Diagonals of a trapezoid. Midline of trapezoid
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- The segment connecting the midpoints of the diagonals of a trapezoid is equal to half the difference of the bases
- Triangles formed by the bases of a trapezoid and the segments of the diagonals up to their point of intersection are similar
- Triangles formed by segments of the diagonals of a trapezoid, the sides of which lie on the lateral sides of the trapezoid - are equal in size (have the same area)
- If you extend the sides of the trapezoid towards the smaller base, then they will intersect at one point with the straight line connecting the midpoints of the bases
- A segment connecting the bases of a trapezoid and passing through the point of intersection of the diagonals of the trapezoid is divided by this point in a proportion equal to the ratio of the lengths of the bases of the trapezoid
- A segment parallel to the bases of the trapezoid and drawn through the point of intersection of the diagonals is divided in half by this point, and its length is equal to 2ab/(a + b), where a and b are the bases of the trapezoid
Properties of a segment connecting the midpoints of the diagonals of a trapezoid
Let's connect the midpoints of the diagonals of the trapezoid ABCD, as a result of which we will have a segment LM.
A segment connecting the midpoints of the diagonals of a trapezoid lies on the midline of the trapezoid.
This segment parallel to the bases of the trapezoid.
The length of the segment connecting the midpoints of the diagonals of a trapezoid is equal to half the difference of its bases.
LM = (AD - BC)/2
or
LM = (a-b)/2
Properties of triangles formed by the diagonals of a trapezoid
Triangles that are formed by the bases of a trapezoid and the point of intersection of the diagonals of the trapezoid - are similar.
Triangles BOC and AOD are similar. Since angles BOC and AOD are vertical, they are equal.
Angles OCB and OAD are internal angles lying crosswise with parallel lines AD and BC (the bases of the trapezoid are parallel to each other) and a secant line AC, therefore they are equal.
Angles OBC and ODA are equal for the same reason (internal crosswise).
Since all three angles of one triangle are equal to the corresponding angles of another triangle, then these triangles are similar.
What follows from this?
To solve problems in geometry, the similarity of triangles is used as follows. If we know the lengths of two corresponding elements of similar triangles, then we find the similarity coefficient (we divide one by the other). From where the lengths of all other elements are related to each other by exactly the same value.
Properties of triangles lying on the lateral side and diagonals of a trapezoid
Consider two triangles lying on the lateral sides of the trapezoid AB and CD. These are triangles AOB and COD. Despite the fact that the sizes of individual sides of these triangles may be completely different, but the areas of the triangles formed by the lateral sides and the point of intersection of the diagonals of the trapezoid are equal, that is, the triangles are equal in size.
If we extend the sides of the trapezoid towards the smaller base, then the point of intersection of the sides will be coincide with a straight line that passes through the middle of the bases.
Thus, any trapezoid can be expanded into a triangle. In this case:
- Triangles formed by the bases of a trapezoid with a common vertex at the point of intersection of the extended sides are similar
- The straight line connecting the midpoints of the bases of the trapezoid is, at the same time, the median of the constructed triangle
Properties of a segment connecting the bases of a trapezoid
If you draw a segment whose ends lie on the bases of a trapezoid, which lies at the point of intersection of the diagonals of the trapezoid (KN), then the ratio of its constituent segments from the side of the base to the point of intersection of the diagonals (KO/ON) will be equal to the ratio of the bases of the trapezoid(BC/AD).
KO/ON = BC/AD
This property follows from the similarity of the corresponding triangles (see above).
