The concept of the midline of the trapezoid
First, let's remember what figure is called a trapezoid.
Definition 1
A trapezoid is a quadrilateral in which two sides are parallel and the other two are not parallel.
In this case, parallel sides are called the bases of the trapezoid, and not parallel - the sides of the trapezoid.
Definition 2
The midline of a trapezoid is a line segment that connects the midpoints of the sides of the trapezoid.
Trapezium midline theorem
We now introduce the theorem on the midline of a trapezoid and prove it by the vector method.
Theorem 1
The median line of the trapezoid is parallel to the bases and equal to half their sum.
Proof.
Let us be given a trapezoid $ABCD$ with bases $AD\ and\ BC$. And let $MN$ be the midline of this trapezoid (Fig. 1).
Figure 1. The middle line of the trapezoid
Let us prove that $MN||AD\ and\ MN=\frac(AD+BC)(2)$.
Consider the vector $\overrightarrow(MN)$. Next, we use the polygon rule for vector addition. On the one hand, we get that
On the other side
Adding the last two equalities, we get
Since $M$ and $N$ are the midpoints of the sides of the trapezoid, we have
We get:
Hence
From the same equality (since $\overrightarrow(BC)$ and $\overrightarrow(AD)$ are codirectional and, therefore, collinear), we get that $MN||AD$.
The theorem has been proven.
Examples of tasks on the concept of the midline of a trapezoid
Example 1
The sides of the trapezoid are $15\cm$ and $17\cm$ respectively. The perimeter of the trapezoid is $52\cm$. Find the length of the midline of the trapezoid.
Solution.
Denote the midline of the trapezoid by $n$.
The sum of the sides is
Therefore, since the perimeter is $52\ cm$, the sum of the bases is
Hence, by Theorem 1, we obtain
Answer:$10\cm$.
Example 2
The ends of the circle's diameter are $9$ cm and $5$ cm respectively from its tangent. Find the diameter of this circle.
Solution.
Let us be given a circle with center $O$ and diameter $AB$. Draw the tangent $l$ and construct the distances $AD=9\ cm$ and $BC=5\ cm$. Let's draw the radius $OH$ (Fig. 2).
Figure 2.
Since $AD$ and $BC$ are the distances to the tangent, then $AD\bot l$ and $BC\bot l$ and since $OH$ is the radius, then $OH\bot l$, hence $OH |\left|AD\right||BC$. From all this we get that $ABCD$ is a trapezoid, and $OH$ is its midline. By Theorem 1, we get
A quadrilateral with only two parallel sides is called trapeze.
The parallel sides of a trapezoid are called its grounds, and those sides that are not parallel are called sides. If the sides are equal, then such a trapezoid is isosceles. The distance between the bases is called the height of the trapezoid.
Middle line of the trapezium
The median line is a segment connecting the midpoints of the sides of the trapezoid. The midline of a trapezoid is parallel to its bases.
Theorem:
If a line intersecting the middle of one side is parallel to the bases of the trapezoid, then it bisects the second side of the trapezoid.
Theorem:
The length of the midline is equal to the arithmetic mean of the lengths of its bases
MN || AB || DCAM=MD; BN=NC
MN midline, AB and CD - bases, AD and BC - sides
MN=(AB+DC)/2
Theorem:
The length of the midline of a trapezoid is equal to the arithmetic mean of the lengths of its bases.
The main task: Prove that the midline of a trapezoid bisects a segment whose ends lie in the middle of the bases of the trapezoid.
Middle Line of the Triangle
The line segment connecting the midpoints of the two sides of a triangle is called the midline of the triangle. It is parallel to the third side and its length is half the length of the third side.
Theorem: If a line intersecting the midpoint of one side of a triangle is parallel to the other side of the given triangle, then it bisects the third side.
AM = MC and BN = NC =>
Applying Triangle and Trapezoid Midline Properties
Dividing a segment by a certain amount equal parts.
Task: Divide segment AB into 5 equal parts.
Solution:
Let p be a random ray whose origin is point A and which does not lie on line AB. We sequentially set aside 5 equal segments on p AA 1 = A 1 A 2 = A 2 A 3 = A 3 A 4 = A 4 A 5
We connect A 5 to B and draw lines through A 4 , A 3 , A 2 and A 1 that are parallel to A 5 B. They intersect AB at B 4 , B 3 , B 2 and B 1 respectively. These points divide segment AB into 5 equal parts. Indeed, from the trapezoid BB 3 A 3 A 5 we see that BB 4 = B 4 B 3 . In the same way, from the trapezoid B 4 B 2 A 2 A 4 we get B 4 B 3 = B 3 B 2
While from the trapezoid B 3 B 1 A 1 A 3 , B 3 B 2 = B 2 B 1 .
