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At first glance, this question seems to be very clueless, and I feel that there are many letters in it, and it is easy to fall into the mistake of random calculation. And if we start from the conditions, we can see the relationship between the letters. So you can use a way to reduce the variables. The details are as follows:

Let a 2 = b 3 = c 4 = n

Then n is used to represent a, b, and c, so that there is a=2n, b=3n, and c=4n

Substituting the above two formulas can be obtained.

a+b+c) b=(2n+3n+4n) 3n=3 The same is true.

a+3b-2c)/(a+c)=1/2

Let's give you two examples.

1.Knowing that a 2 = b 3 = c 4 and 2a + b + c = 33, find the values of a, b, c.

2.Knowing x 4=y 5=z 6, find the value of (x-y+3z) (3x+2y).

Untie. 1, let a 2 = b 3 = c 4 = k, then a = 2k, b = 3k, c = 4k, so 2a + b + c = 2 * 2k + 3k + 4k = 11k = 33, so k = 3, so a = 6, b = 9, c = 12

The solution to question 2 is the same as the question in your question. I won't go into details.

Hope it helps!

First, let ka 2=b 3=c 4=k

a=2k b=3k c=4k

The first question came out.

2k+3k+4k）\3k

Eliminate k to get 3

The second question is the same.

a+b+c) b is one-third.

a+3b-2c) (a+c) is one-half.

(1) ab=de ah=dg angle ahb = angle dge = 90° ahb dge

BH=eg is the same: ahc dgf

hc=gfbh+hc=eg+gf

bc=ef and ab=de, ac=df

abc≌△def

2) It can be discussed on a case-by-case basis.

1) When they are all acute triangles.

The height on the third side falls on the third side.

As in the case of the first question, the conclusion holds.

2) When an obtuse triangle is formed.

When the opposite angle of the third side is obtuse, then the height on it also falls on it, which is also the same as in the case of the first question.

When the opposite angle of the third side is an acute angle.

then the height above it falls on its extension.

The same method can be used to prove this in this case.

ahb≌△dge

ahc≌△dgf

then bh=eg

hc=gf, so bh-hc=ge-gf

So bc=ef

The ABC DEF can still be obtained by SSS

Therefore, the conclusion is valid.

Solution(1)ab=de,ah=dg,abc def(hl)

2) No, take the point m, mg=fg on the eg, then dg=df, meet ab=de, ac=df, ah, dg are high, and ah=dg

But the conclusion does not hold.

Therefore, the statement that the height of the two sides and the third side corresponds to the equal equivalence of two triangles is not true.

1...Connect.

The angle AHC, BGF is a right angle.

The sum of squares of the right-angled sides of a right-angled triangle equals the squares of the hypotenuse to obtain hc=bf

In the same way, we can get bh=eg

hc+bh=bf+eg

The three sides are equal to each other, so the two triangles are congruent.

In def' and abc, there is ab=de, ac=df', ah, dg are high, and ah=dg

Obviously, def' is not perfect with ABC.

Start by doing CF vertical AB

Because de is also vertical AB.

The quadrilateral ABCD is an isosceles trapezoid.

So dc=ef=6 meters.

Because the angle a = 60 degrees, de=2 root number 3, according to tan60° = opposite side: hypotenuse tan60° = root number 3, so de:ae=2 root number 3:ae

ae=2 can be obtained

The same can be obtained for CF=AE=2

So ab=2+2+6=10

S trapezoidal ABCD = (6 + 10) x 2 root number 3 divided by 2 = 16 root number 3 is complete!

Hello, this question is about isosceles trapezoid.

ae=2√3/tan60=2，ab=2+2+6=10。

Trapezoidal area formula: (upper bottom + lower bottom) * height 2 = (6 + 10) * 2 3 2 = 16 3

(1) Point A is the center of the circle ae=ac, c= aec, an ec, c+ can=90°, eo bm, aec+ emo=90°, can= emo, and can= cbn+ anb, ebo= mbn+ mnb

cbn=∠mbn

anb= mnb,bc is the diameter bec=90°, an ce, an be, anb= nbe, mnb= nbe, be=ne;

2) From (1) know an be, point a is the midpoint of bc, point n is the midpoint of ce, tanc=be ce=1 2

3) The line y=kx+3 intersects the y-axis at the point a, oa=3, bo 2+3 2=r 2

An be, s abe=s nbe, bo ae=be en, i.e., bor=be 2, ce=2be, 5be 2=bc 2, i.e., be 2=

bo=，∴(r=5

Draw an image roughly

Read the question again.

Maybe you'll be able to find inspiration for the solution.