Properties of a segment parallel to the bases of a trapezoid
If we draw a segment parallel to the bases of the trapezoid and passing through the point of intersection of the trapezoid’s diagonals, then it will have the following properties:
- Specified distance (KM) bisected by the intersection point of the trapezoid's diagonals
- Section length passing through the point of intersection of the diagonals of the trapezoid and parallel to the bases is equal to KM = 2ab/(a + b)
Formulas for finding the diagonals of a trapezoid
a, b- trapezoid bases
c, d- sides of the trapezoid
d1 d2- diagonals of a trapezoid
α β - angles with a larger base of the trapezoid
Formulas for finding the diagonals of a trapezoid through the bases, sides and angles at the base
The first group of formulas (1-3) reflects one of the main properties of trapezoid diagonals:
1. The sum of the squares of the diagonals of a trapezoid is equal to the sum of the squares of the sides plus twice the product of its bases. This property of trapezoid diagonals can be proven as a separate theorem
2 . This formula is obtained by transforming the previous formula. The square of the second diagonal is thrown through the equal sign, after which the square root is extracted from the left and right sides of the expression.
3 . This formula for finding the length of the diagonal of a trapezoid is similar to the previous one, with the difference that another diagonal is left on the left side of the expression
The next group of formulas (4-5) are similar in meaning and express a similar relationship.
The group of formulas (6-7) allows you to find the diagonal of a trapezoid if the larger base of the trapezoid, one side side and the angle at the base are known.
Formulas for finding the diagonals of a trapezoid through height
Task.
The diagonals of the trapezoid ABCD (AD | | BC) intersect at point O. Find the length of the base BC of the trapezoid if the base AD = 24 cm, length AO = 9 cm, length OS = 6 cm.
Solution.
The solution to this problem is ideologically absolutely identical to the previous problems.
Triangles AOD and BOC are similar in three angles - AOD and BOC are vertical, and the remaining angles are pairwise equal, since they are formed by the intersection of one line and two parallel lines.
Since the triangles are similar, all their geometric dimensions are related to each other, just like the geometric dimensions of the segments AO and OC known to us according to the conditions of the problem. That is
AO/OC = AD/BC
9 / 6 = 24 / BC
BC = 24 * 6 / 9 = 16
Answer: 16 cm
Task .
In the trapezoid ABCD it is known that AD=24, BC=8, AC=13, BD=5√17. Find the area of the trapezoid. 
Solution .
To find the height of a trapezoid from the vertices of the smaller base B and C, we lower two heights to the larger base. Since the trapezoid is unequal, we denote the length AM = a, length KD = b ( not to be confused with the notation in the formula finding the area of a trapezoid). Since the bases of the trapezoid are parallel, and we dropped two heights perpendicular to the larger base, then MBCK is a rectangle.
Means
AD = AM+BC+KD
a + 8 + b = 24
a = 16 - b
Triangles DBM and ACK are rectangular, so their right angles are formed by the altitudes of the trapezoid. Let us denote the height of the trapezoid by h. Then, by the Pythagorean theorem
H 2 + (24 - a) 2 = (5√17) 2
And
h 2 + (24 - b) 2 = 13 2
Let's take into account that a = 16 - b, then in the first equation
h 2 + (24 - 16 + b) 2 = 425
h 2 = 425 - (8 + b) 2
Let's substitute the value of the square of the height into the second equation obtained using the Pythagorean Theorem. We get:
425 - (8 + b) 2 + (24 - b) 2 = 169
-(64 + 16b + b) 2 + (24 - b) 2 = -256
-64 - 16b - b 2 + 576 - 48b + b 2 = -256
-64b = -768
b = 12
So KD = 12
Where
h 2 = 425 - (8 + b) 2 = 425 - (8 + 12) 2 = 25
h = 5
Find the area of the trapezoid through its height and half the sum of the bases
, where a b - the base of the trapezoid, h - the height of the trapezoid
S = (24 + 8) * 5 / 2 = 80 cm 2
Answer: the area of the trapezoid is 80 cm2.
Lesson objectives:
1) introduce students to the concept of the midline of a trapezoid, consider its properties and prove them;
2) teach how to build the midline of the trapezoid;
3) develop students’ ability to use the definition of the midline of a trapezoid and the properties of the midline of a trapezoid when solving problems;
4) continue to develop students’ ability to speak competently, using the necessary mathematical terms; prove your point of view;
5) develop logical thinking, memory, attention.
Lesson progress
1. Homework is checked during the lesson. The homework was oral, remember:
a) definition of a trapezoid; types of trapezoids;
b) determining the midline of the triangle;
c) property of the midline of a triangle;
d) sign of the middle line of the triangle.
