If the trapezoid is isosceles, then the diagonals. Useful properties of trapezoid

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The geometry course for the 8th grade involves the study of the properties and characteristics of convex quadrilaterals. These include parallelograms, special cases of which are squares, rectangles and rhombuses, and trapezoids. And if solving problems on various variations of a parallelogram most often does not cause much difficulty, then figuring out which quadrilateral is called a trapezoid is somewhat more difficult.

Definition and types

Unlike other quadrilaterals studied in the school curriculum, a trapezoid is usually called such a figure, two opposite sides of which are parallel to each other, and the other two are not. There is another definition: it is a quadrilateral with a pair of sides that are unequal and parallel.

The different types are shown in the picture below.

Image number 1 shows an arbitrary trapezoid. Number 2 indicates a special case - a rectangular trapezoid, one of the sides of which is perpendicular to its bases. The last figure is also a special case: it is an isosceles (equilateral) trapezoid, that is, a quadrilateral with equal sides.

The most important properties and formulas

To describe the properties of a quadrilateral, it is customary to highlight certain elements. As an example, consider an arbitrary trapezoid ABCD.

It includes:

  • bases BC and AD - two sides parallel to each other;
  • the sides AB and CD are two non-parallel elements;
  • diagonals AC and BD are segments connecting opposite vertices of the figure;
  • the height of the trapezoid CH is a segment perpendicular to the bases;
  • midline EF - line connecting the midpoints of the lateral sides.

Basic properties of elements

To solve geometry problems or prove any statements, the properties that connect the various elements of a quadrilateral are most often used. They are formulated as follows:

In addition, it is often useful to know and apply the following statements:

  1. A bisector drawn from an arbitrary angle separates a segment at the base, the length of which is equal to the side of the figure.
  2. When drawing diagonals, 4 triangles are formed; Of these, 2 triangles formed by the bases and segments of the diagonals are similar, and the remaining pair has the same area.
  3. Through the point of intersection of the diagonals O, the midpoints of the bases, as well as the point at which the extensions of the sides intersect, a straight line can be drawn.

Calculation of perimeter and area

The perimeter is calculated as the sum of the lengths of all four sides (similar to any other geometric figure):

P = AD + BC + AB + CD.

Inscribed and circumscribed circle

A circle can be described around a trapezoid only if the sides of the quadrilateral are equal.

To calculate the radius of a circumscribed circle, you need to know the lengths of the diagonal, side, and larger base. Magnitude p, used in the formula is calculated as half the sum of all the above elements: p = (a + c + d)/2.

For an inscribed circle, the condition will be as follows: the sum of the bases must coincide with the sum of the sides of the figure. Its radius can be found through the height, and it will be equal to r = h/2.

Special cases

Let's consider a frequently encountered case - an isosceles (equilateral) trapezoid. Its signs are the equality of the lateral sides or the equality of opposite angles. All statements apply to her, which are characteristic of an arbitrary trapezoid. Other properties of an isosceles trapezoid:

The rectangular trapezoid is not found very often in problems. Its signs are the presence of two adjacent angles equal to 90 degrees, and the presence of a side perpendicular to the bases. The height in such a quadrilateral is also one of its sides.

All the properties and formulas considered are usually used to solve planimetric problems. However, they also have to be used in some problems from a stereometry course, for example, when determining the surface area of ​​a truncated pyramid that looks like a volumetric trapezoid.

\[(\Large(\text(Free trapezoid)))\]

Definitions

A trapezoid is a convex quadrilateral in which two sides are parallel and the other two sides are not parallel.

The parallel sides of a trapezoid are called its bases, and the other two sides are called its lateral sides.

The height of a trapezoid is the perpendicular drawn from any point of one base to another base.

Theorems: properties of a trapezoid

1) The sum of the angles at the side is \(180^\circ\) .

2) The diagonals divide the trapezoid into four triangles, two of which are similar, and the other two are equal in size.

Proof

1) Because \(AD\parallel BC\), then the angles \(\angle BAD\) and \(\angle ABC\) are one-sided for these lines and the transversal \(AB\), therefore, \(\angle BAD +\angle ABC=180^\circ\).

2) Because \(AD\parallel BC\) and \(BD\) are a secant, then \(\angle DBC=\angle BDA\) lie crosswise.
Also \(\angle BOC=\angle AOD\) as vertical.
Therefore, at two angles \(\triangle BOC \sim \triangle AOD\).

