Two math problems in the second year of junior high school, find a complete proof process, thank you

3.Proof of ADC BCE can be obtained (hint, certificate ACD= BCE, DC=CE, BC=AC).

4.Prove be=em and cf=fm (hint, bem is an isosceles triangle, fcm is an isosceles triangle).

1.Proof: Because the midpoint of one set of opposite sides of a parallelogram is parallel to the other set of opposite sides, the intersection point with the diagonal must be the midpoint of the diagonal.

Therefore, a parallelogram must be bisected with the diagonal line.

2.Proof is that the diagonal that connects the quadrilaterals, because the midpoints of the two adjacent sides of the quadrilateral are parallel and equal to a diagonal, so the quadrilateral obtained by connecting the midpoints of the sides of any quadrilateral in turn is a parallelogram.

(1) Proof that the point D is made as dg ab and bc is handed over to g, and a b 60° can be known from the question

dg∥ab∠cdg＝∠cgd＝60°

gdf＝∠e

CDG is also an equilateral triangle.

dg＝cd＝be

dgf≌△ebf

aas）df＝ef

2) Solution: from a b 10a 6b 34 0, get (a 5) b 3) 0

a－5）²≥

0（b－3）²≥0∴

a＝5b＝3

Namely: BC 5

cg＝be＝3

dgf ebf

bf＝gfbf＝(bc－cg)＝(5－3)＝1

3) Solution: cd x, bf y

BC 5 and BF (BC CG) (BC CD).

5－x)y＝1/2（5-x）

The value range of the independent variable x is as follows:

0＜x＜5

(1) Proof: passing the point d as dg ab crosses bc to g, abc is an equilateral triangle, and a b 60° dg ab can be obtained

cdg＝∠cgd＝60°

gdf＝∠e

CDG is also an equilateral triangle.

According to the definition of equilateral triangles: dg cd be

dgf≌△ebf

aas）df＝ef

--The corresponding sides of congruent triangles are equal.

2) Solution: from a b 10a 6b 34 0, get (a 5) b 3) 0

a－5）²≥

0，（b－3）²≥0∴

Solution: a 5

b 3 i.e. BC 5

cg＝be＝3

dgf ebf

bf gfbf (bc cg) 2 (5 3) 2 1(3) solution:

cd＝x，bf＝y

BC 5 and BF (BC CG) (BC CD).

5－x)y＝1/2（5-x）

The value range of the independent variable x is as follows:

0＜x＜5

Proof: Link AO

In ABO ae=be om=bm

EM is a median line of ABO.

em = 1 2ao and em ao

In the same way, we get dn=1, 2ao, and dn ao

em//dn em=dn

Even ao, em 1 2(ao) dn em dn em=dn

Certificate: Lian AO, in ABO.

EM is the median line.

em‖ao em=（1/2）ao

The same goes for dn ao dn=ao 2 em dn em em=dn

It can be proved by the reversibility of the optical path and the symmetry of the graph.

Or: extend c to intersect b with a point, the angle = angle 1 (a and b are parallel, the same angle is equal), and angle 1 and angle 4, etc., so the so-called angle = 4

So c d (the inner wrong angles are equal, and the two straight lines are parallel).

7. tane=ad (1+root2)ad

Finally, it is indicated by arctangent!

8、pe=ae,pf=bf

PE+PF=AC2=5*Root2 2