2. Studying new material.
a) The board shows a trapezoid ABCD.
b) The teacher asks you to remember the definition of a trapezoid. Each desk has a hint diagram to help you remember the basic concepts in the topic “Trapezoid” (see Appendix 1). Appendix 1 is issued to each desk.
Students draw the trapezoid ABCD in their notebooks.
c) The teacher asks you to remember in which topic the concept of a midline was encountered (“Midline of a triangle”). Students recall the definition of the midline of a triangle and its properties.
e) Write down the definition of the midline of the trapezoid, drawing it in a notebook.
Middle line A trapezoid is a segment connecting the midpoints of its sides.
The property of the midline of a trapezoid remains unproven at this stage, so the next stage of the lesson involves working on proving the property of the midline of a trapezoid.
Theorem. Middle line of a trapezoid is parallel to its bases and equal to their half-sum.
Given: ABCD – trapezoid,
MN – middle line ABCD
Prove, What:
1. BC || MN || A.D.
2. MN = (AD + BC).
We can write down some corollaries that follow from the conditions of the theorem:
AM = MB, CN = ND, BC || A.D.
It is impossible to prove what is required based on the listed properties alone. The system of questions and exercises should lead students to the desire to connect the midline of a trapezoid with the midline of some triangle, the properties of which they already know. If there are no proposals, then you can ask the question: how to construct a triangle for which the segment MN would be the midline?
Let us write down an additional construction for one of the cases.
Let us draw a straight line BN intersecting the continuation of side AD at point K.
Additional elements appear - triangles: ABD, BNM, DNK, BCN. If we prove that BN = NK, then this will mean that MN is the midline of ABD, and then we can use the property of the midline of a triangle and prove the necessary.
Proof:
1. Consider BNC and DNK, they contain:
a) CNB =DNK (property of vertical angles);
b) BCN = NDK (property of internal cross-lying angles);
c) CN = ND (by corollary to the conditions of the theorem).
This means BNC =DNK (by the side and two adjacent angles).
Q.E.D.
The proof can be done orally in class, and reconstructed and written down in a notebook at home (at the teacher’s discretion).
It is necessary to say about other possible ways of proving this theorem:
1. Draw one of the diagonals of the trapezoid and use the sign and property of the triangle’s midline.
2. Carry out CF || BA and consider the parallelogram ABCF and DCF.
3. Carry out EF || BA and consider the equality of FND and ENC.
g) At this stage it is specified homework: paragraph 84, textbook ed. Atanasyan L.S. (proof of the property of the midline of a trapezoid using a vector method), write it down in your notebook.
h) We solve problems using the definition and properties of the midline of a trapezoid using ready-made drawings (see Appendix 2). Appendix 2 is given to each student, and the solution to the problems is written out on the same sheet in a short form.
In this article we will try to reflect the properties of a trapezoid as fully as possible. In particular, we will talk about 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 the isosceles and rectangular trapezoid.
An example of solving a problem using the properties discussed will help you sort it out in your head and better remember the material.
Trapeze and all-all-all
To begin with, let us briefly recall what a trapezoid is and what other concepts are associated with it.
So, a trapezoid is a quadrilateral figure, two of whose sides are parallel to each other (these are the bases). And the two are not parallel - these are the sides.
In a trapezoid, the height can be lowered - perpendicular to the bases. The center line and diagonals are drawn. It is also possible to draw a bisector from any angle of the trapezoid.
We will now talk about the various properties associated with all these elements and their combinations.
Properties of trapezoid diagonals
To make it clearer, while you are reading, sketch out the trapezoid ACME 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 HT lies on the midline. And its length can be obtained by dividing the difference of the bases by two: ХТ = (a – b)/2.
- Before us is the same trapezoid ACME. The diagonals intersect at point O. Let's look at the triangles AOE and MOK, formed by segments of the diagonals together with the bases of the trapezoid. These triangles are similar. The similarity coefficient k of triangles is expressed through the ratio of the bases of the trapezoid: k = AE/KM.
