Ladies and gentlemen!! Help!! A junior high school geometry proof question!!

On the third floor, your topic is simple.

That's such a strange question, it needs to be added.

Is it for us to give you a question??? /

is a parallelogram. First of all, we need to connect bf, de, mf, ne The first step is to prove: triangle AED = triangle cbf (with corner edges) Part 2: because triangle AED = triangle cbf, so ed = fb.

Part 3: Because AE=CF, DF=EB. Because of the parallelogram ABCD. So DF is parallel to EB. So DF is parallel and equal EB, so DFBE is a parallelogram.

Part 4: Because m and n are the midpoints of de and bf respectively, and because of the parallelogram dfbe. So nf is parallel and equal me, so parallel and equal.

Hello, (1).a=90°, diamond-shaped ABCD, so ABCD is square.

The perpendicular lines of BC and CD are made by O, and the perpendicular feet are M and N. respectively

Triangle OPM and triangle OEN congruence. (asa) so of=oe.

2）.In the same way, the perpendicular line of BC and DE is done by O, and the perpendicular triangle OFM is similar to the triangle OEN.

oe/of=on/om=(1/2*dg)/(1/2*bc-1/2*cg)

dg/(bc-cg)

Root number 3 does not understand how to send a letter Thank you

To put it simply:

1) Connect the OB, OC ABCD is a square.

OB=OC angle ABF and angle FCE are both right angles, and both angle OBA and angle OCF are 45°, so angle OB and angle OCE are equal.

Looking at the ofce graph, two opposite triangles, you can get that the angle ofc and the angle oec are equal According to these three conditions, the triangle ofb and the triangle OCE are equal, so of=oe

(1) Connect ME and MD, ME=MD=1 2AB, and the triangle EMD is an isosceles triangle.

n is the midpoint of ed, so mn de;

2) C=60°, AD BC, so CD=1 2AC, in the same way, CD=1 2AC, triangle ABC and triangle CDE are similar ed=1 2AB;

In the right triangle enm, en=1 4ab, em=1 2ab em en=2 so mn = en, which is 3 times the root number.

mn:de=root number 3:2 to solve.

Link MD, ME

AEB 90°, m is the midpoint of AB.

The same can be said for me 1 2AB (the middle line on the hypotenuse of a right triangle is equal to half of the hypotenuse) is proved: me 1 2AB

md men is the midpoint of de.

MN DE (the height on the base edge of the isosceles triangle coincides with the midline on the bottom edge) ADC BEC 90°

adc∽△bec

cd/ce＝ac/bc

ecd＝∠bca

cde∽△cab

de/ab＝cd/ac

c＝60°，∠adc＝90°

cd/ac=1/2

de/ab＝1/2

de＝1/2ab

md＝me＝1/2ab

md＝me＝de

mn/de＝mn/md＝√3/2

(1) Proof: Because AD is perpendicular to BC to D

So the angle adb = 90 degrees.

So the triangle adb is a right triangle.

Because be is perpendicular to e

So the angle AEB = 90 degrees.

So the triangle AEB is a right triangle.

Because m is the midpoint of ab.

So EM and DM are the midlines of the right triangle AEB and the right triangle AEB respectively, so EM=1 2AB

dm=1/2ab

So em=dm

So the triangle MDE is an isosceles triangle.

Because n is the midpoint of de.

So MN is the midline of the isosceles triangle MDE.

So MN is the perpendicular line of the isosceles triangle MDE (isosceles triangle three lines in one), so MN is perpendicular to DE

2) Solution: Let ad and be intersect at the point o

Because the angle dob = angle oab + angle oba

Because the angle AEB = angle ADB = 90 degrees.

So a, b, d, and e are all round.

So angular deb = angular oab

Because the angle c = 60 degrees, the angle adc = angle ADB = 90 degrees because the angle c + angle ADC + angle DAC = 180 degrees.

So the angle dac = 30 degrees.

Because angle DAC + angle AEB + angle AOE = 180 degrees, angle AOE = 60 degrees.

Because the angle aoe = the angle dob

Because M is the midpoint of AB in the right-angled triangle AEB.

So me=mb

So angular meb = angular mbe

So the angle dob = angle deb + j meb = 60 degrees because mn is perpendicular de (proven).