Let's prove that \(S_(\triangle AOB)=S_(\triangle COD)\). Let \(h\) be the height of the trapezoid. Then \(S_(\triangle ABD)=\frac12\cdot h\cdot AD=S_(\triangle ACD)\). Then: \

Definition

The midline of a trapezoid is a segment connecting the midpoints of the sides.

Theorem

The midline of the trapezoid is parallel to the bases and equal to their half-sum.


Proof*

1) Let's prove parallelism.


Let us draw through the point \(M\) the straight line \(MN"\parallel AD\) (\(N"\in CD\) ). Then, according to Thales’ theorem (since \(MN"\parallel AD\parallel BC, AM=MB\)) point \(N"\) is the middle of the segment \(CD\). This means that the points \(N\) and \(N"\) will coincide.

2) Let's prove the formula.

Let's do \(BB"\perp AD, CC"\perp AD\) . Let \(BB"\cap MN=M", CC"\cap MN=N"\).


Then, by Thales' theorem, \(M"\) and \(N"\) are the midpoints of the segments \(BB"\) and \(CC"\), respectively. This means that \(MM"\) is the middle line of \(\triangle ABB"\) , \(NN"\) is the middle line of \(\triangle DCC"\) . That's why: \

Because \(MN\parallel AD\parallel BC\) and \(BB", CC"\perp AD\), then \(B"M"N"C"\) and \(BM"N"C\) are rectangles. According to Thales' theorem, from \(MN\parallel AD\) and \(AM=MB\) it follows that \(B"M"=M"B\) . Hence, \(B"M"N"C"\) and \(BM"N"C\) are equal rectangles, therefore, \(M"N"=B"C"=BC\) .

Thus:

\ \[=\dfrac12 \left(AB"+B"C"+BC+C"D\right)=\dfrac12\left(AD+BC\right)\]

Theorem: property of an arbitrary trapezoid

The midpoints of the bases, the point of intersection of the diagonals of the trapezoid and the point of intersection of the extensions of the lateral sides lie on the same straight line.


Proof*
It is recommended that you familiarize yourself with the proof after studying the topic “Similarity of triangles”.

1) Let us prove that the points \(P\) , \(N\) and \(M\) lie on the same line.


Let's draw a straight line \(PN\) (\(P\) is the point of intersection of the extensions of the lateral sides, \(N\) is the middle of \(BC\)). Let it intersect the side \(AD\) at the point \(M\) . Let us prove that \(M\) is the midpoint of \(AD\) .

Consider \(\triangle BPN\) and \(\triangle APM\) . They are similar at two angles (\(\angle APM\) – general, \(\angle PAM=\angle PBN\) as corresponding at \(AD\parallel BC\) and \(AB\) secant). Means: \[\dfrac(BN)(AM)=\dfrac(PN)(PM)\]

Consider \(\triangle CPN\) and \(\triangle DPM\) . They are similar at two angles (\(\angle DPM\) – general, \(\angle PDM=\angle PCN\) as corresponding at \(AD\parallel BC\) and \(CD\) secant). Means: \[\dfrac(CN)(DM)=\dfrac(PN)(PM)\]

From here \(\dfrac(BN)(AM)=\dfrac(CN)(DM)\). But \(BN=NC\) therefore \(AM=DM\) .

2) Let us prove that the points \(N, O, M\) lie on the same line.


Let \(N\) be the midpoint of \(BC\) and \(O\) be the point of intersection of the diagonals. Let's draw a straight line \(NO\) , it will intersect the side \(AD\) at the point \(M\) . Let us prove that \(M\) is the midpoint of \(AD\) .

\(\triangle BNO\sim \triangle DMO\) along two angles (\(\angle OBN=\angle ODM\) lying crosswise at \(BC\parallel AD\) and \(BD\) secant; \(\angle BON=\angle DOM\) as vertical). Means: \[\dfrac(BN)(MD)=\dfrac(ON)(OM)\]

Likewise \(\triangle CON\sim \triangle AOM\). Means: \[\dfrac(CN)(MA)=\dfrac(ON)(OM)\]

From here \(\dfrac(BN)(MD)=\dfrac(CN)(MA)\). But \(BN=CN\) therefore \(AM=MD\) .

\[(\Large(\text(Isosceles trapezoid)))\]

Definitions

A trapezoid is called rectangular if one of its angles is right.

A trapezoid is called isosceles if its sides are equal.