Then from B 2 AA 2 it follows that B 2 B 1 = B 1 A. In conclusion, we get:
AB 1 = B 1 B 2 = B 2 B 3 = B 3 B 4 = B 4 B
It is clear that in order to divide the segment AB into another number of equal parts, we need to project the same number of equal segments onto the ray p. And then continue in the manner described above.
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middle line figures in planimetry - a segment connecting the midpoints of the two sides of a given figure. The concept is used for the following figures: triangle, quadrilateral, trapezium.
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Subtitles
Middle line of the triangle
Properties
- the midline of a triangle is parallel to the base and equal to half of it.
- at the intersection of all three middle lines, 4 equal triangles are formed, similar (even homothetic) to the original one with a coefficient of 1/2.
- the middle line cuts off a triangle that is similar to the given one, and its area is equal to one fourth of the area of the original triangle.
- The three middle lines of the triangle divide it into 4 equal (identical) triangles similar to the original triangle. All 4 such identical triangles are called medial triangles. The central one of these 4 identical triangles is called the complementary triangle.
signs
- if the segment is parallel to one of the sides of the triangle and connects the midpoint of one side of the triangle with a point lying on the other side of the triangle, then this is the midline.
Middle line of the quadrilateral
Middle line of the quadrilateral A line segment that joins the midpoints of opposite sides of a quadrilateral.
Properties
The first line connects 2 opposite sides. The second connects 2 other opposite sides. The third one connects the centers of the two diagonals (not in all quadrilaterals the diagonals are bisected by the intersection point).
- If in a convex quadrilateral the midline forms equal angles with the diagonals of the quadrilateral, then the diagonals are equal.
- The length of the midline of a quadrilateral is less than or equal to half the sum of the other two sides if these sides are parallel, and only in this case.
- The midpoints of the sides of an arbitrary quadrilateral are the vertices of a parallelogram. Its area is equal to half the area of the quadrilateral, and its center lies at the point of intersection of the median lines. This parallelogram is called the Varignon parallelogram;
- The last point means the following: In a convex quadrilateral, four middle lines of the second kind. Middle lines of the second kind- four segments inside the quadrilateral passing through its midpoints adjacent parties parallel to the diagonals. Four middle lines of the second kind convex quadrilateral cut it into four triangles and one central quadrilateral. This central quadrilateral is a parallelogram of Varignon.
- The point of intersection of the midlines of the quadrilateral is their common midpoint and bisects the segment connecting the midpoints of the diagonals. In addition, she is
In this article, we will try to reflect the properties of the trapezoid as fully as possible. In particular, we will talk about the general signs and properties of a trapezoid, as well as about the properties of an inscribed trapezoid and about a circle inscribed in a trapezoid. We will also touch on the properties of an isosceles and rectangular trapezoid.
An example of solving a problem using the considered properties will help you sort things out in your head and better remember the material.
Trapeze and all-all-all
To begin with, let's briefly recall what a trapezoid is and what other concepts are associated with it.
So, a trapezoid is a quadrilateral figure, two of the sides of which are parallel to each other (these are the bases). And two are not parallel - these are the sides.
In a trapezoid, the height can be omitted - perpendicular to the bases. The middle line and diagonals are drawn. And also from any angle of the trapezoid it is possible to draw a bisector.
About the various properties associated with all these elements and their combinations, we will now talk.
Properties of the diagonals of a trapezoid
To make it clearer, while reading, sketch out the ACME trapezoid on a piece of paper and draw diagonals in it.
- If you find the midpoints of each of the diagonals (let's call these points X and T) and connect them, you get a segment. One of the properties of the diagonals of a trapezoid is that the segment XT lies on the midline. And its length can be obtained by dividing the difference of the bases by two: XT \u003d (a - b) / 2.
- Before us is the same ACME trapezoid. The diagonals intersect at point O. Let's consider the triangles AOE and IOC formed by the segments of the diagonals together with the bases of the trapezoid. These triangles are similar. The similarity coefficient of k triangles is expressed in terms of the ratio of the bases of the trapezoid: k = AE/KM.
The ratio of the areas of triangles AOE and IOC is described by the coefficient k 2 . - All the same trapezoid, the same diagonals intersecting at point O. Only this time we will consider triangles that the diagonal segments formed together with the sides of the trapezoid. The areas of triangles AKO and EMO are equal - their areas are the same.
- Another property of a trapezoid includes the construction of diagonals. So, if we continue the sides of AK and ME in the direction of the smaller base, then sooner or later they will intersect to some point. Next, draw a straight line through the midpoints of the bases of the trapezoid. It intersects the bases at points X and T.