The ratio of the areas of triangles AOE and MOK is described by the coefficient k 2 . - The same trapezoid, the same diagonals intersecting at point O. Only this time we will consider the triangles that the segments of the diagonals formed together with the sides of the trapezoid. The areas of triangles AKO and EMO are equal in size - their areas are the same.
- Another property of a trapezoid involves the construction of diagonals. So, if you continue the sides of AK and ME in the direction of the smaller base, then sooner or later they will intersect at a certain point. Next, draw a straight line through the middle 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 connect together the point of intersection of the diagonals of the trapezoid O, the point at which the extensions of the sides and the middle of the bases X and T intersect. - Through the point of intersection of the diagonals we will draw a segment that will connect the bases of the trapezoid (T lies on the smaller base KM, X on the larger AE). The intersection point of the diagonals divides this segment in the following ratio: TO/OX = KM/AE.
- Now, through the point of intersection of the diagonals, we will 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 the segment using the formula 2ab/(a + b).
Properties of the midline of a trapezoid
Draw the middle line in the trapezoid 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.
Trapezoid Bisector Property
Select any angle of the trapezoid and draw a bisector. Let's take, for example, the angle KAE of our trapezoid ACME. Having completed the construction yourself, you can easily verify 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.
Properties of trapezoid angles
- Whichever of the two pairs of angles adjacent to the side you choose, the sum of the angles in the pair is always 180 0: α + β = 180 0 and γ + δ = 180 0.
- Let's 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 segment TX can be easily calculated based on the difference in the lengths of the bases, divided in half: TX = (AE – KM)/2.
- If parallel lines are drawn through the sides of a trapezoid angle, they will divide the sides of the angle into proportional segments.
Properties of an isosceles (equilateral) trapezoid
- IN isosceles trapezoid the angles are equal for any of the bases.
- Now build a trapezoid again to make it easier to imagine what we're talking about. Look carefully at the base AE - the vertex of the opposite base 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 middle line 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 around an isosceles trapezoid can a circle be described, since the sum of the opposite angles of a quadrilateral is 180 0 - prerequisite for this.
- The property of an isosceles trapezoid follows from the previous paragraph - if a circle can be described near the trapezoid, it is isosceles.
- From the features of an isosceles trapezoid follows the property of the height of a trapezoid: 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.
- Again, draw the segment TX 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 the height from the opposite vertex of the trapezoid onto the larger base (let's call it a). You will get two segments. 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 us dwell on this issue in more detail. In particular, on where the center of the circle is in relation to the trapezoid. Here, too, it is recommended that you take the time to pick up a pencil and draw what will be discussed below. This way 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, the diagonal may extend from the top of the 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).
- The diagonal and the side can also 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 larger base, if there is an obtuse angle between the diagonal of the trapezoid and the side.
- The angle formed by the diagonal and the large base of the trapezoid ACME (inscribed angle) is half the central angle that corresponds to it: MAE = ½MOE.
- Briefly about two ways to find the radius of a circumscribed circle. Method one: look carefully at your drawing - what do you see? You can easily notice that the diagonal splits the trapezoid into two triangles. The radius can be found by the ratio of the side of the triangle to the sine of the opposite angle, multiplied by two. For example, R = AE/2*sinAME. The formula can be written in a similar way for any of the sides of both triangles.
- Method two: find the radius of the circumscribed circle through the area of the triangle formed by the diagonal, side and base of the trapezoid: R = AM*ME*AE/4*S AME.
Properties of a trapezoid circumscribed about a circle
You can fit a circle into a trapezoid if one condition is met. Read 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 a trapezoid whose sum of bases is equal to the sum of its sides.
- The tangent point of a circle with radius r inscribed in a trapezoid divides the 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. To avoid confusion, draw this example yourself too. We have the good old trapezoid ACME, described around a circle. It contains diagonals that intersect at point O. The triangles AOK and EOM formed by the segments of the diagonals and the lateral sides are rectangular.
