A junior high school math problem, please tell me as soon as possible tonight! Thank!

Let the symmetry point A1 with respect to DG fall on BC, then AA1 DG, the higher AA1 on the side of BC of AA1 BCBC

aa1 to dg to k

So ak=a1k=1 2aa1

If the quadrilateral defg is a square.

then de=a1k, de=dg

Because d and g are the midpoints of ab and ac.

So dg=1 2bc

So aa1=bc

i.e. the height on the BC side of ABC is equal to BC.

The bottom edge is equal to the height on the bottom edge.

As am, perpendicular to BC, BC crosses BC at point M, BC=AM because DG is the median line of the triangle ABC.

So df = 1 2bc

Because the rectangular defg is a square.

So dg=de, so de=1 2bc

Because de=1 2am

So am=bc

Isosceles triangle, b,c is one side of the 6x6 grid, and a is at the midpoint of the other side;

The length of the side of BC = the length of the perpendicular line from point A to the edge of BC.

If A is perpendicular to BC, then E is the midpoint of BH and F is the midpoint of HC.

d and g are the midpoints of ab and ac, respectively.

2009 Beijing Chaoyang High School Entrance Examination 22 Mathematics Questions Find the answer yourself!

The height on the side of BC should be bisected by BC and equal to BC. Do you want to prove the process?

The isosceles triangle handle must be satisfied that both sides are equal.

The length of the base edge BC is equal to the height of the side of the triangle BC.

Suppose A climbs a mountain with x points.

According to the meaning of the topic, 30 minutes on the second night is to go x-30 minutes and arrive at the listed equation at the same time.

10*x=15*(x-30)

>10x=15x-450

>x=90 (points).

That is, A climbs the mountain with 90 minutes.

The height of the mountain is.

90*10=900 meters.

If the speed of the electric locomotive is x kilometers, then the maglev speed is (5x+20) kilometers (x+5x+20).

When the speed of the electric locomotive is 96 km, the maglev speed is 500 km, the speed of A is set to be x km-h, the speed of B is ykm-h, and the distance of ab is s, so s=36+2*(x+y)=4*(x+y)-36 The equation is solved: x+y=36 km-h.

s = 108 km.

A: The distance between the two places is 108 kilometers.

Solution: The distance from the school to the army is X kilometers. (x=

Solution: The time from the school to the army is xh

14x=5x=2×

x=11×5+

Solution: Set the distance to xkm

x÷5=（x+

x=11×5+

Question 1 ;

1) b = 64, c is equal to 56

If we also know AD BC, we can see that ADB ADC is 90 triangle inner angles and 180

It can be obtained: dab=180 - 90 - 64 = 26, dac=180 - 90 - 56 = 34

2) DE bisects ADB (forgot to mark E), we know ADE = EDB = 90 2 = 45

Also know that dab = 26, triangle inner angle and 180 aed = 180 - 45 - 26 = 109

Solution: (1) y=kx+b, when y=0, x=-b k is the coordinate of point c (-b k,0).

y=kx+b intersects y=4 x at point a, and y=4 x is substituted into y=kx+b to get kx 2+bx-4=0, since there is only one solution, so =0 is b 2=-4k s boc=1 2 丨ob丨 丨oc丨=1 2 b (-b k)=-b 2 (2k)=2

2）s△oab=s△oac

Substituting y=k x into y=-bx a+b gives bx 2-abx+ak=0, =0 gives k=ab 4, x=a 2 y=2k a, i.e., point a (a 2, 2k a).

When y=-bx a+b, y=0, x=a, i.e., point c(a,0) s oab=1 2 b xa=ab 4=ks oac=1 2 xc ya=k

s△oab=s△oac