What Is the Trapezium?

Trapezoid is a quadrilateral with only one set of opposite sides parallel ^[1] . The two parallel sides are called the bottom of the trapezoid: the longer one is called the bottom, the shorter one is called the upper bottom; the other two are called the waist; A trapezoid with a waist perpendicular to the bottom is called a right trapezoid. The trapezoids with equal waists are called isosceles trapezoid.

(The symbols used for the following properties are shown in Figure 1)

1. One set of opposite sides is parallel, and the other set of opposite sides is not trapezoidal.

2. A set of quadrilaterals with parallel and unequal sides is trapezoidal ^[3]

1. Make high (determined according to the actual problem);

1 Ladder Example 1

Figure 5 Example 1

As shown in Figure 5, in ABC, AB = AC, BD and CE are bisectors of ABC and ACB, respectively. Verification: The quadrilateral EBCD is an isosceles trapezoid.

Analysis: To prove that the quadrilateral EBCD is an isosceles trapezoid, the solution is to prove ED // BC, BE = CD, and it is easy to prove from known conditions BCD CBE to obtain EB = DC, so AE = AD, using an isosceles triangle The nature can be proved by ED // BC.

Proof :

AB = AC,

ABC = ACB,

DBC = ECB = 1 / 2ABC,

EBC DCB (A.S.A),

BE = CD,

AB-BE = AC-CD, that is, AE = AD.

ABC = AED, ED // BC,

Also, EB and DC intersect at point A, that is, EB and DC are not parallel,

The quadrilateral EBCD is trapezoidal, and BE = DC,

Quadrilateral EBCD is an isosceles trapezoid.

Hint: The key to solving this problem is to prove ED // BC, EB = DC, and the easy point is to ignore the proof that EB and DC are not parallel ^[6] .

2 Ladder example 2

Figure 6 Example 2

As shown in FIG. 6, in the known quadrangular ABCD, AB = DC and AC = DB. It is proved that the quadrilateral ABCD is an isosceles trapezoid.

Proof :

Pass A, make AEDC, cross BC, and then E.

AB = CD, AC = DB,

ABC DCB ABC = DCB

AEDC again,

AEB = DCB

ABC = AEB, AB = AE,

The quadrangle AECD is a parallelogram.

ADBC.

AB = DC and AD BC,

Quadrilateral ABCD is isosceles trapezoid.

Hint: Determine an arbitrary quadrilateral as an isosceles trapezoid. If the determination theorem of an isosceles trapezoid cannot be used directly, the general method is to use auxiliary lines to decompose the quadrilateral into familiar polygons. This example is to make the quadrilateral by parallel lines Decomposed into a parallelogram and an isosceles triangle ^[6] .

3 Ladder Example 3

Figure 7 Example 3

As shown in Fig. 7, P is a point on the lower bottom BC of the isosceles trapezoid ABCD, PM AB, PNCD, M, N are hanging feet, BECD, E are hanging feet. Verify: BE = PM + PN.

Proof :

Pass point P as PHBE and point H.

BECD, PNCD,

The quadrilateral PHEN is rectangular.

HE = PN, ENPH.

BPH = C.

The quadrilateral ABCD is an isosceles trapezoid,

ABC = C.

MBP = HPB.

And PMAB, BP public,

Rt MBPRt HPB.

PM = BH.

BE = BH + HE = PM + PN.

Hint: To prove the sum and difference of line segments, you can usually consider using the "truncated length method" or "shortening method". This example uses the "truncated length method" ^[6] .

4 Ladder Example 4

Figure 8 Example 4

As shown in Figure 8, in the trapezoidal ABCD, ADBC, and AB = AD + BC, M is the midpoint of DC. Verification: AMBM.

Proof :

Extend the extension line of AM to BC at point N. M is the DC midpoint, ADBC,

ADM NCM.

AD = CN, AM = MN.

AB = AD + BC = BN.

It is known from the isosceles triangle " ^[6] Three Lines in One" that BMAM.

Tip: According to the needs of the certificate, the two trapezoidal bottoms are also commonly used to add auxiliary lines. In this example, you can also extend BC to N first, so that BN = AB, and then prove that A, M, and N are in line ^[6] .