The heights of these triangles, lowered to the hypotenuses (i.e., the lateral sides of the trapezoid), coincide with the radii of the inscribed circle. And the height of the trapezoid coincides with the diameter of the inscribed circle.
Properties of a rectangular trapezoid
A trapezoid is called rectangular if one of its angles is right. And its properties stem from this circumstance.
- A rectangular trapezoid has one of its sides perpendicular to its base.
- Height and lateral side of the trapezoid adjacent to 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 diagonals of a trapezoid already described above are relevant.
Evidence of some properties of the trapezoid
Equality of angles at the base of an isosceles trapezoid:
- You probably already guessed that here we will need the AKME trapezoid again - draw an isosceles trapezoid. Draw a straight 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 does 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 trapezoid ACME is isosceles:
- First, let’s draw a straight line MX – MX || KE. We obtain a parallelogram KMHE (base – MX || KE and KM || EX).
∆AMX is isosceles, since AM = KE = MX, and MAX = MEA.
MH || KE, KEA = MXE, therefore MAE = MXE.
It turned out that the triangles AKE and EMA are equal to each other, since AM = KE and AE are the common side of the two triangles. And also MAE = MXE. We can conclude that AK = ME, and from this it follows that the trapezoid AKME is isosceles.
Review task
The bases of the trapezoid ACME are 9 cm and 21 cm, the side side KA, equal to 8 cm, forms an angle of 150 0 with the 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. This means that in total they give 180 0. Therefore, KAN = 30 0 (based on the property of trapezoidal angles).
Let us now consider the rectangular ∆ANC (I believe this point is obvious to readers without additional evidence). From it we will find the height of the trapezoid KH - in a triangle it is the leg that lies opposite the angle of 30 0. Therefore, KN = ½AB = 4 cm.
We find the area of the trapezoid using the formula: S ACME = (KM + AE) * KN/2 = (9 + 21) * 4/2 = 60 cm 2.
Afterword
If you carefully and thoughtfully studied this article, were not too lazy to draw trapezoids for all the given 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 have seen that the difference is huge.
Now you have a detailed summary of all general properties trapezoids. As well as specific properties and characteristics 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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The concept of the midline of the trapezoid
First, let's remember what kind of 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, the parallel sides are called the bases of the trapezoid, and the non-parallel sides are called the lateral sides of the trapezoid.
Definition 2
The midline of a trapezoid is a segment connecting the midpoints of the lateral sides of the trapezoid.
Trapezoid midline theorem
Now we introduce the theorem about the midline of a trapezoid and prove it using the vector method.
Theorem 1
The midline of the trapezoid is parallel to the bases and equal to their half-sum.
Proof.
Let us be given a trapezoid $ABCD$ with bases $AD\ and\ BC$. And let $MN$ be the middle line of this trapezoid (Fig. 1).
Figure 1. Midline of trapezoid
Let us prove that $MN||AD\ and\ MN=\frac(AD+BC)(2)$.
Consider the vector $\overrightarrow(MN)$. We next use the polygon rule to add vectors. On the one hand, we get that
On the other side
Let's add the last two equalities and get
Since $M$ and $N$ are the midpoints of the lateral sides of the trapezoid, we will have
We get:
Hence
From the same equality (since $\overrightarrow(BC)$ and $\overrightarrow(AD)$ are codirectional and, therefore, collinear) we obtain that $MN||AD$.
The theorem has been proven.
Examples of problems on the concept of the midline of a trapezoid
Example 1
The lateral 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.
Let us denote the midline of the trapezoid by $n$.
The sum of the sides is equal to
Therefore, since the perimeter is $52\ cm$, the sum of the bases is equal to
So, by Theorem 1, we get
Answer:$10\cm$.
Example 2
The ends of the circle's diameter are $9$ cm and $5$ cm away from its tangent, respectively. Find the diameter of this circle.
Solution.
Let us be given a circle with center at point $O$ and diameter $AB$. Let's draw a 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 $OH\bot l$, therefore, $OH |\left|AD\right||BC$. From all this we get that $ABCD$ is a trapezoid, and $OH$ is its midline. By Theorem 1, we get
