Eight math problems, hurrying !!!!!!! 5

ab=be=1 2ab=5, f is the midpoint of the bc edge.

The distance between AB and CD is 8: the distance from point D to the straight line AB (i.e., the height on the bottom edge of the triangle ADE) is 8, then the area of the triangle ADE is 1 2 5 8=20

Since F is the midpoint of the BC side, the distance from the F point to the be (i.e., the height on the bottom side of the triangle BEF) is 4, then the area of the triangle BEF is 1 2 5 4=10

In the same way, the area of the triangle DCF is 1 2 10 4=20

The total area of the parallelogram is 10 8=80, and the area of the triangle is subtracted from the above three triangles to get the area of the triangle sought.

The answer is: 30

Let's start with a picture. I do it by the area method. Let the area of def be the midpoint of ab and cd, respectively, where def is the midpoint of s of ab and cd. The area of the parallelogram is 10*8=80

Let half of BC be h=x 40 (calculate the area from another angle), so the S triangle AED+S triangle EBF+S triangle fcd+s=805*8 2+(x*H2) 2+(x*h) 2+s=80x*h=s 2=40 (pay special attention, it is the key, x) 20+10+20+s=80

s=30

For sketching, EF is parallel to the median line and AC2, and the BEF is half as high as ABC.

The original SABCD is AC* (the sum of the heights of ABC and ACD) DEF is the sum of the heights of ACD and ACD - half the height of ABC) 2 = 1 2 *3 4 *1 2 *AC* (the sum of the heights of ABC and ACD) = 3 16SABCD

Because sabcd=80 so sdef=15

The methodology may be miscalculated.

I'm wrong*o*

30 The area of ABCD in a parallelogram is 80 AED is a quarter of ABCD and 20 FCD is also 20 BEF is an area of ABCD and an eighth is 10, so the area of DEF is 30

The ADE area is 5 8 2 20, the BEF area is 5 4 2 10, and the CDF area is 10 4 2 20

The parallelogram ABCD area is 10 8 = 80

The area of the def is 80-20-20-10=30

What about the figure??? Otherwise, it may be different from what you say.

12. Blind x≠ 1 time;

or 5; Grinding ear base.

16、（2s/h）-a；Clan Liang.

y^2-4y+1=0；,1；

It is known that in Figure 1: in the isosceles RT ACB, the ACB RT, the point M is any point on the edge of BC, the point M is the perpendicular line of AM, the perpendicular line of the point B is AB, and the two perpendicular lines intersect at the point N

1) Try to judge the relationship between AM and MN and prove it.

2) If the condition "point M is any point on the edge of BC" is changed to "point M is any point on the extension line of the side of CB", the other conditions remain unchanged, is it still true? And prove.

3) If the condition "point M is any point on the edge of BC" is changed to "point M is any point of the reverse extension line of the side of CB", the other conditions remain unchanged, is it still true? And prove.

Crossing the point M as the perpendicular line of am, and passing the point b as the perpendicular line of ab The question is wrong, right?

Solution: From the meaning of the question, it can be seen that the point (3,-2) is on the straight line y=cx+5, then:

3x+5=-2 gives :c=-7 3

While the points (3,-2) and points (3 4,1 4) are on the line y=ax+b, then:

Substituting yields: a=-3, b=1

The resulting line is: y=-3x+1 y=-7 3x+5

Have you ever learned congruent triangles?

Δbce δfde is judged first

CBE= F and BE=FE , E is the midpoint of BF.

The bisector of baf, a coincides with the high of the bf side, isosceles, abe= f cbe= abe, as evidenced.

This method is very good, first from the angle bisector and the middle line of the triangle ABF AE is the triangle ABF isosceles, so the angle F is equal to the angle ABF, and the angle F = angle CBF (internal error), so it can be proved.

You can do this way, you can prove dee bec first, and you can get def becdf bc, abf def bec abfBe bisected by ABC

Extending be to f and then connecting df is still a good method. It can be verified. You just need to use EDF and ECB congruence first, and then use AE to bisect the opposite side BE, FE to prove that ABE and EFE are congruent.

Point E is the midpoint of DC, de=CE.

ad‖bc，∴∠fdc=∠bcd、∠f=∠fbc。

def≌△bce。∴be=ef。

and ae bisect bad, ae=ae, bae fae. ∴∠f=∠abe。∴∠fbc=∠abe。

Be bisected by ABC

Take the midpoint g of ab and join eg, then eg ad bc, so aeg= ead= eab, so eg=ag=gb, so abe= beg= ebc

Of course you can do that, good way, and wait for it or something.

It's best to look at reference books, this one is not suitable for online communication.

Incongruity, ABP and PCD only AB=CD, the other corners are not necessarily equal, do not meet the condition.

The area is equal. The angle bisector, so the distance from the point p to om and on (i.e., the height on the edge of the triangle ab and cd) is equal.

ab=cd again

Triangle area = base * height.

The bottom and height are equal, and the area is of course equal.

Solution: (1) Solution: Not necessarily congruent. pa is not necessarily equal to cp, not necessarily equal to cp, dp.

2) Solution: Equal. ABP and CDP are equal in the area of the same floor and height.

(1) ABP and PCD are not necessarily congruent, they are congruent only when OA=OC.

2) Their area must be equal because the two triangles are equal in base and height.