If we now extend the line XT, then it will join together the point of intersection of the diagonals of the trapezoid O, the point at which the extensions of the sides and the midpoints of the bases of X and T intersect. - Through the point of intersection of the diagonals, we draw a segment that will connect the bases of the trapezoid (T lies on the smaller base of KM, X - on the larger AE). The intersection point of the diagonals divides this segment in the following ratio: TO/OH = KM/AE.
- And now through the point of intersection of the diagonals we draw a segment parallel to the bases of the trapezoid (a and b). The intersection point will divide it into two equal parts. You can find the length of a segment using the formula 2ab/(a + b).
Properties of the midline of a trapezoid
Draw the middle line in the trapezium parallel to its bases.
- The length of the midline of a trapezoid can be calculated by adding the lengths of the bases and dividing them in half: m = (a + b)/2.
- If you draw any segment (height, for example) through both bases of the trapezoid, the middle line will divide it into two equal parts.
Property of the bisector of a trapezoid
Pick any angle of the trapezoid and draw a bisector. Take, for example, the angle KAE of our trapezoid ACME. Having completed the construction on your own, you can easily see that the bisector cuts off from the base (or its continuation on a straight line outside the figure itself) a segment of the same length as the side.
Trapezoid angle properties
- Whichever of the two pairs of angles adjacent to the side you choose, the sum of the angles in a pair is always 180 0: α + β = 180 0 and γ + δ = 180 0 .
- Connect the midpoints of the bases of the trapezoid with a segment TX. Now let's look at the angles at the bases of the trapezoid. If the sum of the angles for any of them is 90 0, the length of the TX segment is easy to calculate based on the difference in the lengths of the bases, divided in half: TX \u003d (AE - KM) / 2.
- If parallel lines are drawn through the sides of the angle of a trapezoid, they will divide the sides of the angle into proportional segments.
Properties of an isosceles (isosceles) trapezoid
- In an isosceles trapezoid, the angles at any of the bases are equal.
- Now build a trapezoid again to make it easier to imagine what it is about. Look carefully at the base of AE - the vertex of the opposite base of M is projected to a certain point on the line that contains AE. The distance from vertex A to the projection point of vertex M and the midline of an isosceles trapezoid are equal.
- A few words about the property of the diagonals of an isosceles trapezoid - their lengths are equal. And also the angles of inclination of these diagonals to the base of the trapezoid are the same.
- Only near an isosceles trapezoid can a circle be described, since the sum of the opposite angles of a quadrilateral 180 0 is a prerequisite for this.
- The property of an isosceles trapezoid follows from the previous paragraph - if a circle can be described near a trapezoid, it is isosceles.
- From the features of an isosceles trapezoid, the property of the height of a trapezoid follows: if its diagonals intersect at right angles, then the length of the height is equal to half the sum of the bases: h = (a + b)/2.
- Draw the line TX again through the midpoints of the bases of the trapezoid - in an isosceles trapezoid it is perpendicular to the bases. And at the same time, TX is the axis of symmetry of an isosceles trapezoid.
- This time lower to the larger base (let's call it a) the height from the opposite vertex of the trapezoid. You will get two cuts. The length of one can be found if the lengths of the bases are added and divided in half: (a+b)/2. We get the second one when we subtract the smaller one from the larger base and divide the resulting difference by two: (a – b)/2.
Properties of a trapezoid inscribed in a circle
Since we are already talking about a trapezoid inscribed in a circle, let's dwell on this issue in more detail. In particular, where is the center of the circle in relation to the trapezoid. Here, too, it is recommended not to be too lazy to pick up a pencil and draw what will be discussed below. So you will understand faster, and remember better.
- The location of the center of the circle is determined by the angle of inclination of the diagonal of the trapezoid to its side. For example, a diagonal may emerge from the top of a trapezoid at right angles to the side. In this case, the larger base intersects the center of the circumscribed circle exactly in the middle (R = ½AE).
- Diagonal and side can meet under acute angle- then the center of the circle is inside the trapezoid.
- The center of the circumscribed circle may be outside the trapezoid, beyond its large base, if there is an obtuse angle between the diagonal of the trapezoid and the lateral side.
- The angle formed by the diagonal and the large base of the trapezoid ACME (inscribed angle) is half that central corner, which corresponds to it: MAE = ½MY.
- Briefly about two ways to find the radius of the circumscribed circle. Method one: look carefully at your drawing - what do you see? You will easily notice that the diagonal splits the trapezoid into two triangles. The radius can be found through the ratio of the side of the triangle to the sine of the opposite angle, multiplied by two. For example, R \u003d AE / 2 * sinAME. Similarly, the formula can be written for any of the sides of both triangles.
