Mathematics proof questions in the second year of junior high school

n,k n, then the coordinates of point e are: 0, and the coordinates of k m f point are: n,0 , and the equation of the straight line can be found from two points:

mn:y= k mn x k n 1 mn, ef:y= k mn x k m, which is obtained by the equal coefficients of x of two straight lines:

These two straight lines are parallel.

It is proved that if the coordinates of m and n are (x1,y1),(x2,y2), then the slope of Mn is (y1-y2) (x1-x2) and the slope of EF is y1 x2

Because y=k x, y1=k x1, y2=k x2 is substituted by mn and ef slopes, which can be obtained.

y1-y2) (x1-x2) y1 x2 k x1x2 hence mn ef

M point (x1, k x1), n point (x2, k x2), e point (0, k x1), f point (x2, 0), the slope of the straight line Mn is (K x2-k x1) (x2-x1) = -k (x1x2), the slope of the straight line EF is -k (x1x2), so the two lines are parallel.

Let the coordinates of m n be (x1, k x1) (x2, k x2) me perpendicular y-axis, nf perpendicular x-axis, and the perpendicular feet are e and f respectively, and the coordinates of e f are (0, k x1) (x2, 0) respectively. Proven.

1.If m and n are in the same quadrant, let m(x1,k x1),n (x2, k x2) e(0,k x1)f(x2,0).

The slopes of Mn and EF are found separately, and the conclusion is that the slopes are equal and the two lines are parallel.

2.If m,n are in different quadrants, m(x1,k x1), e(0,k x1), n(x2, k x2),f(x2,k x2).

The slopes of Mn and EF are found separately, and the conclusion is that the slopes are equal and the two lines are parallel.

Summary: The two lines are parallel.

1) AO bisects BAC

bao=∠cao

In ABO vs. AOC.

ao=ao(common edge).

bao = cao (verified).

ab=ac (known).

ABO congruent AOC (Corner Edge).

2) Because the triangle ABO congruent triangle AOC (proven), the angle ABC = angle ACD (the corresponding angles of the congruent triangle are equal) in the triangle ABE and the triangle ACD.

Angular ABC = Angular ACD (verified).

Angle bae = corner cad (common corner).

ab=ac (known).

So the triangle abe congruent triangle acd (corner corner) 3) because ao is bisectoral.

So ao also bisects the angle doe

So the angle doa = the angle ade

In the triangle ado with aeo.

Angular dao = angular eao (verified).

ao=ao(common edge).

Angular AOD = Angular AOE (verified).

So the triangle AOD congruent triangle aeo (corner edge corner) 4) because the triangle abo congruent triangle aoc

Triangle ado congruent triangle aeo

So triangle abo-triangle ado = triangle aco-triangle aeo, so triangle dbo, congruent triangle eoc

ao bisects bac, ab=ac

abo=△aco

and extend the intersection ac, ab to e, d

ado=△aeo

bod=△coe

abe=△ acd

1) Because de is the median line.

So ade and cde are congruent.

So a= DCE

And because cef= a

So cd balances ef

Because de is the median line.

So de balances bf

So the quadrilateral cdef is a parallelogram.

2) AC=4 BC=3 Pythagorean lawAB=5CD=1 2AB

de=1/2bc

So the parallelogram area = 2 (cd + de) = 2 (1 2ab + 1 2bc) = 2 (

The circumference is 8

1) Because de is the median line, de bf, de=1 2bc, ae=ec=1 2ac, and cef= a, ecf= aed=90°, aed ecf, de=cf, therefore, the quadrilateral cdef is a parallelogram.

2) ab=5, cd=1 2ab=5 2=ef, de=1 2bc=3 2=cf, so, the perimeter is 8

The hall hole proves that because ae is the bisector of the slippery bac, let the answer caf = eab.

And because acb=90°, cd is high, acf=eba.

So afc= bea, which can be proved that acf abe because acf abe so ae af=ab ac ae=5cm

Proof idea: The diamond-shaped diagonals are perpendicular to each other, connecting the other diagonals.

It can be seen that the midpoint line of each tomb is a triangular median line of ruler rock cultivation.

It can be proved that the quadrilateral jujube rock efgh is a parallelogram.

Then in the proof there is an angle that is a right angle.

It can be concluded that the quadrilateral efgh is rectangular.

It has been known that the quadrilateral ABCD is a diamond.

So ab=ad

BC = CD angle A = angle C

Angle b = angle search d

g and h are the midpoints of each side of the parallel omission, respectively.

So ae=af=cg=ch

be=bg=dh=df

ef=gheg=fh

Discussed in two cases:

1）△b'fc is similar to (it should be represented by a symbol here, but it cannot be typed) abc.

Let bf=b'f=x, then cf=4-x.

Due to b'FC is similar to ABC, so B'f ab=cf bc, i.e., x 3=(4-x) 4, and the solution is x=bf=12 7

2）△b'FC is similar to BAC.

The same is done with bf=b'f=x, then cf=4-x.

Due to b'Fc is similar to BAC and B'f ab=cf ac, i.e., x 3 = (4-x) 3, and the solution is x = bf = 2

In summary, bf = 12 7 or 2

Set b'f=x, then bf=x, cf=4-x, because they are similar, so x 3=(4-x) 4, and solve x=12 7

So b'f=bf=12/7

Because of the triangle B'FC is similar to the triangular ABC, so AE is parallel to B'f, angle fb'C = angular EAB', so triangle aeb'Similar to b'fc, so ae eb'=3 4, so ae=9 7, be=12 7, ab'=9/7,b'c=12/7,fc=16/7，bf=12/7.