of the circle, i.e., coincide with point E, which means that a circle can be described around the quadrilateral ABCD.
Consequences. 1. A circle can be described around any rectangle.
2. A circle can be described around an isosceles trapezoid.
In both cases, the sum of opposite angles is 180°.
Theorem 3. In a circumscribed quadrilateral, the sums of opposite sides are equal. Let the quadrilateral ABCD be described about a circle (Fig. 415), that is, its sides AB, BC, CD and DA are tangent to this circle.
It is required to prove that AB + CD = AD + BC. Let us denote the points of tangency by the letters M, N, K, P. Based on the properties of tangents drawn to a circle from one point (§ 75), we have:
AR = AK;
VR = VM;
DN = DK;
CN = CM.
Let us add these equalities term by term. We get:
AR + BP + DN + CN = AK + VM + DK + SM,
i.e. AB + CD = AD + BC, which is what needed to be proven.
Exercises.
1. In an inscribed quadrilateral, two opposite angles are in the ratio 3:5,
and the other two are in the ratio 4:5. Determine the magnitude of these angles.
2. In the described quadrilateral, the sum of two opposite sides is 45 cm. The remaining two sides are in the ratio 0.2: 0.3. Find the length of these sides.
