In the quadrilateral ABCD, angle a = angle C. angle B = angle D. try to explain that ABCD is a parallelogram

In the quadrilateral ABCD, angle a = angle C. angle B = angle D. try to explain that ABCD is a parallelogram

prove:
∵∵∵ a = C, ∵ B = D, and
∠A+∠B+∠C+∠D＝360°,
In addition, it can be seen that ﹥ a + ﹥ B = 180 ° and ﹥ B + ﹥ C = 180 °,
∥ BC ∥ ad, and ab ∥ CD,
ABCD is a parallelogram

In quadrilateral ABCD, ∠ a = ∠ C, ∠ B = ∠ D, it is proved that ABCD is a parallelogram according to the definition of parallelogram
Look at the title and say the following steps

∵∠A=∠C,∠B=∠D
∴∠A+∠B=∠C+∠D=180°
∴AD‖BC,AB‖CD
The quadrilateral ABCD is a parallelogram

How to prove that a quadrilateral ABCD is a parallelogram
In the quadrilateral ABCD, how to prove that it is a parallelogram

The vertical line through B as ad intersects the extension line of Da with E, and the vertical line through D as BC intersects the extension line of BC with F

It is proved that the quadrilateral ABCD is a parallelogram
In known quadrilateral ABCD, is ad / / BC, ∠ a = ∠ C, AB equal to CD
How to prove it
Hope to understand people can give me a complete answer process

(1) "Side" is called "SAS" (2) "side" is called "ASA" (3) "side" is called "SSS" (4) "side" is called "AAS" (5) "hypotenuse, right angle" is called "HL" (right triangle). Note: in the congruence judgment, there is no AAA and SSA

In quadrilateral ABCD, ab = CD. If you want quadrilateral ABCD to be parallelogram, you should add a condition (just add a condition)______ ．

According to the judgment of parallelogram, we can add ab ‖ CD (the answer is not unique), so the answer is: ab ‖ CD (or ad = BC)

As shown in the figure, in quadrilateral ABCD, ab ‖ CD, ∠ B = ∠ D. proof: quadrilateral ABCD is parallelogram

It is proved that: ∵ ab ‖ CD, ≌ △ ACD (AAS), ∥ AB = CD, ∥ AB = CD, ab ‖ CD, ∥ quadrilateral ABCD is parallelogram in △ ABC and △ ACD

In the parallelogram ABCD, e and F are the midpoint of AB and CD respectively

Prove; ∵ in parallelogram ABCD
AB‖CD
∴AB=CD
And ∵ E and F are the midpoint of AB and CD respectively
∴AE=CF
Again, CF
A quadrilateral aecf is a parallelogram

As shown in the figure, it is known that the length of the major axis ab of the ellipse C: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > O, b > o) is 4, the eccentricity e = two-thirds root sign 3, O is the origin of the coordinate, passing through B
The line L passing through point B is perpendicular to the X axis. P is any point on the ellipse different from a and B, pH is perpendicular to the X axis, h is the perpendicular foot, extend HP to point Q so that HP = PQ, connect AQ, extend the intersection line L to point m, and N is the midpoint of MB.
(1) The equation of finding ellipse
(2) It is proved that Q is on a circle with diameter ab
(3) Try to judge the position relationship between line QN and circle o

According to the meaning of the title
2a=4
a=2
e=c/a=√3/2
c=√3
b²=a²-c²=4-3=1
b=1
Elliptic equation: X & sup2 / 4 + B & sup2; = 1
(2) Point P (2cosa, Sina)
Then q (2cosa, 2sina)
Kqa×Kqb=2sina/(2cosa-2)*2sina/(2cosa+2)
=4sin²a/(4cos²a-4)=sin²a/(cos²a-1)=sin²a/(-sin²a)=-1
So AQ ⊥ BQ
Angle AQB = 90 degrees, AB is the diameter
It's over
(3) Because ∠ BQM = 90 degrees, n is the midpoint of BM
QN=NB
∠BQN=∠NBQ
∠ NBq = ∠ MAb (∠ NBq is tangent angle, MB is tangent)
Therefore, bqn = mAb
Then QN is the tangent of circle o

It is known that the four numbers a, B, C and D are in direct proportion, and A.D is the external term. It is proved that (a, d) (C, d) and the coordinate origin o are on the same line

Let a (a, b) B (C, d),
The slope of OA is B / A, the slope of ob is D / C,
From the four numbers of ABCD, B / a = D / C is obtained,
So the slopes of OA and ob are equal,
So o, a, B are on the same line

If the four numbers a, B, C and D are proportional, i.e. a / b = C / D, prove ad = BC

It's very simple. Multiply BD on both sides of the equation, and then you can get ad = BC