Mathematics and geometry questions, 20 points per question

Parallel lines that cross E to do AF, cross the extension line of MN at H, and connect the HF angle EHA = angle HAF = angle ACB

Angular EAH = Angular ABC

ea=ab

So the triangle EHA is congruent with ABC.

So eh=af

EH parallel AF again

So the parallelogram EHFA

n is the intersection of two diagonals, so n is the ef midpoint.

Proof that the perpendicular line of Mn over e intersects Mn at X, and the perpendicular line of Mn over F intersects Mn at Y, then the triangle ABM is equal to the triangle Eao (AAS or ASA) So, EO=AM

In the same way, fy=am

So, eo=fy

So, the triangle exn is equal to the triangle fyn

Proof: Connecting OB, OA, OC, let the height of abc be h, then equilateral abc is divided into three small stools

Respectively OBC, OAB, OAC

s△obcbc*oe/2

s△oabab*od/2

s△oacac*of/2

S bac= obc + oab + oacs bacab*h 2 = (ab 2)*h

obc+△oab+△oac

bc*oe/2+ab*od/2+ac*of/2(ab/2)*(oe+od+of)

So h=oe+od+of

That is, the sum of OD+OE+OF is equal to the high jujube bothic rock of equilateral triangle ABC. That is.

Connecting the people to pick up the ao

BOCO, divided into three triangular circles and conjugated with an area of 1 2*(ab*od+bc*oe+of+ac).

Since the triangle is equilateral, ab=bc=ac

The area is 1 2 * ab * high.

od+oe+of=high.

Cut the triangle into three parts to find the area.

1 dong song 2od ab + 1 2od bc

The Hall of AC is an equilateral triangle.

So. Above equation = 1 2

ab×（od+oe+of）

So od+oe+of is the height of the triangle.

Directly use the cosine theorem to do, this method can avoid a lot of trouble, and it doesn't matter if you don't know how much cos105 is, just use bf = root number (18-8 root number 2 * cos105) perfunctory.

In the triangle ABF, AB=4, AF=Root 2, and Angle BAF=60+45=105 degrees.

Then bf 2 = ab 2 + af 2-2 * ab * af * cos (angle baf) = 4 2 + 2-8 root number 2 * cos105

18-8 root No. 2*cos105

18-8 root number 2 * (cos45 * cos60-sin45 * sin60) = 18-8 root number 2 * root number 2 2 (1 2-root number 3 2) = 14 + 4 root number 3

So bf = root number (14 + 4 root number 3).

The answer looks at the picture, I don't know if it's right.

The main points are at the bottom, and the answer is attached to the right, the standard answer.

The answer is the brother downstairs, I'm sure.

I didn't want to give the answer directly when I didn't want the process.

The intersection point of BF and AC can be O, and B is the intersection of BG perpendicular to AC and AC to G, and G is the midpoint of AC.

It can be proved that the triangle bgo is similar to the triangle feo, ef=1, bg=2*root number 3 to obtain the similarity ratio, and eo+og=eg=1 2ac-ae=1, and eo go=ef gb, eo, go can be obtained, and then the pythagorean theorem can be obtained fo,bo, then bf=bo+of. It's not hard.

There are many more ways to do it, think about it yourself.

As the title suggests, 1 4 represents the workload of 1 hour of cooperation between A and B, and 1 5 indicates the workload of 1 hour of cooperation between B and C.

1 4+1 5) 2 9 10 indicates the amount of work that A has done for 2 hours, B has done 4 hours, and C has done 2 hours.

According to "After A and C do 2 hours together, the remaining B needs to do 6 hours to complete", it can be seen that A does 2 hours, B does 6 hours, and C does 2 hours, and the total workload is 1.

So 1 9 10 1 10 means that B does 6-4 2 hours of work.

1 10 2 1 20 indicates B's work efficiency.

1 1 20 20 hours means that B needs 20 hours to complete alone.

First, connect FC, find FC by solving the right triangle, then find the angle FCE by solving the right triangle, and then make the auxiliary line FD cross BC to D, and D is not perpendicular. Find fd, then fb

I do it by the area method.