- Method two: we find the radius of the circumscribed circle through the area of the triangle formed by the diagonal, side and base of the trapezoid: R \u003d AM * ME * AE / 4 * S AME.
Properties of a trapezoid circumscribed about a circle
You can inscribe a circle in a trapezoid if one condition is met. More about it below. And together this combination of figures has a number of interesting properties.
- If a circle is inscribed in a trapezoid, the length of its midline can be easily found by adding the lengths of the sides and dividing the resulting sum in half: m = (c + d)/2.
- For a trapezoid ACME, circumscribed about a circle, the sum of the lengths of the bases is equal to the sum of the lengths of the sides: AK + ME = KM + AE.
- From this property of the bases of a trapezoid, the converse statement follows: a circle can be inscribed in that trapezoid, the sum of the bases of which is equal to the sum of the sides.
- The tangent point of a circle with radius r inscribed in a trapezoid divides the lateral side into two segments, let's call them a and b. The radius of a circle can be calculated using the formula: r = √ab.
- And one more property. In order not to get confused, draw this example yourself. We have the good old ACME trapezoid, circumscribed around a circle. Diagonals are drawn in it, intersecting at the point O. The triangles AOK and EOM formed by the segments of the diagonals and the sides are rectangular.
The heights of these triangles, lowered to the hypotenuses (i.e., the sides of the trapezoid), coincide with the radii of the inscribed circle. And the height of the trapezoid is the same as the diameter of the inscribed circle.
Properties of a rectangular trapezoid
A trapezoid is called rectangular, one of the corners of which is right. And its properties stem from this circumstance.
- A rectangular trapezoid has one of the sides perpendicular to the bases.
- The height and side of the trapezoid adjacent to the right angle are equal. This allows you to calculate the area of a rectangular trapezoid ( general formula S = (a + b) * h/2) not only through the height, but also through the side adjacent to the right angle.
- For a rectangular trapezoid, the general properties of the trapezoid diagonals already described above are relevant.
Proofs of some properties of a trapezoid
Equality of angles at the base of an isosceles trapezoid:
- You probably already guessed that here we again need the ACME trapezoid - draw an isosceles trapezoid. Draw a line MT from vertex M parallel to the side of AK (MT || AK).
The resulting quadrilateral AKMT is a parallelogram (AK || MT, KM || AT). Since ME = KA = MT, ∆ MTE is isosceles and MET = MTE.
AK || MT, therefore MTE = KAE, MET = MTE = KAE.
Where AKM = 180 0 - MET = 180 0 - KAE = KME.
Q.E.D.
Now, based on the property of an isosceles trapezoid (equality of diagonals), we prove that trapezium ACME is isosceles:
- To begin with, let's draw a straight line МХ – МХ || KE. We get a parallelogram KMHE (base - MX || KE and KM || EX).
∆AMH is isosceles, since AM = KE = MX, and MAX = MEA.
MX || KE, KEA = MXE, therefore MAE = MXE.
It turned out that the triangles AKE and EMA are equal to each other, because AM \u003d KE and AE is the common side of the two triangles. And also MAE \u003d MXE. We can conclude that AK = ME, and hence it follows that the trapezoid AKME is isosceles.
Task to repeat
The bases of the trapezoid ACME are 9 cm and 21 cm, the side of the KA, equal to 8 cm, forms an angle of 150 0 with a smaller base. You need to find the area of the trapezoid.
Solution: From vertex K we lower the height to the larger base of the trapezoid. And let's start looking at the angles of the trapezoid.
Angles AEM and KAN are one-sided. Which means they add up to 1800. Therefore, KAN = 30 0 (based on the property of the angles of the trapezoid).
Consider now the rectangular ∆ANK (I think this point is obvious to readers without further proof). From it we find the height of the trapezoid KH - in a triangle it is a leg, which lies opposite the angle of 30 0. Therefore, KN \u003d ½AB \u003d 4 cm.
The area of the trapezoid is found by the formula: S AKME \u003d (KM + AE) * KN / 2 \u003d (9 + 21) * 4/2 \u003d 60 cm 2.
Afterword
If you carefully and thoughtfully studied this article, were not too lazy to draw trapezoids for all the above properties with a pencil in your hands and analyze them in practice, you should have mastered the material well.
Of course, there is a lot of information here, varied and sometimes even confusing: it is not so difficult to confuse the properties of the described trapezoid with the properties of the inscribed one. But you yourself saw that the difference is huge.
Now you have a detailed summary of all common properties trapezoid. As well as specific properties and features of isosceles and rectangular trapezoids. It is very convenient to use to prepare for tests and exams. Try it yourself and share the link with your friends!
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