5 Ladder example 5

Figure 9

As shown in Fig. 9, in trapezoidal ABCD, ADBC, diagonal ACBD, and AC = 5cm, BD = 12cm, find the sum of the upper and lower bases of the trapezoid.

Solution :

D. Make the extension line of DEAC and BC at point E.

ADCE, DE = AC = 5cm, AD = CE.

ACBD,

DEBD.

In Rt BDE,

AD + BC = CE + BC = BE = 13cm.

Hint: Make a diagonal parallel line through the vertices, concentrate the quantity and position relationship of the two diagonals into one triangle, and convert the length of the upper and lower bases of the trapezoid to the length of the hypotenuse of the right triangle ^[6] .

6 Ladder Example 6

Figure 10 Example 6

As shown in Fig. 10, in the isosceles trapezoid ABCD, ADBC, AB = DC, and ACBD, AF is the height of the trapezoid, and the area of the trapezoid is 49cm2. Find the height of the trapezoid.

Solution 1 :

As shown in Figure 10 (a), cross the CB extension line at AEDB at point E.

ACBD,

ACAE.

ADEB,

AE = BD, EB = AD.

The quadrangular ABCD is an isosceles trapezoid,

AC = BD.

AE = AC.

AEC is an isosceles right-angled triangle.

And AF is high on the hypotenuse, so AF is also the midline on the hypotenuse.

AF = 7cm

Solution 2 :

Let the two diagonal lines of trapezoid ABCD intersect at point O, pass O as OHBC at point H, and extend HO to AD at point G (see Figure 10 (B)).

ADBC,

HGAD.

AB = DC, AC = DB, BC common,

ABC DCB.

2 = 1.

ACBD,

BOC is an isosceles right-angled triangle.

The same reason.

The following answer process is the same as solution 1.

Solution 3 :

Pass D for DM and BC at point M (see Figure 10 (c)).

trapezoid ABCD is an isosceles trapezoid,

AC = DB, ABC = DCB.

Then AF = DM,

Rt AFCRt DMB,

DBC = ACB.

ACBD,

DBM = ACF = 45 °.

AFC and DMB are both isosceles right-angled triangles. AF = FC, DM = MB,

The following solution process is the same as solution 1.

Hint: The three solutions to this problem are to use the properties of isosceles right triangles or congruent triangles to prove that the height of the trapezoid is equal to the length of the median line of the trapezoid. Therefore, in an isosceles trapezoid, if the two diagonal lines are perpendicular, the height of this trapezoid is equal to the length of the median line, and the area of the trapezoid is equal to the square of the height ^[6] .

7 Ladder Example 7

Figure 11 Example 7

As shown in Figure 11, in the trapezoidal ABCD, AD // BC, AB = DC, points E, F, and G are on edges AB, BC, and CD, respectively, and AE = GF = GC.

(1) Verify that the quadrilateral AEFG is a parallelogram;

(2) When FGC = 2EFB, verify that the quadrilateral AEFG is rectangular.

Analysis : This question examines comprehensive proofs about triangles and quadrilaterals. It involves the properties of isosceles trapezoids, the determination and properties of parallelograms, and the properties of isosceles triangles. Pay attention to the standardization of proof format and reasoning method in the answering process.

Proof :

(1) In trapezoidal ABCD, AB = DC,

B = C.

GF = GC, C = GFC,

B = GFC

AB // GF, namely AE // GF.

AE = GF

The quadrilateral AEFG is a parallelogram.

(2) Solution : Pass G as GHFC, and foot as H.

GF = GC,

FGH = 1 / 2FGC.

FGC = 2EFB

FGH = EFB.

FGH + GFH = 90 °

EFB + GFH = 90 °

EFG = 90 °

The quadrilateral AEFG is a parallelogram,

The quadrilateral AEFG is rectangular.

Note: The bottom corner of the trapezoid can refer to any one of the trapezoids, so it is incorrect to say that a trapezoid with the same bottom angle is an isosceles trapezoid. ^[6]