So the angle mne = 90 degrees.

So tan60=mn ne

So mn ne = root number 3

Because n is the midpoint of de.

So ne=1 2de

So mn:de=root number 3:2

The circumferential angle of the circle is 90 degrees from the diameter of the circle.

D and E are all on the circumference of the circle with AB as the diameter, M is the midpoint of AB, M is the center of this circle to connect ME and MD, ME and MD are the radius of the circle, the equal triangle MED is an isosceles triangle, the angle C is 60 degrees, then the circumferential angle EAD is 30 degrees, and the corresponding central angle EMD is 60 degrees.

From this, it can be seen that the triangle EMD is an equilateral triangle.

The three-phase coincidence of an equilateral triangle proves mn de

mn:de=mn:2ne=sin60 2=root number three 2

(1) Connection, ME=MD=1 2AB, so the triangle EMD is an isosceles triangle, and N is the midpoint of ED, so MN DE

2) C=60°, AD BC, so CD=1 2AC, in the same way, CD=1 2AC, triangle ABC and triangle CDE are similar ed=1 2AB;

In the right triangle enm, en=1 4ab, em=1 2ab em en=2 so mn = en, which is 3 times the root number.

mn:de=root number 3:2

Dear ......Tomorrow morning, oh ......I've been thinking about it for a long time, and I've ......The head rusted during the summer vacation said ......

It is known that BCE is an equilateral triangle (using the mutual congruence theorem); So d=60°, let's do it yourself.

Hey, hey, do you want a format, wait... Now I'm going to sleep. I'll give it to you.

Take AC as the axis to do the axis reflection, turn the triangle ACD over, and use Menelaus's theorem and Seva's theorem to prove that F is on ED, and end.

1.The blackboard reads 1, 2, 3 ,......1997, 1998, the 1998 numbers, do the following to do this: erase three of the numbers, and then write the single digit of the sum of the three numbers on the blackboard.

Columns such as: erase 5, 13, 1998, add 6; Another example is to erase 6, 6, 38, add 0, and so on. If, after 998 operations, there are only two numbers left on the blackboard, one is 25, ask how much is the other?

2.**On segment AB, mark 0 at point A and 2002 at point B, which is called the first operation; Then mark (0+2002) 2=1001 at the midpoint C of AB, which is called the second operation; And mark half of the sum of the numbers marked at both ends of the corresponding line segment at the midpoint D and E of the obtained line segment AC and BC, that is, (0 + 1001) 2 and (1001 + 2002) 2, which is called the third operation, and so on, then after 11 operations, what is the sum of all the numbers marked on the ** segment AB?

3.It is known that x, y, z satisfies :

x+［y］+﹛z﹜=

x］+﹛y﹜+z=

x﹜+y+［z］=

Where: for the number a, [a] represents the largest integer not greater than a, {a}=a-[a], find the value of x, y, z.

4.Driver Xiao Li drove on the highway at an average speed, he saw that the number on the milestone was two digits, 1 hour later, he saw that the number on the milestone was exactly the first time to see the opposite number of two digits, and after another hour, he saw that the number on the milestone was the first time to see the two digits plus a 0 in the middle, to find the number that Xiao Li saw on the milestone every time.

5.Someone is going to get 1,2....A few of these numbers are entered into the electricity to find the average. When he finished typing, the computer showed that only (n-1) numbers had been entered, with an average of 35 and 5/7. Ask what is the number entered at the end.

6.Find all the primes so that the power of 8p +1 is prime.

7.It is known that the two sides of an isosceles triangle are and the angles between the two sides are to find the length of the third side!

(2) Take the point M in AD to make MD=ED, connect EM, prove that the triangle Mde is an equilateral triangle, do MN vertical AE, DS vertical HE, and then think about it yourself.

(1) Extend AE and Cd at point M

mce∽△abc

cm=1/2ab,∠amc=∠bam

bam=∠mag

AMG is isosceles

ag=gmag=cm+cg=1/2ab+cg

2) The length of ab and bc in the figure does not match the question stem.

The intersection of the extension cords can solve the first question, the second question is not neat, you can use a camera to take a photo**, pass it up, what you said is incomplete.

It's not junior high school, is it? What about the figure? 、

If you don't give a diagram for such a complex exercise, who can help you?