It is proved that the quadrilateral ABCD is a parallelogram

(1) "Side" is referred to as "SAS" (2) "side" is referred to as "ASA" (3) "side" is referred to as "SSS" (4) "corner" is referred to as "AAS" (5) "hypotenuse, right angle" is referred to as "HL" (right triangle). Note: in the determination of congruence, there is no AAA and SSA, In both cases, the shape of a triangle can not be uniquely determined. You can prove congruent by using the angle and edge, connect BD, there are triangle ADB and triangle BCD, angle a, angle c are equal, AD / / BC, and the internal stagger angle is equal. I believe you know which is it, and there is a common edge BD

It is known that the quadrilateral ABCD is inscribed in the circle O, and AC is the diameter of the circle, and BCDE is a parallelogram,
If OA vector + ob vector + od vector = xoe vector, then real number x =?

X=1

The perimeter of parallelogram ABCD is 36cm, which is composed of···
The perimeter of parallelogram ABCD is 36cm. Two high De, DF are introduced from obtuse vertex d to AB, BC, and de = 4cm, DF = 5cm?
New curriculum practice and Exploration Series p104 No.8

Solution
Connect BD
∵ parallelogram ABCD
∴∠A＝∠D,AB＝CD,AD＝BC
∵DE⊥AB,DF⊥BC
The Δ DAE is similar to the Δ DCF
∴DA/DE＝DC/DF
∵DE＝4,DF＝5
∴AD/CD＝4/5
∴AD＝4CD/5
The perimeter of the parallelogram ABCD is 36
∴AB+BC＝36/2＝18
∴AB+4AB/5＝18
∴AB＝10
∴BC＝4*10/5＝8
∵DE⊥AB
∴S△ABD＝AB*DE/2＝10*4/2＝20
∵DF⊥BC
∴S△BDC＝BC*DF/2＝8*5/2＝20
Ψ sabcd = s △ abd + s △ BDC = 40 (cm2)

As shown in the figure, in the parallelogram ABCD be vertical CD BF vertical ad angle FBE is 45 degrees AF plus CE is 3 times the root sign 2, find the perimeter of the parallelogram ABCD
Only use the content of quadrilateral in the first chapter of the third semester of junior high school, not trigonometric function, etc

∵ BC ∥ AD and BF ⊥ ad in parallelogram ABCD
∴∠CBF＝∠BFD＝90°
Also, EBF = 45 degrees
∴∠CBE＝∠EBF＝45°
Be CD
∴∠BEC＝90°
∴∠C＝∠CBE＝45°
∴BC＝√2CE
Similarly, Ba = √ 2AF
AF + CE = 3 √ 2
∴√2AF﹢√2CE＝√2×3√2＝6
That is BC + Ba = 6
The perimeter of parallelogram = 2 (BC + BA) = 2 × 6 = 12

In ▱ ABCD, AB: BC = 1:2, perimeter is 18cm, then ab= cm，AD= cm．

The circumference of ABCD is 18cm ▱ 2 (AB + BC) = 18 ∵ AB + BC = 9 ∵ AB: BC = 1:2 ∵ AB = 3cm, ad = BC = 6cm, so the answer is 3,6

▱ in ABCD, if the circumference is 20 & nbsp; cm, ab = 4 & nbsp; cm, then CD= cm，AD= cm．

According to the property of parallelogram AB = CD = 4cm, then 2ad + 2Ab = 20, the solution is ad = 6cm

In the parallelogram ABCD, if AB: BC = 3:5 and the circumference is equal to 48, then AB = (), BC = (), ad = (), CD = ()

9 15 15 9

As shown in the figure, ab ∥ CD, ad ∥ BC, if the perimeter of parallelogram ABCD is 30cm, the distance between AD and BC is 4cm, and the distance between AB and DC is 6cm, calculate the area of parallelogram ABCD

S=AD*4=CD*6
AD=3CD/2
30=2(AD+CD)=2(3CD/2+CD)=5CD
CD=6
So: S = 6 * 6 = 36

It is known that the circumference of the parallelogram ABCD is 36cm, and ab = 10cm. If the distance between AD and BC is 6cm, then the distance between AB and CD is 6cm______ cm．

If a is used as AE ⊥ BC in E and C as CF ⊥ AB in F, then AE = 6cm, the distance between AB and CD is the length of CF, the perimeter of ∵ parallelogram ABCD is 36cm, and ab = 10cm, ∵ AB = CD, ad = BC, AB + CD + ad + BC = 36cm, the solution is: BC = ad = 8cm, CD = AB = 10cm, ∵ s parallelogram ABCD = ad × AE = CD × CF, ∵ 8cm × 6cm = 10cm × CF, ∵ CF = 4.8cm, so the answer is: 4.8

As shown in the figure, the perimeter of the parallelogram ABCD is 80cm. When ad is the bottom, the height is 12cm. When AB is the bottom, the height is 20cm. Calculate the area of the parallelogram ABCD

Ad = x, then AB = (80-2x) / 2 = 40-x
12x=20*(40-x)
x=5/2
Area of parallelogram ABCD:
12 * 5 / 2 = 30 square centimeter

It is proved that the quadrilateral ABCD is a parallelogram