Theorems: properties of an isosceles trapezoid

1) An isosceles trapezoid has equal base angles.

2) The diagonals of an isosceles trapezoid are equal.

3) Two triangles formed by diagonals and a base are isosceles.

Proof

1) Consider the isosceles trapezoid \(ABCD\) .

From the vertices \(B\) and \(C\), we drop the perpendiculars \(BM\) and \(CN\) to the side \(AD\), respectively. Since \(BM\perp AD\) and \(CN\perp AD\) , then \(BM\parallel CN\) ; \(AD\parallel BC\) , then \(MBCN\) is a parallelogram, therefore, \(BM = CN\) .

Consider the right triangles \(ABM\) and \(CDN\) . Since their hypotenuses are equal and the leg \(BM\) is equal to the leg \(CN\) , then these triangles are equal, therefore, \(\angle DAB = \angle CDA\) .

2)

Because \(AB=CD, \angle A=\angle D, AD\)- general, then according to the first sign. Therefore, \(AC=BD\) .

3) Because \(\triangle ABD=\triangle ACD\), then \(\angle BDA=\angle CAD\) . Therefore, the triangle \(\triangle AOD\) is isosceles. Similarly, it is proved that \(\triangle BOC\) is isosceles.

Theorems: signs of an isosceles trapezoid

1) If a trapezoid has equal base angles, then it is isosceles.

2) If a trapezoid has equal diagonals, then it is isosceles.

Proof

Consider the trapezoid \(ABCD\) such that \(\angle A = \angle D\) .


Let's complete the trapezoid to the triangle \(AED\) as shown in the figure. Since \(\angle 1 = \angle 2\) , then the triangle \(AED\) is isosceles and \(AE = ED\) . Angles \(1\) and \(3\) are equal as corresponding angles for parallel lines \(AD\) and \(BC\) and secant \(AB\). Similarly, angles \(2\) and \(4\) are equal, but \(\angle 1 = \angle 2\), then \(\angle 3 = \angle 1 = \angle 2 = \angle 4\), therefore, the triangle \(BEC\) is also isosceles and \(BE = EC\) .

Eventually \(AB = AE - BE = DE - CE = CD\), that is, \(AB = CD\), which is what needed to be proven.

2) Let \(AC=BD\) . Because \(\triangle AOD\sim \triangle BOC\), then we denote their similarity coefficient as \(k\) . Then if \(BO=x\) , then \(OD=kx\) . Similar to \(CO=y \Rightarrow AO=ky\) .


Because \(AC=BD\) , then \(x+kx=y+ky \Rightarrow x=y\) . This means \(\triangle AOD\) is isosceles and \(\angle OAD=\angle ODA\) .

Thus, according to the first sign \(\triangle ABD=\triangle ACD\) (\(AC=BD, \angle OAD=\angle ODA, AD\)– general). So, \(AB=CD\) , why.

Let's consider several directions for solving problems in which a trapezoid is inscribed in a circle.

When can a trapezoid be inscribed in a circle? A quadrilateral can be inscribed in a circle if and only if the sum of its opposite angles is 180º. It follows that You can only fit an isosceles trapezoid into a circle.

The radius of a circle circumscribed by a trapezoid can be found as the radius of a circle circumscribed by one of the two triangles into which the trapezoid is divided by its diagonal.

Where is the center of the circle circumscribed by the trapezoid? It depends on the angle between the diagonal of the trapezoid and its side.

If the diagonal of a trapezoid is perpendicular to its side, then the center of the circle described around the trapezoid lies in the middle of its larger base. The radius of the circle circumscribed about the trapezoid in this case is equal to half of its larger base:

If the diagonal of a trapezoid forms an acute angle with its side, the center of the circle described around the trapezoid lies inside the trapezoid.

If the diagonal of a trapezoid forms an obtuse angle with its side, the center of the circle circumscribed about the trapezoid lies outside the trapezoid, behind the large base.

The radius of a circle circumscribed about a trapezoid can be found by a corollary of the theorem of sines. From triangle ACD

From triangle ABC

Another option to find the radius of the circumscribed circle is

The sines of angle D and angle CAD can be found, for example, from right triangles CFD and ACF:

When solving problems involving a trapezoid inscribed in a circle, you can also use the fact that the inscribed angle is equal to half of its corresponding central angle. For example,

By the way, you can also use the angles COD and CAD to find the area of ​​a trapezoid. Using the formula for finding the area of ​​a quadrilateral using its diagonals