A problem of congruent triangles in elementary and junior high school solves the problem of congruen

I choose BCongruence, based on SAS

By a+ b= c, b'+∠c'=∠a'and a+ b+ c=180, b'+∠c'+∠a'=180

Got, a'=90 and c=90

These are two right triangles.

by b-a=b'-c'（1）

b+a=b'+c'（2）

1) + (2) get.

b=b'1) + (2) get.

a=c'Based on SAS congruence.

It can be known from the relationship of the angles as.

Right-angled triangle.

And add the formulas of the two sides to get it.

b=b', all with right-angled edges.

The formula for the two sides is subtracted to get it.

a=c', also for right-angled edges.

So wait! Based on the corner edge, choose B

This kind of question is generally impossible for a, the main ones should be b and c, because many multiple-choice questions, it will have a certain probability, and the probability of the big one must be (in general) this category, so it is not difficult to see that this question is asking us what basis to prove its congruence.

But I really can't find a conclusion about this question,,, so you can think about it yourself (I'm short on time, I'm sorry, if I can, I'll help you figure it out).

As far as I can speculate, if b= b', then a= c'，∠c=∠a'。。That's all there is to help you!

a+ b= c, abc is c, and c, is 90 degrees.

b'+∠c'=∠a'Note a'It's a triangle of 90 degrees.

b-a=b'-c',b+a=b'+c'Add the two and subtract them to get b=b' a=c'then select B

In ABC and A'b'c'Medium, a+ b= c, b'+∠c'=∠a', and b-a=b'-c',b+a=b'+c', then the two triangles [

a.Not necessarily congruent bCongruence according to SAS CIncompleteness dCongruent, based on ASA

I choose B

I choose BDiscuss it together if it's wrong, hehe

Extend AD to C'Make DC'=ad

then ADC c'db

bc'=ac

The sum of the two sides of the triangle is greater than the third side, and the difference between the two sides of the triangle is less than the third side, so ab+bc'>ac',ac'>bc'-ab i.e.: bc'-ab<2adac-ab<2ad8-4<2ad<8+4

4<2ad<12

2. Happy studying.

The answer is a, and the details are as follows: It is known that A: B: in ABC

c=3:5:10, then a=180*(3 18)=30, in the same way we find b=50, c=100, so bcn=180- c=80, and because mnc abc, acb= mcn=100, so bcm= mcn- bcn=100-80=20, so bcm:

BCN=20:80=1:4, that is, A

Choose A. In the triangle ABC, according to the proportional relationship of a, b, c and the knowledge of the sum of the inner angles of the triangle, we can obtain: a=30°, abc=50°, and from the theorem of the outer angle of the triangle, we can find BCN=30°+50°=80°

Because the triangle mnc is equal to the triangle abc, m=30°, n=50°, from the triangle outer angle theorem, find mca=80°, then mcb=180°-80°-80°=20°

bcm：∠bcn=20°：80°=1：4。So choose A.

Let me write the whole process in detail:

Solution: In ACD and A'c'd', because both are right-angled triangles, and there is ac=a'c'，cd=c'd', so ACD and A'c'd'congruence, (according to HL's theorem) so we get the corresponding angle: angle a = angle a'。

In ABC vs. A'b'c', ac=a'c', angle a = angle a'，ab=a'b', so ABC is the same as A'b'c'Congruence (SAS).

That's all for the reasons and process.

(1) From ab cd and ad bc, abcd is a parallelogram, and the parallelogram is parallel and equal to the opposite sides, so ab=cd

ab cd dca= cab (two straight lines are parallel, and the inner wrong angles are equal.) ）

again be ac, df ac, so aeb= cfd, so by aeb= cfd

dca=∠cab

ab=cd can be known as abe cdf (corner edge determination) be=df

2）∵be∥df ∴∠dfe=∠feb180°-∠dfe=180°-∠feb

cfd=∠aeb

From the first question, we can get the following: cd=ab, acd= cab, abe cdf (corner edge determination) be=df

Are you in a hurry, when I'm free, I'll write to you.

It's easy, it's just that the steps are not easy to write.