Similar triangle math problems, help!

Hello little friend

ADE and CDE are the same height, right? The area ratio is 1:3, so ae:ce=1:3de, parallel to bc, so ad:bc=1:3

If the height of the triangle ADE and ABC is made, the two triangles are similar because DE is parallel to BC, and the ratio of their heights is 1:4

The high of ABC is also the high of DBC.

Therefore, the ratio of the area of the two is 1:12

1:9 is wrong because the ratio of the higher is not 1:3).

Think of ae as the bottom edge of ade, and d as the vertex; Think of CE as the bottom edge of CDE, and D as the vertex. In this way, ade and cde are the same height, and s ade:s cde=ae:

ce=1:3。So point E is a quarter of the line segment AC close to point A.

de bc, so ade abc, has a similarity ratio of 1:4 and an area ratio of 1:16

Since s abc = s ade + s cde + s dbc and let s ade be one unit, then s dbc is (16-1-3=12) units, so s ade:s dbc = 1:12

1 is the same as 9ade CDE.

then ae:ec=1:3

ad:db=1:3

The height of the ADE DBC is also 1:3

Multiply to get 1:9

1): Proof:

The angle DAC is equal to the angle cab and is both in a right triangle with equal cosine values.

i.e. ad ac = ac ab, then ac multiplied by ac equals ad multiplied by ab.

2): From (1), it can be seen that ac is equal to twice the root number six.

And from e as the midpoint, the triangle abc is a right triangle, so that ae is equal to ce is equal to three.

And the angle DAC is equal to the angle CAE is equal to the angle ace

Then the triangular clear or shaped ADF is similar to the triangular CEF

then ad ce is equal to af cf (3).

and af plus cf equals ac(4).

There are (3) and (4) that can solve the fruition that the subject wants.

In ABC, BC:Ca:AB=42:56:70=3:4:5, 3 +4 =5, ABC is a right triangle, and AB is an hypotenuse;

If another triangle is known to be similar to abc, then that triangle must also be a right-angled triangle with a ratio of 3:4 between the two right-angled sides.

The shortest side of another triangle is 24 cm long and is a short right-angled side, and the long right-angled side is 24 3 4 = 32, so the area of the other triangle is s = 24 32 2 = 384 square centimeters.

The triangle ABC is a right triangle. Area s=1 2*42*56 From the ratio of the shortest side, the similarity ratio can be obtained as 42 24=7 4, and the area ratio of the similar triangle is the square of the similarity ratio. S large S small = 49 16s small = 1 2 * 42 * 56 * 16 49 = 384

The area ratio is equal to the square of the side length ratio, you only need to calculate the area of the known triangle to calculate the answer, these three sides, just the proportion in the Pythagorean chord, so the area is 42*56 2=1176

The side length ratio is 4:7, and the area ratio is 16:49, so the area of the triangle is 384 In addition, the other two sides can be calculated, and the three sides of the triangle are: 24:32:40 is the same as the Pythagorean chord, and the area is 24*32 2=384

42 +56 = 70 is a right triangle.

So the other one is also a right triangle.

Let the other right-angled edge be x

42:46=24:x x=184/7

The area of the other triangle = 1 2 * 148 7 * 24

Area 4:1

The circumference ratio is 2:1

The circumference is 40, 20 respectively

Knowing that the area ratio is 4:1, we can know that the perimeter ratio is 2:1, and if the difference is 20, let x, 2x-x=20, and get x=20, the big one is 40, and the small one is 20. There is a formula, remember that if the circumference ratio is k, then the area ratio is k squared.

Similarity means that the two sides are equal in proportion, and the corresponding angles are equal. Here a= b, then we only need to prove that the proportions of the edges AD and AP are equal to the edges BP and BC.

Assuming that the p-point exists, there are two scenarios:

1. AD BC = AP BP

Knowing ab=7 ad=2 bc=3, bring it into the equation, get 2 3=ap (7-ap) and sort out the result ap=14 5=.

2. AD BP = AP BC

Bring in a known length and get 2 (7-ap)=ap 3 to get ap=1 or ap=6.

In summary, point p exists, and there are three points.

1.isosceles trapezoidal so b= c; At the same time, the sum of the three angles at the point p is 180 degrees, which is equal to the sum of the three inner angles of abp, and since ape= b, epc= bap.

So the two sets of angles are equal, and ABP is similar to PCE

The problem requires PE to cross dc to e, so the maximum value of ce=y is dc length that is, ab length is 5, and the calculation shows that x=5 at this time. In addition, it is obvious that x is greater than zero, so 0 x 5 (whether the equal sign is taken here...) It is debatable whether E and D coincide as PE and DC intersect, but it does not seem to be very important...

2.Discuss on a case-by-case basis.

The simplest case: Is EC=PC possible? The answer is yes ... The maximum value of y in question 1 is this case (which can be calculated by bringing in y=10-x), ec=pc=5, and bp=5.

And then about the other two cases ... I haven't seen it yet... lz give me the distribution first, I'll figure it out and make it up for you.

Well, I discussed the other two cases, ep=ec is possible, in this case bp=25 6;EP=PC is not possible according to the meaning of the title (it cannot be taken within the range of X values).

The problem solving process is a bit complicated... lz give me the points first, and then I'll help you fill in the detailed process if you need it. Otherwise, I'm too lazy to write a large paragraph of analysis...

Let AE=X, make EF perpendicular BC to F, DC= (5+X) 5 25 13

ac=x+5

In the triangle cde and cab, cd ca=5 13=ce cb and they have a common angle c, so they are similar, so de ab=5 13, I use the unknown to find the similarity relationship, which is also a method.