Let the intersection point be bf and the intersection point of ac is o, the area of the triangle oaf perpendicular to ab is oa times ef, and the area of the triangle oab is ab times om (the knowledge point used to find om is in a right triangle, and the side opposite by 30 degrees is half of the hypotenuse).

The sum of the two areas is the area of the triangle ABF (higher than the EF equilength).

Crossing the point b to do bh ac, let the intersection of ac and bf be o

Proof of EFO HOB

fo bo=ef bh=eo ho=1 root number 3ch=ah=2,ae=1,eh=1

eo=(root number 3-1) 2, ho=(3-root number3) 2 Then the Pythagorean theorem finds fo, and bo is added at the end.

I don't know if it's right or not, it's a verbal calculation, and you can do the math yourself. But the amount of calculation seems to be a bit too much.

Incorrect.

Let the total be m, and estimate the total to be an b, which means n m=b an m is a fixed value, but b a is not fixed, so this estimation is not possible.

Take a and the probability that there is b marked in it is random b=n*a m is one of the cases, and the probability is not necessarily the largest.

It doesn't make sense, because there is no pattern to the fishermen fishing from the ponds, there may be more or less.

Pyramid height = (

1.The spire of the triangle is sharpened into degrees = arcsin(5 2..)The degree of the edge of the triangle = arctan[(5 2)

I'll break it down for you:

Draw a square ABCD with two diagonal lines intersecting at the point E. The diagonal line divides the square into four small isosceles triangles, with an angle of 90 degrees between the two waists.

Okay, so this is a floor plan, how do you convert it to a three-dimensional shape?

Lift the point e vertically! At this time, AE, BE, CE, DE are the four sides of the vertebral body.

During the lifting process, it can be found that the higher the vertebral body height, the smaller the angle between the two waists of the isosceles triangle.

That is, when the angle between the two waists of the 0 < isosceles triangle is < 90, the four isosceles triangles can be spliced together to form a vertebral body.

But you're asking about the angle between two triangles, and using the limit principle, the height of the vertebral body is variable.

Available: 90 "Angle< 180

When the angle is 180, it is a floor plan, the height is infinite, and the angle tends to be close to 90, but not 90).

If it is thick, it can't be put together into a cone, because no matter how you put it together, you can't spell the sharp corner on it.

It is necessary to know the proportional relationship between the height of your isosceles triangle and the waist to get the answer, the waist is the same, and the angle obtained by different heights is different.

Bae is equal to δcde

So. be=ce

Again. fg parallel is equal to 1 2ec

GH parallel is equal to 1 2BE

So parallelograms.

efgh and be=ce

Old. fg=gh

The state is suspected to be a diamond-shaped efgh

Because the dust of the Zen type is bc=2AD

So sδbae=sδcde=1 4sδbce let sδbce=k

Rule. k+1 4k+1 4k=s trapezoidal abcd=30so. k=20

So s rhombic efgh = 1 2

sδbce=10

Connect the EF and extend the EF and AD to the point G

In triangular EFC and GFD.

Angle EFC = GFD, angle GDF = angle C = 90 degrees, DF = FC so the triangle EFC and GFD are congruent.

So dg=ce, ef=fg

So ae=cd+ce=ad+dg=ag

In the triangle AEG, ae=ag, ef=fg

So AF bisected angle EAG, i.e., AF bisector angle DAE.

Extend AD to G so that DG=CE, AE=AG;

angle aef=g,d=c=90;

df=cf triangle dfg==efc

ef=fg;

AF bisector EAD

Is ABC an isosceles triangle?

In the case of an isosceles triangle, abc 50 acb aec 90 bac 80 ad is the angular bisector, then bad 40 aoe cod 50 aoc eod (360 aoe cod) 2 130

Because the angle b=50 AD is the height on the edge of bc, so the angle adb=90, so the angle bad=40, and because CE is the height on the ab, the angle AEC=90, so the angle AOC=angle AEC+angle BAD=90+40=130